Re: [Tutor] regular expression query
On 08Jun2019 22:27, Sean Murphy wrote:
Windows 10 OS, Python 3.6
Thanks for this.
I have a couple of queries in relation to extracting content using
regular expressions. I understand [...the regexp syntax...]
The challenge I am finding is getting a pattern to
extract specific word(s). Trying to identify the best method to use and how
to use the \1 when using forward and backward search pattern (Hoping I am
using the right term). Basically I am trying to extract specific phrases or
digits to place in a dictionary within categories. Thus if "ROYaL_BANK
123123123" is found, it is placed in a category called transfer funds. Other
might be a store name which likewise is placed in the store category.
I'll tackle your specific examples lower down, and make some
suggestions.
Note, I have found a logic error with "ROYAL_BANK 123123123", but that
isn't a concern. The extraction of the text is.
Line examples:
Royal_bank M-BANKING PAYMENT TRANSFER 123456 to 9922992299
Royal_bank M-BANKING PAYMENT TRANSFER 123456 FROM 9922992299
PAYMENT TO SARWARS-123123123
ROYAL_BANK INTERNET BANKING BPAY Kangaroo Store {123123123}
EFTPOS Amazon
PAY/SALARY FROM foo bar 123123123
PAYMENT TO Tax Man 666
Thanks.
Assuming the below is a cut/paste accident from some code:
result = re.sub(r'ROYAL_BANK INTERNET BANKING FUNDS TFER TRANSFER \d+ TO ',
'ROYAL_BANK ', line)
r'ROYAL_BANK INTERNET BANKING TRANSFER Mouth in foot
And other similar structures. Below is the function I am currently using.
Not sure if the sub, match or search is going to be the best method. The
reason why I am using a sub is to delete the unwanted text. The
searchmatch/findall could do the same if I use a group. Also I have not
used any tests in the below and logically I think I should. As the code will
override the results if not found in the later tests. If there is a more
elegant way to do it then having:
If line.startswith('text string to match'):
Regular expression
el If line.startswith('text string to match'):
regular expression
return result
There is. How far you take it depends on how variable your input it.
Banking statement data I would expect to have relatively few formats
(unless the banking/financ industry is every bit as fragmented as I
sometimes believe, in which case the structure might be less driven by
_your_ bank and instead arbitrarily garbled according the the various
other entities due to getting ad hoc junk as the description).
I would like to know. The different regular expressions I have used
are:
# this sometimes matches and sometimes does not. I want all the text up to
the from or to, to be replaced with "ROYAL_BANK". Ending up with ROYAL_BANK
123123123
result= re.sub(r'ROYAL_BANK M-BANKING PAYMENT TRANSFER \d+ (TO|FROM) ',
'ROYAL_BANK ', line)
Looks superficially ok. Got an example input line where it fails? Not
that the above is case sentitive, so if "to" etc can be in lower case
(as in your example text earlier) this will fail. See the re.I modifier.
# the below returns from STARWARS and it shouldn't. I should just get
STARWARS.
result = re.match(r'PAYMENT TO (SARWARS)-\d+ ', line)
Well, STARWARS seems misseplt above. And you should get a "match"
object, with "STARWARS" in .group(1).
So earlier you're getting a str in result, and here you're getting an
re.match object (or None for a failed match).
# the below should (doesn't work the last time I tested it) should
return the words between the (.)
result = re.match(r'ROYAL_BANK INTERNET BANKING BPAY (.*) [{].*$', '\1',
line)
"should" what? It would help to see the input line you expect this to
match. And re.match is not an re.sub - it looks like you have these
confused here, based on the following '\`',line parameters.
# the below patterns should remove the text at the beginning of the string
result = re.sub(r'ROYAL_BANK INTERNET BANKING FUNDS TFER TRANSFER \d+ TO ',
'ROYAL_BANK ', line)
result = re.sub(r'ROYAL_BANK INTERNET BANKING TRANSFER ', '', line)
result = re.sub(r'EFTPOS ', '', line)
Sure. Got an example line where this does not happen?
# The below does not work and I am trying to use the back or forward
search feature. Is this syntax wrong or the pattern wrong? I cannot work it out
from the information I have read.
result = re.sub(r'PAY/SALARY FROM (*.) \d+$', '\1', line)
result = re.sub(r'PAYMENT TO (*.) \d+', '\1', line)
You've got "*." You probably mean ".*"
Main issues:
1: Your input data seems to be mixed case, but all your regexps are case
sensitive. They will not match if the case is different eg "Royal_Bank"
vs "ROYAL_BANK", "to" vs "TO", etc. Use the re.I modified to make your
regexps case insensitive.
2: You're using re.sub a lot. I'd be inclined to always use re.match and
to pull information from the match object you get back. Untested example
sketch:
m = re.match('(ROYAL_BANK|COMMONER_CREDIT_UNION) INTERNET BANKING FUNDS TFER
TRANSFER (\d+) TO
Re: [Tutor] (regular expression)
Hello Isaac,
This second posting you have made has provided more information
about what you are trying to accomplish and how (and also was
readable, where the first one looked like it got mangled by your
mail user agent; it's best to try to post only plain text messages
to this sort of mailing list).
I suspect that we can help you a bit more, now.
If we knew even more about what you were looking to do, we might be
able to help you further (with all of the usual remarks about how we
won't do your homework for you, but all of us volunteers will gladly
help you understand the tools, the systems, the world of Python and
anything else we can suggest in the realm of computers, computer
science and problem solving).
I will credit the person who assigned this task for you, as this is
not dissimilar from the sort of problem that one often has when
facing a new practical computing problem. Often (and in your case)
there is opaque structure and hidden assumptions in the question
which need to be understood. See further below
These were your four lines of code:
>with
>urllib.request.urlopen("https://www.sdstate.edu/electrical-engineering-and-computer-science;)
> as cs:
>cs_page = cs.read()
>soup = BeautifulSoup(cs_page, "html.parser")
>print(len(soup.body.find_all(string = ["Engineering","engineering"])))
The fourth line is an impressive attempt at compressing all of the
searching, finding, counting and reporting steps into a single line.
Your task (I think), is more complicated than that single line can
express. So, that will need to be expanded to a few more lines of
code.
You may have heard these aphorisms before:
* brevity is the soul of wit
* fewer lines of code are better
* prefer a short elegant solution
But, when complexity intrudes into brevity, the human mind
struggles. As a practitioner, I will say that I spend more of my
time reading and understanding code than writing it, so writing
simple, self-contained and understandable units of code leads to
intelligibility for humans and composability for systems.
Try this at a Python console [1].
import this
>i used control + f on the link in the code and i get 11 for ctrl +
>f and 3 for the code
Applause! Look at the raw data! Study the raw data! That is an
excellent way to start to try to understand the raw data. You must
always go back to the raw input data and then consider whether your
tooling or the data model in your program matches what you are
trying to extract/compute/transform.
The answer (for number of occurrences of the word 'engineering',
case-insensitive) that I get is close to your answer when searching
with control + f, but is a bit larger than 11.
Anyway, here are my thoughts. I will start with some tips that are
relevant to your 4-line pasted program:
* BeautifulSoup is wonderfully convenient, but also remember it
is another high-level tool; it is often forgiving where other
tools are more rigorous, however it is excellent for learning
and (I hope you see below) that it is a great tool for the
problem you are trying to solve
* in your code, soup.body is a handle that points to the
tag of the HTML document you have fetched; so why can't you
simply find_all of the strings "Engineering" and "engineering"
in the text and count them?
- find_all is a method that returns all of the tags in the
structured document below (in this case) soup.body
- your intent is not to count tags with the string
'engineering' but rather , you are looking for that string
in the text (I think)
* it is almost always a mistake to try to process HTML with
regular expressions, however, it seems that you are trying to
find all matches of the (case-insensitive) word 'engineering' in
the text of this document; that is something tailor-made for
regular expressions, so there's the Python regular expression
library, too: 'import re'
* and on a minor note, since you are using urllib.request.open()
in a with statement (using contexts this way is wonderful), you
could collect the data from the network socket, then drop out of
the 'with' block to allow the context to close, so if your block
worked as you wanted, you could adjust it as follows:
with urllib.request.urlopen(uri as cs:
cs_page = cs.read()
soup = BeautifulSoup(cs_page, "html.parser")
print(len(soup.body.find_all(string = ["Engineering","engineering"])))
* On a much more minor point, I'll mention that urllib / urllib2
are available with the main Python releases but there are other
libraries for handling fetching; I often recommend the
third-party requests [0] library, as it is both very Pythonic,
reasonably high-level and frightfully flexible
So, connecting the Zen of Python [1] to your problem, I would
suggest making shorter, simpler lines and separating the
Re: [Tutor] (regular expression)
this is the real code
with
urllib.request.urlopen("https://www.sdstate.edu/electrical-engineering-and-computer-science;)
as cs:
cs_page = cs.read()
soup = BeautifulSoup(cs_page, "html.parser")
print(len(soup.body.find_all(string = ["Engineering","engineering"])))
i used control + f on the link in the code and i get 11 for ctrl + f and 3 for
the code
THanks
From: Tutor on behalf of Bob
Gailer
Sent: Saturday, December 10, 2016 7:54 PM
To: Tetteh, Isaac - SDSU Student
Cc: Python Tutor
Subject: Re: [Tutor] (no subject)
On Dec 10, 2016 12:15 PM, "Tetteh, Isaac - SDSU Student" <
isaac.tet...@jacks.sdstate.edu> wrote:
>
> Hello,
>
> I am trying to find the number of times a word occurs on a webpage so I
used bs4 code below
>
> Let assume html contains the "html code"
> soup = BeautifulSoup(html, "html.parser")
> print(len(soup.find_all(string=["Engineering","engineering"])))
> But the result is different from when i use control + f on my keyboard to
find
>
> Please help me understand why it's different results. Thanks
> I am using Python 3.5
>
What is the URL of the web page?
To what are you applying control-f?
What are the two different counts you're getting?
Is it possible that the page is being dynamically altered after it's loaded?
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Re: [Tutor] Regular expression on python
On 15/04/15 09:24, Peter Otten wrote: function call. I've never seen (or noticed?) the embedded form, and don't see it described in the docs anywhere Quoting https://docs.python.org/dev/library/re.html: (?aiLmsux) (One or more letters from the set 'a', 'i', 'L', 'm', 's', 'u', 'x'.) The group matches the empty string; the letters set the corresponding flags: Aha. The trick is knowing the correct search string... I tried 'flag' and 'verbose' but missed this entry. Again, where is that described? (?#...) A comment; the contents of the parentheses are simply ignored. OK, I missed that too. Maybe I just wasn't awake enough this morning! :-) Thanks Peter. -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On 2015-04-14 16:49, Alan Gauld wrote: New one on me. Where does one find out about verbose mode? I don't see it in the re docs? This is where I go whenever I find myself having to (re)learn the details of regex: https://docs.python.org/3/howto/regex.html I believe a '2' can be substituted for the '3' but I've not found any difference between the two. (I submit this not so much for Alan (tutor) as for those like me who are learning.) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On 15/04/15 02:02, Steven D'Aprano wrote: New one on me. Where does one find out about verbose mode? I don't see it in the re docs? or embed the flag in the pattern. The flags that I know of are: (?x) re.X re.VERBOSE The flag can appear anywhere in the pattern and applies to the whole pattern, but it is good practice to put them at the front, and in the future it may be an error to put the flags elsewhere. I've always applied flags as separate params at the end of the function call. I've never seen (or noticed?) the embedded form, and don't see it described in the docs anywhere (although it probably is). But the re module descriptions of the flags only goive the re.X/re.VERBOSE options, no mention of the embedded form. Maybe you are just supposed to infer the (?x) form from the re.X... However, that still doesn't explain the difference in your comment syntax. The docs say the verbose syntax looks like: a = re.compile(r\d + # the integral part \.# the decimal point \d * # some fractional digits, re.X) Whereas your syntax is like: a = re.compile(r(?x) (?# turn on verbose mode) \d + (?# the integral part) \.(?# the decimal point) \d * (?# some fractional digits)) Again, where is that described? -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On Tue, 4/14/15, Peter Otten __pete...@web.de wrote:
Subject: Re: [Tutor] Regular expression on python
To: tutor@python.org
Date: Tuesday, April 14, 2015, 4:37 PM
Steven D'Aprano wrote:
On Tue, Apr 14, 2015 at 10:00:47AM +0200, Peter Otten
wrote:
Steven D'Aprano wrote:
I swear that Perl has been a blight on an
entire generation of
programmers. All they know is regular
expressions, so they turn every
data processing problem into a regular
expression. Or at least they
*try* to. As you have learned, regular
expressions are hard to read,
hard to write, and hard to get correct.
Let's write some Python code instead.
[...]
The tempter took posession of me and dictated:
pprint.pprint(
... [(k, int(v)) for k, v in
...
re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall(line)])
[('Input Read Pairs', 2127436),
('Both Surviving', 1795091),
('Forward Only Surviving', 17315),
('Reverse Only Surviving', 6413),
('Dropped', 308617)]
Nicely done :-)
Yes, nice, but why do you use
re.compile(regex).findall(line)
and not
re.findall(regex, line)
I know what re.compile is for. I often use it outside a loop and then actually
use the compiled regex inside a loop, I just haven't see the way you use it
before.
I didn't say that it *couldn't* be done with a regex.
I didn't claim that.
Only that it is
harder to read, write, etc. Regexes are good tools, but
they aren't the
only tool and as a beginner, which would you rather
debug? The extract()
function I wrote, or
r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s* ?
I know a rhetorical question when I see one ;)
Oh, and for the record, your solution is roughly 4-5
times faster than
the extract() function on my computer.
I wouldn't be bothered by that. See below if you are.
If I knew the requirements were
not likely to change (that is, the maintenance burden
was likely to be
low), I'd be quite happy to use your regex solution in
production code,
although I would probably want to write it out in
verbose mode just in
case the requirements did change:
r(?x) (?# verbose mode)
personally, I prefer to be verbose about being verbose, ie use the re.VERBOSE
flag. But perhaps that's just a matter of taste. Are there any use cases when
the ?iLmsux operators are clearly a better choice than the equivalent flag? For
me, the mental burden of a regex is big enough already without these operators.
(.+?): (?# capture one or
more character, followed by a colon)
\s+ (?#
one or more whitespace)
(\d+) (?#
capture one or more digits)
(?: (?#
don't capture ... )
\s+
(?# one or more whitespace)
\(.*?\) (?# anything
inside round brackets)
)?
(?# ... and optional)
\s* (?#
ignore trailing spaces)
snip
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Re: [Tutor] Regular expression on python
Alan Gauld wrote: On 15/04/15 02:02, Steven D'Aprano wrote: New one on me. Where does one find out about verbose mode? I don't see it in the re docs? or embed the flag in the pattern. The flags that I know of are: (?x) re.X re.VERBOSE The flag can appear anywhere in the pattern and applies to the whole pattern, but it is good practice to put them at the front, and in the future it may be an error to put the flags elsewhere. I've always applied flags as separate params at the end of the function call. I've never seen (or noticed?) the embedded form, and don't see it described in the docs anywhere (although it probably is). Quoting https://docs.python.org/dev/library/re.html: (?aiLmsux) (One or more letters from the set 'a', 'i', 'L', 'm', 's', 'u', 'x'.) The group matches the empty string; the letters set the corresponding flags: re.A (ASCII-only matching), re.I (ignore case), re.L (locale dependent), re.M (multi-line), re.S (dot matches all), and re.X (verbose), for the entire regular expression. (The flags are described in Module Contents.) This is useful if you wish to include the flags as part of the regular expression, instead of passing a flag argument to the re.compile() function. Note that the (?x) flag changes how the expression is parsed. It should be used first in the expression string, or after one or more whitespace characters. If there are non-whitespace characters before the flag, the results are undefined. But the re module descriptions of the flags only goive the re.X/re.VERBOSE options, no mention of the embedded form. Maybe you are just supposed to infer the (?x) form from the re.X... However, that still doesn't explain the difference in your comment syntax. The docs say the verbose syntax looks like: a = re.compile(r\d + # the integral part \.# the decimal point \d * # some fractional digits, re.X) Whereas your syntax is like: a = re.compile(r(?x) (?# turn on verbose mode) \d + (?# the integral part) \.(?# the decimal point) \d * (?# some fractional digits)) Again, where is that described? (?#...) A comment; the contents of the parentheses are simply ignored. Let's try it out: re.compile(\d+(?# sequence of digits)).findall(alpha 123 beta 456) ['123', '456'] re.compile(\d+# sequence of digits).findall(alpha 123 beta 456) [] re.compile(\d+# sequence of digits, re.VERBOSE).findall(alpha 123 beta 456) ['123', '456'] So (?#...)-style comments work in non-verbose mode, too, and Steven is wearing belt and braces (almost, the verbose flag is still necessary to ignore the extra whitespace). ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
Albert-Jan Roskam wrote:
On Tue, 4/14/15, Peter Otten __pete...@web.de wrote:
pprint.pprint(
... [(k, int(v)) for k, v in
...
re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall(line)])
[('Input Read Pairs', 2127436),
('Both Surviving', 1795091),
('Forward Only Surviving', 17315),
('Reverse Only Surviving', 6413),
('Dropped', 308617)]
Yes, nice, but why do you use
re.compile(regex).findall(line)
and not
re.findall(regex, line)
I know what re.compile is for. I often use it outside a loop and then
actually use the compiled regex inside a loop, I just haven't see the way
you use it before.
What you describe here is how I use regular expressions most of the time.
Also, re.compile() behaves the same over different Python versions while the
shortcuts for the pattern methods changed signature over time.
Finally, some have a gotcha. Compare:
re.compile(a, re.IGNORECASE).sub(b, aAAaa)
'b'
re.sub(a, b, aAAaa, re.IGNORECASE)
'bAAba'
Did you expect that? Congrats for thorough reading of the docs ;)
personally, I prefer to be verbose about being verbose, ie use the
re.VERBOSE flag. But perhaps that's just a matter of taste. Are there any
use cases when the ?iLmsux operators are clearly a better choice than the
equivalent flag? For me, the mental burden of a regex is big enough
already without these operators.
I pass flags separately myself, but
re.sub((?i)a, b, aAAaa)
'b'
might serve as an argument for inlined flags.
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Re: [Tutor] Regular expression on python
Steven D'Aprano wrote:
On Mon, Apr 13, 2015 at 02:29:07PM +0200, jarod...@libero.it wrote:
Dear all.
I would like to extract from some file some data.
The line I'm interested is this:
Input Read Pairs: 2127436 Both Surviving: 1795091 (84.38%) Forward
Only Surviving: 17315 (0.81%) Reverse Only Surviving: 6413 (0.30%)
Dropped: 308617 (14.51%)
Some people, when confronted with a problem, think I know, I'll
use regular expressions. Now they have two problems.
-- Jamie Zawinski
I swear that Perl has been a blight on an entire generation of
programmers. All they know is regular expressions, so they turn every
data processing problem into a regular expression. Or at least they
*try* to. As you have learned, regular expressions are hard to read,
hard to write, and hard to get correct.
Let's write some Python code instead.
def extract(line):
# Extract key:number values from the string.
line = line.strip() # Remove leading and trailing whitespace.
words = line.split()
accumulator = [] # Collect parts of the string we care about.
for word in words:
if word.startswith('(') and word.endswith('%)'):
# We don't care about percentages in brackets.
continue
try:
n = int(word)
except ValueError:
accumulator.append(word)
else:
accumulator.append(n)
# Now accumulator will be a list of strings and ints:
# e.g. ['Input', 'Read', 'Pairs:', 1234, 'Both', 'Surviving:', 1000]
# Collect consecutive strings as the key, int to be the value.
results = {}
keyparts = []
for item in accumulator:
if isinstance(item, int):
key = ' '.join(keyparts)
keyparts = []
if key.endswith(':'):
key = key[:-1]
results[key] = item
else:
keyparts.append(item)
# When we have finished processing, the keyparts list should be empty.
if keyparts:
extra = ' '.join(keyparts)
print('Warning: found extra text at end of line %s.' % extra)
return results
Now let me test it:
py line = ('Input Read Pairs: 2127436 Both Surviving: 1795091'
... ' (84.38%) Forward Only Surviving: 17315 (0.81%)'
... ' Reverse Only Surviving: 6413 (0.30%) Dropped:'
... ' 308617 (14.51%)\n')
py
py print(line)
Input Read Pairs: 2127436 Both Surviving: 1795091 (84.38%) Forward
Only Surviving: 17315 (0.81%) Reverse Only Surviving: 6413 (0.30%)
Dropped: 308617 (14.51%)
py extract(line)
{'Dropped': 308617, 'Both Surviving': 1795091, 'Reverse Only Surviving':
6413, 'Forward Only Surviving': 17315, 'Input Read Pairs': 2127436}
Remember that dicts are unordered. All the data is there, but in
arbitrary order. Now that you have a nice function to extract the data,
you can apply it to the lines of a data file in a simple loop:
with open(255.trim.log) as p:
for line in p:
if line.startswith(Input ):
d = extract(line)
print(d) # or process it somehow
The tempter took posession of me and dictated:
pprint.pprint(
... [(k, int(v)) for k, v in
... re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall(line)])
[('Input Read Pairs', 2127436),
('Both Surviving', 1795091),
('Forward Only Surviving', 17315),
('Reverse Only Surviving', 6413),
('Dropped', 308617)]
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Re: [Tutor] Regular expression on python
On Tue, Apr 14, 2015 at 10:00:47AM +0200, Peter Otten wrote:
Steven D'Aprano wrote:
I swear that Perl has been a blight on an entire generation of
programmers. All they know is regular expressions, so they turn every
data processing problem into a regular expression. Or at least they
*try* to. As you have learned, regular expressions are hard to read,
hard to write, and hard to get correct.
Let's write some Python code instead.
[...]
The tempter took posession of me and dictated:
pprint.pprint(
... [(k, int(v)) for k, v in
... re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall(line)])
[('Input Read Pairs', 2127436),
('Both Surviving', 1795091),
('Forward Only Surviving', 17315),
('Reverse Only Surviving', 6413),
('Dropped', 308617)]
Nicely done :-)
I didn't say that it *couldn't* be done with a regex. Only that it is
harder to read, write, etc. Regexes are good tools, but they aren't the
only tool and as a beginner, which would you rather debug? The extract()
function I wrote, or r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s* ?
Oh, and for the record, your solution is roughly 4-5 times faster than
the extract() function on my computer. If I knew the requirements were
not likely to change (that is, the maintenance burden was likely to be
low), I'd be quite happy to use your regex solution in production code,
although I would probably want to write it out in verbose mode just in
case the requirements did change:
r(?x)(?# verbose mode)
(.+?): (?# capture one or more character, followed by a colon)
\s+ (?# one or more whitespace)
(\d+) (?# capture one or more digits)
(?: (?# don't capture ... )
\s+ (?# one or more whitespace)
\(.*?\) (?# anything inside round brackets)
)?(?# ... and optional)
\s* (?# ignore trailing spaces)
That's a hint to people learning regular expressions: start in verbose
mode, then de-verbose it if you must.
--
Steve
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Re: [Tutor] Regular expression on python
Steven D'Aprano wrote:
On Tue, Apr 14, 2015 at 10:00:47AM +0200, Peter Otten wrote:
Steven D'Aprano wrote:
I swear that Perl has been a blight on an entire generation of
programmers. All they know is regular expressions, so they turn every
data processing problem into a regular expression. Or at least they
*try* to. As you have learned, regular expressions are hard to read,
hard to write, and hard to get correct.
Let's write some Python code instead.
[...]
The tempter took posession of me and dictated:
pprint.pprint(
... [(k, int(v)) for k, v in
... re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall(line)])
[('Input Read Pairs', 2127436),
('Both Surviving', 1795091),
('Forward Only Surviving', 17315),
('Reverse Only Surviving', 6413),
('Dropped', 308617)]
Nicely done :-)
I didn't say that it *couldn't* be done with a regex.
I didn't claim that.
Only that it is
harder to read, write, etc. Regexes are good tools, but they aren't the
only tool and as a beginner, which would you rather debug? The extract()
function I wrote, or r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s* ?
I know a rhetorical question when I see one ;)
Oh, and for the record, your solution is roughly 4-5 times faster than
the extract() function on my computer.
I wouldn't be bothered by that. See below if you are.
If I knew the requirements were
not likely to change (that is, the maintenance burden was likely to be
low), I'd be quite happy to use your regex solution in production code,
although I would probably want to write it out in verbose mode just in
case the requirements did change:
r(?x)(?# verbose mode)
(.+?): (?# capture one or more character, followed by a colon)
\s+ (?# one or more whitespace)
(\d+) (?# capture one or more digits)
(?: (?# don't capture ... )
\s+ (?# one or more whitespace)
\(.*?\) (?# anything inside round brackets)
)?(?# ... and optional)
\s* (?# ignore trailing spaces)
That's a hint to people learning regular expressions: start in verbose
mode, then de-verbose it if you must.
Regarding the speed of the Python approach: you can easily improve that by
relatively minor modifications. The most important one is to avoid the
exception:
$ python parse_jarod.py
$ python3 parse_jarod.py
The regex for reference:
$ python3 -m timeit -s from parse_jarod import extract_re as extract
extract()
10 loops, best of 3: 18.6 usec per loop
Steven's original extract():
$ python3 -m timeit -s from parse_jarod import extract_daprano as extract
extract()
1 loops, best of 3: 92.6 usec per loop
Avoid raising ValueError (This won't work with negative numbers):
$ python3 -m timeit -s from parse_jarod import extract_daprano2 as extract
extract()
1 loops, best of 3: 44.3 usec per loop
Collapse the two loops into one, thus avoiding the accumulator list and the
isinstance() checks:
$ python3 -m timeit -s from parse_jarod import extract_daprano3 as extract
extract()
1 loops, best of 3: 29.6 usec per loop
Ok, this is still slower than the regex, a result that I cannot accept.
Let's try again:
$ python3 -m timeit -s from parse_jarod import extract_py as extract
extract()
10 loops, best of 3: 15.1 usec per loop
Heureka? The winning code is brittle and probably as hard to understand as
the regex. You can judge for yourself if you're interested:
$ cat parse_jarod.py
import re
line = (Input Read Pairs: 2127436
Both Surviving: 1795091 (84.38%)
Forward Only Surviving: 17315 (0.81%)
Reverse Only Surviving: 6413 (0.30%)
Dropped: 308617 (14.51%))
_findall = re.compile(r(.+?):\s+(\d+)(?:\s+\(.*?\))?\s*).findall
def extract_daprano(line=line):
# Extract key:number values from the string.
line = line.strip() # Remove leading and trailing whitespace.
words = line.split()
accumulator = [] # Collect parts of the string we care about.
for word in words:
if word.startswith('(') and word.endswith('%)'):
# We don't care about percentages in brackets.
continue
try:
n = int(word)
except ValueError:
accumulator.append(word)
else:
accumulator.append(n)
# Now accumulator will be a list of strings and ints:
# e.g. ['Input', 'Read', 'Pairs:', 1234, 'Both', 'Surviving:', 1000]
# Collect consecutive strings as the key, int to be the value.
results = {}
keyparts = []
for item in accumulator:
if isinstance(item, int):
key = ' '.join(keyparts)
keyparts = []
if key.endswith(':'):
key = key[:-1]
results[key] = item
else:
keyparts.append(item)
# When we have finished processing, the keyparts list should be empty.
if keyparts:
extra = ' '.join(keyparts)
print('Warning: found
Re: [Tutor] Regular expression on python
On 14/04/15 13:21, Steven D'Aprano wrote: although I would probably want to write it out in verbose mode just in case the requirements did change: r(?x)(?# verbose mode) (.+?): (?# capture one or more character, followed by a colon) \s+ (?# one or more whitespace) (\d+) (?# capture one or more digits) (?: (?# don't capture ... ) \s+ (?# one or more whitespace) \(.*?\) (?# anything inside round brackets) )?(?# ... and optional) \s* (?# ignore trailing spaces) That's a hint to people learning regular expressions: start in verbose mode, then de-verbose it if you must. New one on me. Where does one find out about verbose mode? I don't see it in the re docs? I see an re.X flag but while it seems to be similar in purpose yet it is different to your style above (no parens for example)? -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On 15/04/2015 00:49, Alan Gauld wrote: On 14/04/15 13:21, Steven D'Aprano wrote: although I would probably want to write it out in verbose mode just in case the requirements did change: r(?x)(?# verbose mode) (.+?): (?# capture one or more character, followed by a colon) \s+ (?# one or more whitespace) (\d+) (?# capture one or more digits) (?: (?# don't capture ... ) \s+ (?# one or more whitespace) \(.*?\) (?# anything inside round brackets) )?(?# ... and optional) \s* (?# ignore trailing spaces) That's a hint to people learning regular expressions: start in verbose mode, then de-verbose it if you must. New one on me. Where does one find out about verbose mode? I don't see it in the re docs? I see an re.X flag but while it seems to be similar in purpose yet it is different to your style above (no parens for example)? https://docs.python.org/3/library/re.html#module-contents re.X and re.VERBOSE are together. -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On Wed, Apr 15, 2015 at 12:49:26AM +0100, Alan Gauld wrote: New one on me. Where does one find out about verbose mode? I don't see it in the re docs? I see an re.X flag but while it seems to be similar in purpose yet it is different to your style above (no parens for example)? I presume it is documented in the main docs, but I actually found this in the Python Pocket Reference by Mark Lutz :-) All of the regex flags have three forms: - a numeric flag with a long name; - the same numeric flag with a short name; - a regular expression pattern. So you can either do: re.compile(pattern, flags) or embed the flag in the pattern. The flags that I know of are: (?i) re.I re.IGNORECASE (?L) re.L re.LOCALE (?M) re.M re.MULTILINE (?s) re.S re.DOTALL (?x) re.X re.VERBOSE The flag can appear anywhere in the pattern and applies to the whole pattern, but it is good practice to put them at the front, and in the future it may be an error to put the flags elsewhere. When provided as a separate argument, you can combine flags like this: re.I|re.X -- Steve ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On 13/04/15 13:29, jarod...@libero.it wrote: Input Read Pairs: 2127436 Both Surviving: 1795091 (84.38%) Forward Only Surviving: 17315 (0.81%) Reverse Only Surviving: 6413 (0.30%) Dropped: 308617 (14.51%) Its not clear where the tabs are in this line. But if they are after the numbers, like so: Input Read Pairs: 2127436 \t Both Surviving: 1795091 (84.38%) \t Forward Only Surviving: 17315 (0.81%) \t Reverse Only Surviving: 6413 (0.30%) \t Dropped: 308617 (14.51%) Then you may not need to use regular expressions. Simply split by tab then split by : And if the 'number' contains parens split again by space with open(255.trim.log,r) as p: for i in p: lines= i.strip(\t) lines is a bad name here since its only a single line. In fact I'd lose the 'i' variable and just use for line in p: if lines.startswith(Input): tp = lines.split(\t) print re.findall(Input\d,str(tp)) Input is not followed by a number. You need a more powerful pattern. Which is why I recommend trying to solve it as far as possible without using regex. So I started to find : from the row: with open(255.trim.log,r) as p: for i in p: lines= i.strip(\t) if lines.startswith(Input): tp = lines.split(\t) print re.findall(:,str(tp[0])) Does finding the colons really help much? Or at least, does it help any more than splitting by colon would? And I'm able to find, but when I try to take the number using \d not work. Someone can explain why? Because your pattern doesn't match the string. HTH -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On 13/04/15 19:42, Alan Gauld wrote: if lines.startswith(Input): tp = lines.split(\t) print re.findall(Input\d,str(tp)) Input is not followed by a number. You need a more powerful pattern. Which is why I recommend trying to solve it as far as possible without using regex. I also just realised that you call split there then take the str() of the result. That means you are searching the string representation of a list, which doesn't seem to make much sense? -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression on python
On Mon, Apr 13, 2015 at 02:29:07PM +0200, jarod...@libero.it wrote:
Dear all.
I would like to extract from some file some data.
The line I'm interested is this:
Input Read Pairs: 2127436 Both Surviving: 1795091 (84.38%) Forward
Only Surviving: 17315 (0.81%) Reverse Only Surviving: 6413 (0.30%)
Dropped: 308617 (14.51%)
Some people, when confronted with a problem, think I know, I'll
use regular expressions. Now they have two problems.
-- Jamie Zawinski
I swear that Perl has been a blight on an entire generation of
programmers. All they know is regular expressions, so they turn every
data processing problem into a regular expression. Or at least they
*try* to. As you have learned, regular expressions are hard to read,
hard to write, and hard to get correct.
Let's write some Python code instead.
def extract(line):
# Extract key:number values from the string.
line = line.strip() # Remove leading and trailing whitespace.
words = line.split()
accumulator = [] # Collect parts of the string we care about.
for word in words:
if word.startswith('(') and word.endswith('%)'):
# We don't care about percentages in brackets.
continue
try:
n = int(word)
except ValueError:
accumulator.append(word)
else:
accumulator.append(n)
# Now accumulator will be a list of strings and ints:
# e.g. ['Input', 'Read', 'Pairs:', 1234, 'Both', 'Surviving:', 1000]
# Collect consecutive strings as the key, int to be the value.
results = {}
keyparts = []
for item in accumulator:
if isinstance(item, int):
key = ' '.join(keyparts)
keyparts = []
if key.endswith(':'):
key = key[:-1]
results[key] = item
else:
keyparts.append(item)
# When we have finished processing, the keyparts list should be empty.
if keyparts:
extra = ' '.join(keyparts)
print('Warning: found extra text at end of line %s.' % extra)
return results
Now let me test it:
py line = ('Input Read Pairs: 2127436 Both Surviving: 1795091'
... ' (84.38%) Forward Only Surviving: 17315 (0.81%)'
... ' Reverse Only Surviving: 6413 (0.30%) Dropped:'
... ' 308617 (14.51%)\n')
py
py print(line)
Input Read Pairs: 2127436 Both Surviving: 1795091 (84.38%) Forward
Only Surviving: 17315 (0.81%) Reverse Only Surviving: 6413 (0.30%)
Dropped: 308617 (14.51%)
py extract(line)
{'Dropped': 308617, 'Both Surviving': 1795091, 'Reverse Only Surviving':
6413, 'Forward Only Surviving': 17315, 'Input Read Pairs': 2127436}
Remember that dicts are unordered. All the data is there, but in
arbitrary order. Now that you have a nice function to extract the data,
you can apply it to the lines of a data file in a simple loop:
with open(255.trim.log) as p:
for line in p:
if line.startswith(Input ):
d = extract(line)
print(d) # or process it somehow
--
Steven
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Re: [Tutor] Regular expression
On Tue, Sep 23, 2014 at 11:40:25AM +0200, jarod...@libero.it wrote:
Hi there!!
I need to read this file:
pippo.count :
10566 ZXDC
2900 ZYG11A
7909 ZYG11B
3584 ZYX
9614 ZZEF1
17704 ZZZ3
How can extract only the number and the work in array? Thanks for any help
There is no need for the nuclear-powered bulldozer of regular
expressions just to crack this peanut.
with open('pippo.count') as f:
for line in f:
num, word = line.split()
num = int(num)
print num, word
Or, if you prefer the old-fashioned way:
f = open('pippo.count')
for line in f:
num, word = line.split()
num = int(num)
print num, word
f.close()
but the first way with the with-statement is better.
--
Steven
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Re: [Tutor] Regular expression - I
Steve,
i am trying to under r - raw string notation. Am i understanding it wrong.
Rather than using \, it says we can use the r option.
http://docs.python.org/2/library/re.html
Check the first paragraph for the above link.
Thanks,
santosh
On Tue, Feb 18, 2014 at 11:33 PM, Steve Willoughby st...@alchemy.comwrote:
Because the regular expression H* means “match an angle-bracket
character, zero or more H characters, followed by a close angle-bracket
character” and your string does not match that pattern.
This is why it’s best to check that the match succeeded before going ahead
to call group() on the result (since in this case there is no result).
On 18-Feb-2014, at 09:52, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call
last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
Thanks,
santosh
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RHCE | SCSA
+91-9703206361
Every task has a unpleasant side .. But you must focus on the end result
you are producing.
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Re: [Tutor] Regular expression - I
The problem is not the use of the raw string, but rather the regular expression
inside it.
In regular expressions, the * means that whatever appears before it may be
repeated zero or more times. So if you say H* that means zero or more H’s in a
row. I think you mean an H followed by any number of other characters which
would be H.* (the . matches any single character, so .* means zero or more of
any characters).
On the other hand, H\* means to match an H followed by a literal asterisk
character.
Does that help clarify why one matched and the other doesn’t?
steve
On 18-Feb-2014, at 10:09, Santosh Kumar rhce@gmail.com wrote:
Steve,
i am trying to under r - raw string notation. Am i understanding it wrong.
Rather than using \, it says we can use the r option.
http://docs.python.org/2/library/re.html
Check the first paragraph for the above link.
Thanks,
santosh
On Tue, Feb 18, 2014 at 11:33 PM, Steve Willoughby st...@alchemy.com wrote:
Because the regular expression H* means “match an angle-bracket character,
zero or more H characters, followed by a close angle-bracket character” and
your string does not match that pattern.
This is why it’s best to check that the match succeeded before going ahead to
call group() on the result (since in this case there is no result).
On 18-Feb-2014, at 09:52, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
Thanks,
santosh
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RHCE | SCSA
+91-9703206361
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are producing.
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Re: [Tutor] Regular expression - I
Because the regular expression H* means “match an angle-bracket character,
zero or more H characters, followed by a close angle-bracket character” and
your string does not match that pattern.
This is why it’s best to check that the match succeeded before going ahead to
call group() on the result (since in this case there is no result).
On 18-Feb-2014, at 09:52, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
Thanks,
santosh
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Re: [Tutor] Regular expression - I
Hi Santosh,
On Tue, Feb 18, 2014 at 9:52 AM, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
It is working as expected, but you're not expecting the right thing
;). Raw strings don't escape anything, they just prevent backslash
escapes from expanding. Case I works because \* is not a special
character to Python (like \n or \t), so it leaves the backslash in
place:
'H\*'
'H\*'
The equivalent raw string is exactly the same in this case:
r'H\*'
'H\*'
The raw string you provided doesn't have the backslash, and Python
will not add backslashes for you:
r'H*'
'H*'
The purpose of raw strings is to prevent Python from recognizing
backslash escapes. For example:
path = 'C:\temp\new\dir' # Windows paths are notorious...
path # it looks mostly ok... [1]
'C:\temp\new\\dir'
print(path) # until you try to use it
C: emp
ew\dir
path = r'C:\temp\new\dir' # now try a raw string
path # Now it looks like it's stuffed full of backslashes [2]
'C:\\temp\\new\\dir'
print(path) # but it works properly!
C:\temp\new\dir
[1] Count the backslashes in the repr of 'path'. Notice that there is
only one before the 't' and the 'n', but two before the 'd'. \d is
not a special character, so Python didn't do anything to it. There
are two backslashes in the repr of \d, because that's the only way
to distinguish a real backslash; the \t and \n are actually the
TAB and LINE FEED characters, as seen when printing 'path'.
[2] Because they are all real backslashes now, so they have to be
shown escaped (\\) in the repr.
In your regex, since you're looking for, literally, H*, you'll
need to backslash escape the * since it is a special character *in
regular expressions*. To avoid having to keep track of what's special
to Python as well as regular expressions, you'll need to make sure the
backslash itself is escaped, to make sure the regex sees \*, and the
easiest way to do that is a raw string:
re.match(r'H\*', string).group()
'H*'
I hope this makes some amount of sense; I've had to write it up
piecemeal and will never get it posted at all if I don't go ahead and
post :). If you still have questions, I'm happy to try again. You
may also want to have a look at the Regex HowTo in the Python docs:
http://docs.python.org/3/howto/regex.html
Hope this helps,
--
Zach
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Re: [Tutor] Regular expression - I
On 18/02/2014 18:03, Steve Willoughby wrote:
Because the regular expression H* means “match an angle-bracket character,
zero or more H characters, followed by a close angle-bracket character” and your
string does not match that pattern.
This is why it’s best to check that the match succeeded before going ahead to
call group() on the result (since in this case there is no result).
On 18-Feb-2014, at 09:52, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
Thanks,
santosh
Please do not top post on this list.
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
---
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Re: [Tutor] Regular expression - I
On Tue, Feb 18, 2014 at 11:39 AM, Zachary Ware zachary.ware+py...@gmail.com wrote: snip 'H\*' 'H\*' The equivalent raw string is exactly the same in this case: r'H\*' 'H\*' Oops, I mistyped both of these. The repr should be 'H\\*' in both cases. Sorry for the confusion! -- Zach ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression - I
does any one know how to use 2to3 program to convert 2.7 coding 3.X please i need help sorry On Tuesday, 18 February 2014, 19:50, Zachary Ware zachary.ware+py...@gmail.com wrote: On Tue, Feb 18, 2014 at 11:39 AM, Zachary Ware zachary.ware+py...@gmail.com wrote: snip 'H\*' 'H\*' The equivalent raw string is exactly the same in this case: r'H\*' 'H\*' Oops, I mistyped both of these. The repr should be 'H\\*' in both cases. Sorry for the confusion! -- Zach ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression - I
_
From: Steve Willoughby st...@alchemy.com
To: Santosh Kumar rhce@gmail.com
Cc: python mail list tutor@python.org
Sent: Tuesday, February 18, 2014 7:03 PM
Subject: Re: [Tutor] Regular expression - I
Because the regular expression H* means “match an angle-bracket character,
zero or more H characters, followed by a close angle-bracket character” and
your string does not match that pattern.
This is why it’s best to check that the match succeeded before going ahead to
call group() on the result (since in this case there is no result).
On 18-Feb-2014, at 09:52, Santosh Kumar rhce@gmail.com wrote:
You also might want to consider making it a non-greedy match. The explanation
http://docs.python.org/2/howto/regex.html covers an example almost identical to
yours:
Greedy versus Non-Greedy
When repeating a regular expression, as in a*, the resulting action is to
consume as much of the pattern as possible. This fact often bites you when
you’re trying to match a pair of balanced delimiters, such as the angle brackets
surrounding an HTML tag. The naive pattern for matching a single HTML tag
doesn’t work because of the greedy nature of .*.
s = 'htmlheadtitleTitle/title' len(s) 32 print
re.match('.*', s).span() (0, 32) print re.match('.*', s).group()
htmlheadtitleTitle/title
The RE matches the '' in html, and the .* consumes the rest of
the string. There’s still more left in the RE, though, and the can’t
match at the end of the string, so the regular expression engine has to
backtrack character by character until it finds a match for the . The
final match extends from the '' in html to the '' in /title, which isn’t
what you want.
In this case, the solution is to use the non-greedy qualifiers *?, +?, ??, or
{m,n}?, which match as little text as possible. In the above
example, the '' is tried immediately after the first '' matches, and
when it fails, the engine advances a character at a time, retrying the '' at
every step. This produces just the right result:
print re.match('.*?', s).group() html
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Re: [Tutor] Regular expression - I
On 2/18/2014 11:42 AM, Mark Lawrence wrote: On 18/02/2014 18:03, Steve Willoughby wrote: Because the regular expression H* means “match an angle-bracket SNIP Please do not top post on this list. Appropriate trimming is also appreciated. Emile ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression - I
On 02/18/2014 08:39 PM, Zachary Ware wrote:
Hi Santosh,
On Tue, Feb 18, 2014 at 9:52 AM, Santosh Kumar rhce@gmail.com wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
It is working as expected, but you're not expecting the right thing
;). Raw strings don't escape anything, they just prevent backslash
escapes from expanding. Case I works because \* is not a special
character to Python (like \n or \t), so it leaves the backslash in
place:
'H\*'
'H\*'
The equivalent raw string is exactly the same in this case:
r'H\*'
'H\*'
The raw string you provided doesn't have the backslash, and Python
will not add backslashes for you:
r'H*'
'H*'
The purpose of raw strings is to prevent Python from recognizing
backslash escapes. For example:
path = 'C:\temp\new\dir' # Windows paths are notorious...
path # it looks mostly ok... [1]
'C:\temp\new\\dir'
print(path) # until you try to use it
C: emp
ew\dir
path = r'C:\temp\new\dir' # now try a raw string
path # Now it looks like it's stuffed full of backslashes [2]
'C:\\temp\\new\\dir'
print(path) # but it works properly!
C:\temp\new\dir
[1] Count the backslashes in the repr of 'path'. Notice that there is
only one before the 't' and the 'n', but two before the 'd'. \d is
not a special character, so Python didn't do anything to it. There
are two backslashes in the repr of \d, because that's the only way
to distinguish a real backslash; the \t and \n are actually the
TAB and LINE FEED characters, as seen when printing 'path'.
[2] Because they are all real backslashes now, so they have to be
shown escaped (\\) in the repr.
In your regex, since you're looking for, literally, H*, you'll
need to backslash escape the * since it is a special character *in
regular expressions*. To avoid having to keep track of what's special
to Python as well as regular expressions, you'll need to make sure the
backslash itself is escaped, to make sure the regex sees \*, and the
easiest way to do that is a raw string:
re.match(r'H\*', string).group()
'H*'
I hope this makes some amount of sense; I've had to write it up
piecemeal and will never get it posted at all if I don't go ahead and
post :). If you still have questions, I'm happy to try again. You
may also want to have a look at the Regex HowTo in the Python docs:
http://docs.python.org/3/howto/regex.html
In addition to all this:
* You may confuse raw strings with regex escaping (a tool func that escapes
special regex characters for you).
* For simplicity, always use raw strings for regex formats (as in your second
example); this does not prevent you to escape special characters, but you only
have to do it once!
d
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Re: [Tutor] Regular expression - I
Thank you all. I got it. :)
I need to read more between lines .
On Wed, Feb 19, 2014 at 4:25 AM, spir denis.s...@gmail.com wrote:
On 02/18/2014 08:39 PM, Zachary Ware wrote:
Hi Santosh,
On Tue, Feb 18, 2014 at 9:52 AM, Santosh Kumar rhce@gmail.com
wrote:
Hi All,
If you notice the below example, case I is working as expected.
Case I:
In [41]: string = H*testH*
In [42]: re.match('H\*',string).group()
Out[42]: 'H*'
But why is the raw string 'r' not working as expected ?
Case II:
In [43]: re.match(r'H*',string).group()
---
AttributeErrorTraceback (most recent call
last)
ipython-input-43-d66b47f01f1c in module()
1 re.match(r'H*',string).group()
AttributeError: 'NoneType' object has no attribute 'group'
In [44]: re.match(r'H*',string)
It is working as expected, but you're not expecting the right thing
;). Raw strings don't escape anything, they just prevent backslash
escapes from expanding. Case I works because \* is not a special
character to Python (like \n or \t), so it leaves the backslash in
place:
'H\*'
'H\*'
The equivalent raw string is exactly the same in this case:
r'H\*'
'H\*'
The raw string you provided doesn't have the backslash, and Python
will not add backslashes for you:
r'H*'
'H*'
The purpose of raw strings is to prevent Python from recognizing
backslash escapes. For example:
path = 'C:\temp\new\dir' # Windows paths are notorious...
path # it looks mostly ok... [1]
'C:\temp\new\\dir'
print(path) # until you try to use it
C: emp
ew\dir
path = r'C:\temp\new\dir' # now try a raw string
path # Now it looks like it's stuffed full of backslashes [2]
'C:\\temp\\new\\dir'
print(path) # but it works properly!
C:\temp\new\dir
[1] Count the backslashes in the repr of 'path'. Notice that there is
only one before the 't' and the 'n', but two before the 'd'. \d is
not a special character, so Python didn't do anything to it. There
are two backslashes in the repr of \d, because that's the only way
to distinguish a real backslash; the \t and \n are actually the
TAB and LINE FEED characters, as seen when printing 'path'.
[2] Because they are all real backslashes now, so they have to be
shown escaped (\\) in the repr.
In your regex, since you're looking for, literally, H*, you'll
need to backslash escape the * since it is a special character *in
regular expressions*. To avoid having to keep track of what's special
to Python as well as regular expressions, you'll need to make sure the
backslash itself is escaped, to make sure the regex sees \*, and the
easiest way to do that is a raw string:
re.match(r'H\*', string).group()
'H*'
I hope this makes some amount of sense; I've had to write it up
piecemeal and will never get it posted at all if I don't go ahead and
post :). If you still have questions, I'm happy to try again. You
may also want to have a look at the Regex HowTo in the Python docs:
http://docs.python.org/3/howto/regex.html
In addition to all this:
* You may confuse raw strings with regex escaping (a tool func that
escapes special regex characters for you).
* For simplicity, always use raw strings for regex formats (as in your
second example); this does not prevent you to escape special characters,
but you only have to do it once!
d
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RHCE | SCSA
+91-9703206361
Every task has a unpleasant side .. But you must focus on the end result
you are producing.
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Re: [Tutor] regular expression wildcard search
Hi there
Do you want your * to represent a single letter, or what is your intent?
If you want only a single letter between the V and VVP, use \w
instead of *.
re.search('v\wVVP',myseq)
Emma
On Tue, Dec 11, 2012 at 8:54 AM, Hs Hs ilhs...@yahoo.com wrote:
Dear group:
I have 50 thousand lists. My aim is to search a pattern in the
alphabetical strings (these are protein sequence strings).
MMSASRLAGTLIPAMAFLSCVRPESWEPC VEVVP
NITYQCMELNFYKIPDNLPFSTKNLDLSFNPLRHLGSYSFFSFPELQVLDLSRCEIQTIED
my aim is to find the list of string that has V*VVP.
myseq = 'MMSASRLAGTLIPAMAFLSCVRPESWEPC VEVVP
NITYQCMELNFYKIPDNLPFSTKNLDLSFNPLRHLGSYSFFSFPELQVLDLSRCEIQTIED'
if re.search('V*VVP',myseq):
print myseq
the problem with this is, I am also finding junk with just VVP or VP etc.
How can I strictly search for V*VVP only.
Thanks for help.
Hs
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Re: [Tutor] regular expression wildcard search
On Tue, Dec 11, 2012 at 10:54 AM, Hs Hs ilhs...@yahoo.com wrote:
Dear group:
Please send mail as plain text. It is easier to read
I have 50 thousand lists. My aim is to search a pattern in the
alphabetical strings (these are protein sequence strings).
MMSASRLAGTLIPAMAFLSCVRPESWEPC VEVVP
NITYQCMELNFYKIPDNLPFSTKNLDLSFNPLRHLGSYSFFSFPELQVLDLSRCEIQTIED
my aim is to find the list of string that has V*VVP.
Asterisk
The * matches 0 or more instances of the previous element.
I am not sure what you want, but I don't think it is this. Do you want V
then any characters followed by VVP? In that case perhaps
V.+VP
There are many tutorials about how to create regular expressions
**
**
myseq = 'MMSASRLAGTLIPAMAFLSCVRPESWEPC VEVVP
NITYQCMELNFYKIPDNLPFSTKNLDLSFNPLRHLGSYSFFSFPELQVLDLSRCEIQTIED'
if re.search('V*VVP',myseq):
print myseq
the problem with this is, I am also finding junk with just VVP or VP etc.
How can I strictly search for V*VVP only.
Thanks for help.
Hs
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Re: [Tutor] regular expression wildcard search
On 11/12/12 15:54, Hs Hs wrote:
myseq = 'MMSASRLAGTLIPAMAFLSCVRPESWEPC VEVVP
NITYQCMELNFYKIPDNLPFSTKNLDLSFNPLRHLGSYSFFSFPELQVLDLSRCEIQTIED'
if re.search('V*VVP',myseq):
print myseq
I hope this is just a typo but you are printing your original string not
the things found...
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
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Re: [Tutor] Regular expression grouping insert thingy
re.sub(r'(\d+)x', r'\1*x', input_text) -- I enjoy haiku but sometimes they don't make sense; refrigerator? On Tue, Jun 8, 2010 at 10:11 PM, Lang Hurst l...@tharin.com wrote: This is so trivial (or should be), but I can't figure it out. I'm trying to do what in vim is :s/\([0-9]\)x/\1*x/ That is, find a number followed by an x and put a * in between the number and the x So, if the string is 6443x - 3, I'll get back 6443*x - 3 I won't write down all the things I've tried, but suffice it to say, nothing has done it. I just found myself figuring out how to call sed and realized that this should be a one-liner in python too. Any ideas? I've read a lot of documentation, but I just can't figure it out. Thanks. -- There are no stupid questions, just stupid people. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression grouping insert thingy
Oh. Crap, I knew it would be something simple, but honestly, I don't think that I would have gotten there. Thank you so much. Seriously saved me more grey hair. Matthew Wood wrote: re.sub(r'(\d+)x', r'\1*x', input_text) -- I enjoy haiku but sometimes they don't make sense; refrigerator? On Tue, Jun 8, 2010 at 10:11 PM, Lang Hurst l...@tharin.com mailto:l...@tharin.com wrote: This is so trivial (or should be), but I can't figure it out. I'm trying to do what in vim is :s/\([0-9]\)x/\1*x/ That is, find a number followed by an x and put a * in between the number and the x So, if the string is 6443x - 3, I'll get back 6443*x - 3 I won't write down all the things I've tried, but suffice it to say, nothing has done it. I just found myself figuring out how to call sed and realized that this should be a one-liner in python too. Any ideas? I've read a lot of documentation, but I just can't figure it out. Thanks. -- There are no stupid questions, just stupid people. ___ Tutor maillist - Tutor@python.org mailto:Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor -- There are no stupid questions, just stupid people. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
Hello, The following code returns 'abc123abc45abc789jk'. How do I revise the pattern so that the return value will be 'abc789jk'? In other words, I want to find the pattern 'abc' that is closest to 'jk'. Here the string '123', '45' and '789' are just examples. They are actually quite different in the string that I'm working with. import re s = 'abc123abc45abc789jk' p = r'abc.+jk' lst = re.findall(p, s) print lst[0] I suggest using r'abc.+?jk' instead. the additional ? makes the preceeding '.+' non-greedy so instead of matching as long string as it can it matches as short string as possible. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
Dnia 28 kwietnia 2009 11:16 Andre Engels andreeng...@gmail.com napisał(a): 2009/4/28 Marek spociń...@go2.pl,Poland : Hello, The following code returns 'abc123abc45abc789jk'. How do I revise the pattern so that the return value will be 'abc789jk'? In other words, I want to find the pattern 'abc' that is closest to 'jk'. Here the string '123', '45' and '789' are just examples. They are actually quite different in the string that I'm working with. import re s = 'abc123abc45abc789jk' p = r'abc.+jk' lst = re.findall(p, s) print lst[0] I suggest using r'abc.+?jk' instead. the additional ? makes the preceeding '.+' non-greedy so instead of matching as long string as it can it matches as short string as possible. That was my first idea too, but it does not work for this case, because Python will still try to _start_ the match as soon as possible. To use .+? one would have to revert the string, then use the reverse regular expression on the result, which looks like a rather roundabout way of doing things. I don't have access to python right now so i cannot test my ideas... And i don't really want to give you wrong idea too. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
Le Tue, 28 Apr 2009 11:06:16 +0200,
Marek spociń...@go2.pl, Poland marek...@10g.pl s'exprima ainsi:
Hello,
The following code returns 'abc123abc45abc789jk'. How do I revise the
pattern so that the return value will be 'abc789jk'? In other words, I
want to find the pattern 'abc' that is closest to 'jk'. Here the string
'123', '45' and '789' are just examples. They are actually quite
different in the string that I'm working with.
import re
s = 'abc123abc45abc789jk'
p = r'abc.+jk'
lst = re.findall(p, s)
print lst[0]
I suggest using r'abc.+?jk' instead.
the additional ? makes the preceeding '.+' non-greedy so instead of
matching as long string as it can it matches as short string as possible.
Non-greedy repetition will not work in this case, I guess:
from re import compile as Pattern
s = 'abc123abc45abc789jk'
p = Pattern(r'abc.+?jk')
print p.match(s).group()
==
abc123abc45abc789jk
(Someone explain why?)
My solution would be to explicitely exclude 'abc' from the sequence of chars
matched by '.+'. To do this, use negative lookahead (?!...) before '.':
p = Pattern(r'(abc((?!abc).)+jk)')
print p.findall(s)
==
[('abc789jk', '9')]
But it's not exactly what you want. Because the internal () needed to express
exclusion will be considered by findall as a group to be returned, so that you
also get the last char matched in there.
To avoid that, use non-grouping parens (?:...). This also avoids the need for
parens around the whole format:
p = Pattern(r'abc(?:(?!abc).)+jk')
print p.findall(s)
['abc789jk']
Denis
--
la vita e estrany
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Re: [Tutor] regular expression question
Andre Engels andreengels at gmail.com writes: 2009/4/28 Marek Spociński at go2.pl,Poland marek_sp at 10g.pl: I suggest using r'abc.+?jk' instead. That was my first idea too, but it does not work for this case, because Python will still try to _start_ the match as soon as possible. yeah, i tried the '?' as well and realized it would not work. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
2009/4/28 Marek spociń...@go2.pl,Poland marek...@10g.pl: import re s = 'abc123abc45abc789jk' p = r'abc.+jk' lst = re.findall(p, s) print lst[0] I suggest using r'abc.+?jk' instead. the additional ? makes the preceeding '.+' non-greedy so instead of matching as long string as it can it matches as short string as possible. Did you try it? It doesn't do what you expect, it still matches at the beginning of the string. The re engine searches for a match at a location and returns the first one it finds. A non-greedy match doesn't mean Find the shortest possible match anywhere in the string, it means, find the shortest possible match starting at this location. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
On Tue, Apr 28, 2009 at 4:03 AM, Kelie kf9...@gmail.com wrote: Hello, The following code returns 'abc123abc45abc789jk'. How do I revise the pattern so that the return value will be 'abc789jk'? In other words, I want to find the pattern 'abc' that is closest to 'jk'. Here the string '123', '45' and '789' are just examples. They are actually quite different in the string that I'm working with. import re s = 'abc123abc45abc789jk' p = r'abc.+jk' lst = re.findall(p, s) print lst[0] re.findall() won't work because it finds non-overlapping matches. If there is a character in the initial match which cannot occur in the middle section, change .+ to exclude that character. For example, r'abc[^a]+jk' works with your example. Another possibility is to look for the match starting at different locations, something like this: p = re.compile(r'abc.+jk') lastMatch = None i = 0 while i len(s): m = p.search(s, i) if m is None: break lastMatch = m.group() i = m.start() + 1 print lastMatch Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
2009/4/28 Marek spociń...@go2.pl,Poland marek...@10g.pl: Hello, The following code returns 'abc123abc45abc789jk'. How do I revise the pattern so that the return value will be 'abc789jk'? In other words, I want to find the pattern 'abc' that is closest to 'jk'. Here the string '123', '45' and '789' are just examples. They are actually quite different in the string that I'm working with. import re s = 'abc123abc45abc789jk' p = r'abc.+jk' lst = re.findall(p, s) print lst[0] I suggest using r'abc.+?jk' instead. the additional ? makes the preceeding '.+' non-greedy so instead of matching as long string as it can it matches as short string as possible. That was my first idea too, but it does not work for this case, because Python will still try to _start_ the match as soon as possible. To use .+? one would have to revert the string, then use the reverse regular expression on the result, which looks like a rather roundabout way of doing things. -- André Engels, andreeng...@gmail.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
spir denis.spir at free.fr writes: To avoid that, use non-grouping parens (?:...). This also avoids the need for parens around the whole format: p = Pattern(r'abc(?:(?!abc).)+jk') print p.findall(s) ['abc789jk'] Denis This one works! Thank you Denis. I'll try it out on the actual much longer (multiline) string and see what happens. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression problem
Spencer Parker wrote: I have a python script that takes a text file as an argument. It then loops through the text file pulling out specific lines of text that I want. I have a regular expression that evaluates the text to see if it matches a specific phrase. Right now I have it writing to another text file that output. The problem I am having is that it finds the phrase prints it, but then it continuously prints the statement. There is only 1 entries in the file for the result it finds, but it prints it multiple times...several hundred before it moves onto the next one. But it appends the first one to the next entry...and does this till it finds everything. http://dpaste.com/33982/ Any Help? dedent the 2nd for loop. -- Bob Gailer Chapel Hill NC 919-636-4239 ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression problem
After he said that...I realized where I was being dumb... On Wed, Apr 15, 2009 at 10:29 AM, bob gailer bgai...@gmail.com wrote: Spencer Parker wrote: I have a python script that takes a text file as an argument. It then loops through the text file pulling out specific lines of text that I want. I have a regular expression that evaluates the text to see if it matches a specific phrase. Right now I have it writing to another text file that output. The problem I am having is that it finds the phrase prints it, but then it continuously prints the statement. There is only 1 entries in the file for the result it finds, but it prints it multiple times...several hundred before it moves onto the next one. But it appends the first one to the next entry...and does this till it finds everything. http://dpaste.com/33982/ Any Help? dedent the 2nd for loop. -- Bob Gailer Chapel Hill NC 919-636-4239 ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression oddity
bob gailer a écrit :
Emmanuel Ruellan wrote:
Hi tutors!
While trying to write a regular expression that would split a string
the way I want, I noticed a behaviour I didn't expect.
re.findall('.?', 'some text')
['s', 'o', 'm', 'e', ' ', 't', 'e', 'x', 't', '']
Where does the last string, the empty one, come from?
I find this behaviour rather annoying: I'm getting one group too many.
The ? means 0 or 1 occurrence. I think re is matching the null string at
the end.
Drop the ? and you'll get what you want.
Of course you can get the same thing using list('some text') at lower cost.
I find this fully consistent, for your regex means matching
* either any char
* or no char at all
Logically, you first get n chars, then one 'nothing'. Only after that will
parsing be stopped because of end of string. Maybe clearer:
print re.findall('.?', '')
== ['']
print re.findall('.', '')
== []
denis
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Re: [Tutor] Regular expression oddity
Emmanuel Ruellan wrote:
Hi tutors!
While trying to write a regular expression that would split a string
the way I want, I noticed a behaviour I didn't expect.
re.findall('.?', 'some text')
['s', 'o', 'm', 'e', ' ', 't', 'e', 'x', 't', '']
Where does the last string, the empty one, come from?
I find this behaviour rather annoying: I'm getting one group too many.
The ? means 0 or 1 occurrence. I think re is matching the null string at
the end.
Drop the ? and you'll get what you want.
Of course you can get the same thing using list('some text') at lower cost.
--
Bob Gailer
Chapel Hill NC
919-636-4239
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Re: [Tutor] Regular expression to match \f in groff input?
On Thu, Aug 21, 2008 at 1:40 PM, Bill Campbell [EMAIL PROTECTED] wrote: I've been beating my head against the wall try to figure out how to get regular expressions to match the font change sequences in *roff input (e.g. \fB for bold, \fP to revert to previous font). The re library maps r'\f' to the single form-feed character (as it does other common single-character sequences like r'\n'). Does this example help? ### sampleText = r\This\ has \backslashes\. print sampleText \This\ has \backslashes\. import re re.findall(r\\\w+\\, sampleText) ['\\This\\', '\\backslashes\\'] ### ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression to match \f in groff input?
Bill Campbell [EMAIL PROTECTED] wrote to get regular expressions to match the font change sequences in *roff input (e.g. \fB for bold, \fP to revert to previous font). The re library maps r'\f' to the single form-feed character (as it does other common single-character sequences like r'\n'). I think all you need is an extra \ to escape the \ character in \f This does not work in puthon: s = re.sub(r'\f[1NR]', '/emphasis, sinput) Try s = re.sub(r'\\f[1NR]', '/emphasis, sinput) HTH, -- Alan Gauld Author of the Learn to Program web site http://www.freenetpages.co.uk/hp/alan.gauld ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression
I need to pull the highligted data from a similar file and can't seem to
get
my script to work:
Script:
import re
infile = open(filter.txt,r)
outfile = open(out.txt,w)
patt = re.compile(r~02([\d{10}]))
You have to allow for the characters at the beginning and end too.
Try this.
re.compile(r.*~02(\d{10})~.*)
Also outfile.write(%s\n) literally writes %s\n
You need this I believe
outfile.write(%s\n % m.group(1))
for line in infile:
m = patt.match(line)
if m:
outfile.write(%s\n)
infile.close()
outfile.close()
File:
200~02001491~05070
200~02001777~05070
200~02001995~05090
200~02002609~05090
200~02002789~05070
200~012~02004169~0
200~02004247~05090
200~02008623~05090
200~02010957~05090
200~02 011479~05090
200~0199~02001237~
200~02011600~05090
200~012~02 022305~0
200~02023546~05090
200~02025427~05090
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Re: [Tutor] Regular Expression
You need to pass a parameter to the string in the following line:
outfile.write(%s\n % m.string[m.start():m.end()])
And you need to use m.search, not m.match in the line where you're
actually apply the expression to the string
m = patt.search(line)
--Michael
On 12/21/07, Que Prime [EMAIL PROTECTED] wrote:
I need to pull the highligted data from a similar file and can't seem to get
my script to work:
Script:
import re
infile = open(filter.txt,r)
outfile = open(out.txt,w)
patt = re.compile(r~02([\d{10}]))
for line in infile:
m = patt.match(line)
if m:
outfile.write(%s\n)
infile.close()
outfile.close()
File:
200~02001491~05070
200~02001777~05070
200~02001995~05090
200~02002609~05090
200~02002789~05070
200~012~02004169~0
200~02004247~05090
200~02008623~05090
200~02010957~05090
200~02 011479~05090
200~0199~02001237~
200~02011600~05090
200~012~02 022305~0
200~02023546~05090
200~02025427~05090
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Phone: 404-386-0495
Consulting: http://www.RowdyLabs.com
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Re: [Tutor] Regular Expression help - parsing AppleScript Lists as Strings
Andrew Wu wrote:
pattern3 = '''
^{
(
%s
| {%s} # Possible to have 1 level of nested lists
,?)* # Items are comma-delimited, except for the last item
}$
''' % (pattern2, pattern2)
The above doesn't allow comma after the first instance of pattern2 and
it doesn't allow space after either instance. Here is a version that
passes your tests:
pattern3 = '''
^{
(
(%s
| {%s}) # Possible to have 1 level of nested lists
,?\s*)* # Items are comma-delimited, except for the last item
}$
''' % (pattern2, pattern2)
You might want to look at doing this with pyparsing, I think it will
make it easier to get the data out vs just recognizing the correct pattern.
Kent
PS Please post in plain text, not HTML.
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Re: [Tutor] Regular Expression help - parsing AppleScript Lists as Strings
Kent Johnson wrote:
You might want to look at doing this with pyparsing, I think it will
make it easier to get the data out vs just recognizing the correct pattern.
Here is a pyparsing version that correctly recognizes all of your
patterns and returns a (possibly nested) Python list in case of a match.
Note that this version will parse lists that are nested arbitrarily
deeply. If you don't want that you will have to define two kinds of
lists, a singly-nested list and a non-nested list.
Kent
from pyparsing import *
List = Forward()
T = Literal('true').setParseAction( lambda s,l,t: [ True ] )
F = Literal('false').setParseAction( lambda s,l,t: [ False ] )
String = QuotedString('')
Number = Word(nums).setParseAction( lambda s,l,t: [ int(t[0]) ] )
List Literal('{').suppress() +
delimitedList(T|F|String|Number|Group(List)) + Literal('}').suppress()
def IsASList(s):
# AppleScript lists are bracked by curly braces with items separate
by commas
# Each item is an alphanumeric label(?) or a string enclosed by
# double quotes or a list itself
# e.g. {2, True, hello}
try:
parsed = List.parseString(s)
return parsed.asList()
except Exception, e:
return None
sample_strs = [
'{}', # Empty list
'{a}', # Should not match
'{a, b, c}', # Should not match
'{hello}',
'{hello, kitty}',
'{true}',
'{false}',
'{true, false}',
'{9}',
'{9,10, 11}',
'{93214, true, false, hello, kitty}',
'{{1, 2, 3}}', # This matches
'{{1, 2, cat}, 1}', # This matches
# These don't match:
'{{1,2,3},1,{4,5,6},2}',
'{1, {2, 3, 4}, 3}',
'{{1, 2, 3}, {4, 5, 6}, 1}',
'{1, {1, 2, 3}}', # Should match but doesn't
'{93214, true, false, hello, kitty, {1, 2, 3}}', # Should match
but doesn't
'{label: hello, value: false, num: 2}', # AppleScript dictionary
- should not match
]
for sample in sample_strs:
result = IsASList(sample)
print 'Is AppleScript List: %s; String: %s' % (bool(result), sample)
if result:
print result
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Re: [Tutor] Regular Expression help - parsing AppleScript Lists as Strings
Ah - thanks for the correction! I missed the extra grouping and the
extra spacing ... doh! Sorry about the HTML-formatted e-mail ...
Thanks also for the pyparsing variant as well - I didn't know the
module existed before!
Andrew
On 11/1/07, Kent Johnson [EMAIL PROTECTED] wrote:
Andrew Wu wrote:
pattern3 = '''
^{
(
%s
| {%s} # Possible to have 1 level of nested lists
,?)* # Items are comma-delimited, except for the last item
}$
''' % (pattern2, pattern2)
The above doesn't allow comma after the first instance of pattern2 and
it doesn't allow space after either instance. Here is a version that
passes your tests:
pattern3 = '''
^{
(
(%s
| {%s}) # Possible to have 1 level of nested lists
,?\s*)* # Items are comma-delimited, except for the last item
}$
''' % (pattern2, pattern2)
You might want to look at doing this with pyparsing, I think it will
make it easier to get the data out vs just recognizing the correct pattern.
Kent
PS Please post in plain text, not HTML.
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Re: [Tutor] Regular Expression help
Thanks everyone for the replies all worked well, I adopted the string splitting approach in favour of the regex one as it seemed to miss less of the edge cases. I would like to thank everyone for their help once again -Original Message- From: Kent Johnson [mailto:[EMAIL PROTECTED] Sent: 27 June 2007 14:55 To: tutor@python.org; Gardner, Dean Subject: Re: [Tutor] Regular Expression help Gardner, Dean wrote: Hi I have a text file that I would like to split up so that I can use it in Excel to filter a certain field. However as it is a flat text file I need to do some processing on it so that Excel can correctly import it. File Example: tag descVR VM (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1 What I essentially want is to use python to process this file to give me (0012,0042); Clinical Trial Subject Reading ID; LO; 1 (0012,0050); Clinical Trial Time Point ID; LO; 1 (0012,0051); Clinical Trial Time Point Description; ST; 1 (0012,0060); Clinical Trial Coordinating Center Name; LO; 1 (0018,0010); Contrast/Bolus Agent; LO; 1 (0018,0012); Contrast/Bolus Agent Sequence; SQ ;1 (0018,0014); Contrast/Bolus Administration Route Sequence; SQ; 1 (0018,0015); Body Part Examined; CS; 1 so that I can import to excel using a delimiter. This file is extremely long and all I essentially want to do is to break it into it 'fields' Now I suspect that regular expressions are the way to go but I have only basic experience of using these and I have no idea what I should be doing. This seems to work: data = '''\ (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1'''.splitlines() import re fieldsRe = re.compile(r'^(\(\d+,\d+\)) (.*?) (\w+) (\d+)$') for line in data: match = fieldsRe.match(line) if match: print ';'.join(match.group(1, 2, 3, 4)) I don't think you want the space after the ; that you put in your example; Excel wants a single-character delimiter. Kent DISCLAIMER: Unless indicated otherwise, the information contained in this message is privileged and confidential, and is intended only for the use of the addressee(s) named above and others who have been specifically authorized to receive it. If you are not the intended recipient, you are hereby notified that any dissemination, distribution or copying of this message and/or attachments is strictly prohibited. The company accepts no liability for any damage caused by any virus transmitted by this email. Furthermore, the company does not warrant a proper and complete transmission of this information, nor does it accept liability for any delays. If you have received this message in error, please contact the sender and delete the message. Thank you. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression help
Gardner, Dean wrote: Hi I have a text file that I would like to split up so that I can use it in Excel to filter a certain field. However as it is a flat text file I need to do some processing on it so that Excel can correctly import it. File Example: tag descVR VM (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1 What I essentially want is to use python to process this file to give me (0012,0042); Clinical Trial Subject Reading ID; LO; 1 (0012,0050); Clinical Trial Time Point ID; LO; 1 (0012,0051); Clinical Trial Time Point Description; ST; 1 (0012,0060); Clinical Trial Coordinating Center Name; LO; 1 (0018,0010); Contrast/Bolus Agent; LO; 1 (0018,0012); Contrast/Bolus Agent Sequence; SQ ;1 (0018,0014); Contrast/Bolus Administration Route Sequence; SQ; 1 (0018,0015); Body Part Examined; CS; 1 so that I can import to excel using a delimiter. This file is extremely long and all I essentially want to do is to break it into it 'fields' Now I suspect that regular expressions are the way to go but I have only basic experience of using these and I have no idea what I should be doing. This seems to work: data = '''\ (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1'''.splitlines() import re fieldsRe = re.compile(r'^(\(\d+,\d+\)) (.*?) (\w+) (\d+)$') for line in data: match = fieldsRe.match(line) if match: print ';'.join(match.group(1, 2, 3, 4)) I don't think you want the space after the ; that you put in your example; Excel wants a single-character delimiter. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression help
Gardner, Dean wrote: Hi I have a text file that I would like to split up so that I can use it in Excel to filter a certain field. However as it is a flat text file I need to do some processing on it so that Excel can correctly import it. File Example: tag descVR VM (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1 What I essentially want is to use python to process this file to give me (0012,0042); Clinical Trial Subject Reading ID; LO; 1 (0012,0050); Clinical Trial Time Point ID; LO; 1 (0012,0051); Clinical Trial Time Point Description; ST; 1 (0012,0060); Clinical Trial Coordinating Center Name; LO; 1 (0018,0010); Contrast/Bolus Agent; LO; 1 (0018,0012); Contrast/Bolus Agent Sequence; SQ ;1 (0018,0014); Contrast/Bolus Administration Route Sequence; SQ; 1 (0018,0015); Body Part Examined; CS; 1 so that I can import to excel using a delimiter. This file is extremely long and all I essentially want to do is to break it into it 'fields' Now I suspect that regular expressions are the way to go but I have only basic experience of using these and I have no idea what I should be doing. This seems to work: data = '''\ (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1'''.splitlines() import re fieldsRe = re.compile(r'^(\(\d+,\d+\)) (.*?) (\w+) (\d+)$') for line in data: match = fieldsRe.match(line) if match: print ';'.join(match.group(1, 2, 3, 4)) I don't think you want the space after the ; that you put in your example; Excel wants a single-character delimiter. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression help
Argh... My e-mail program really messed up the threads. I didn't notice that there was already multiple replies to this message. Doh! Mike ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression help
Gardner, Dean wrote: Hi I have a text file that I would like to split up so that I can use it in Excel to filter a certain field. However as it is a flat text file I need to do some processing on it so that Excel can correctly import it. File Example: tag descVR VM (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1 What I essentially want is to use python to process this file to give me (0012,0042); Clinical Trial Subject Reading ID; LO; 1 (0012,0050); Clinical Trial Time Point ID; LO; 1 (0012,0051); Clinical Trial Time Point Description; ST; 1 (0012,0060); Clinical Trial Coordinating Center Name; LO; 1 (0018,0010); Contrast/Bolus Agent; LO; 1 (0018,0012); Contrast/Bolus Agent Sequence; SQ ;1 (0018,0014); Contrast/Bolus Administration Route Sequence; SQ; 1 (0018,0015); Body Part Examined; CS; 1 so that I can import to excel using a delimiter. This file is extremely long and all I essentially want to do is to break it into it 'fields' Now I suspect that regular expressions are the way to go but I have only basic experience of using these and I have no idea what I should be doing. This seems to work: data = '''\ (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1'''.splitlines() import re fieldsRe = re.compile(r'^(\(\d+,\d+\)) (.*?) (\w+) (\d+)$') for line in data: match = fieldsRe.match(line) if match: print ';'.join(match.group(1, 2, 3, 4)) I don't think you want the space after the ; that you put in your example; Excel wants a single-character delimiter. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression help
Gardner, Dean wrote: Hi I have a text file that I would like to split up so that I can use it in Excel to filter a certain field. However as it is a flat text file I need to do some processing on it so that Excel can correctly import it. File Example: tag descVR VM (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1 What I essentially want is to use python to process this file to give me (0012,0042); Clinical Trial Subject Reading ID; LO; 1 (0012,0050); Clinical Trial Time Point ID; LO; 1 (0012,0051); Clinical Trial Time Point Description; ST; 1 (0012,0060); Clinical Trial Coordinating Center Name; LO; 1 (0018,0010); Contrast/Bolus Agent; LO; 1 (0018,0012); Contrast/Bolus Agent Sequence; SQ ;1 (0018,0014); Contrast/Bolus Administration Route Sequence; SQ; 1 (0018,0015); Body Part Examined; CS; 1 so that I can import to excel using a delimiter. This file is extremely long and all I essentially want to do is to break it into it 'fields' Now I suspect that regular expressions are the way to go but I have only basic experience of using these and I have no idea what I should be doing. This seems to work: data = '''\ (0012,0042) Clinical Trial Subject Reading ID LO 1 (0012,0050) Clinical Trial Time Point ID LO 1 (0012,0051) Clinical Trial Time Point Description ST 1 (0012,0060) Clinical Trial Coordinating Center Name LO 1 (0018,0010) Contrast/Bolus Agent LO 1 (0018,0012) Contrast/Bolus Agent Sequence SQ 1 (0018,0014) Contrast/Bolus Administration Route Sequence SQ 1 (0018,0015) Body Part Examined CS 1'''.splitlines() import re fieldsRe = re.compile(r'^(\(\d+,\d+\)) (.*?) (\w+) (\d+)$') for line in data: match = fieldsRe.match(line) if match: print ';'.join(match.group(1, 2, 3, 4)) I don't think you want the space after the ; that you put in your example; Excel wants a single-character delimiter. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression questions
Bernard Lebel wrote:
Hello,
Once again struggling with regular expressions.
I have a string that look like something_shp1.
I want to replace _shp1 by _shp. I'm never sure if it's going to
be 1, if there's going to be a number after _shp.
So I'm trying to use regular expression to perform this replacement.
But I just can't seem to get a match! I always get a None match.
I would think that this would have done the job:
r = re.compile( r(_shp\d)$ )
The only way I have found to get a match, is using
r = re.compile( r(\S+_shp\d)$ )
My guess is you are calling r.match() rather than r.search(). r.match()
only looks for matches at the start of the string; r.search() will find
a match anywhere.
My second question is related more to the actual string replacement.
Using regular expressions, what would be the way to go? I have tried
the following:
newstring = r.sub( '_shp', oldstring )
But the new string is always _shp instead of something_shp.
Because your re matches something_shp.
I think
newstring = re.sub('_shp\d' '_shp', oldstring )
will do what you want.
Kent
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Re: [Tutor] Regular expression questions
Thanks a lot Kent, that indeed solves the issues altogether.
Cheers
Bernard
On 5/3/07, Kent Johnson [EMAIL PROTECTED] wrote:
Bernard Lebel wrote:
Hello,
Once again struggling with regular expressions.
I have a string that look like something_shp1.
I want to replace _shp1 by _shp. I'm never sure if it's going to
be 1, if there's going to be a number after _shp.
So I'm trying to use regular expression to perform this replacement.
But I just can't seem to get a match! I always get a None match.
I would think that this would have done the job:
r = re.compile( r(_shp\d)$ )
The only way I have found to get a match, is using
r = re.compile( r(\S+_shp\d)$ )
My guess is you are calling r.match() rather than r.search(). r.match()
only looks for matches at the start of the string; r.search() will find
a match anywhere.
My second question is related more to the actual string replacement.
Using regular expressions, what would be the way to go? I have tried
the following:
newstring = r.sub( '_shp', oldstring )
But the new string is always _shp instead of something_shp.
Because your re matches something_shp.
I think
newstring = re.sub('_shp\d' '_shp', oldstring )
will do what you want.
Kent
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Re: [Tutor] regular expression
arbaro arbaro wrote:
Hello,
I'm trying to mount an usb device from python under linux.
To do so, I read the kernel log /proc/kmsg and watch for something like:
6 /dev/scsi/host3/bus0/target0/lun0/:7usb-storage: device scan
complete
When I compile a regular expression like:
r = re.compile('\d+\s/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
You should use raw strings for regular expressions that contain \
characters.
It is found. But I don't want the \d+\s or '6 ' in front of the
path, so I tried:
r = re.compile('/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
But this way the usb device path it is not found.
You are using re.match(), which just looks for a match at the start of
the string. Try using re.search() instead.
http://docs.python.org/lib/matching-searching.html
Kent
So what i'm trying to do is:
- find the usb device path from the kernel log with the regular
expression.
- Determine the start and end positions of the match (and add /disc or
/part1 to the match).
- And use that to mount the usb stick on /mnt/usb - mount -t auto
match /mnt/usb
If anyone can see what i'm doing wrong, please tell me, because I
don't understand it anymore.
Thanks.
Below is the code:
# \d+ = 1 or more digits
# \s = an empty space
import re
def findusbdevice():
''' Returns path of usb device '''
# I did a 'cat /proc/kmsg /log/kmsg' to be able to read the kernel
message.
# Somehow I can't read /proc/kmsg directly.
kmsg = open('/log/kmsg', 'r')
r = re.compile('/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
#r = re.compile('\d+\s/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
for line in kmsg:
if 'usb-storage' in line and r.match(line):
print 'Success', line
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Re: [Tutor] regular expression
Hello,
Im just answering my own email, since I just found out what my error was.
From a regular expression howto:
http://www.amk.ca/python/howto/regex/regex.html
The match() function only checks if the RE matches at the beginning of
the string while search() will scan forward through the string for a
match. It's important to keep this distinction in mind. Remember,
match() will only report a successful match which will start at 0; if
the match wouldn't start at zero, match() will not report it.
That was exactly my problem. Replacing r.match(line) for
r.search(line) solved it.
Sorry for having bothered you prematurely.
On 8/3/06, arbaro arbaro [EMAIL PROTECTED] wrote:
Hello,
I'm trying to mount an usb device from python under linux.
To do so, I read the kernel log /proc/kmsg and watch for something like:
6 /dev/scsi/host3/bus0/target0/lun0/:7usb-storage: device scan
complete
When I compile a regular expression like:
r = re.compile('\d+\s/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
It is found. But I don't want the \d+\s or '6 ' in front of the path, so
I tried:
r = re.compile('/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
But this way the usb device path it is not found.
So what i'm trying to do is:
- find the usb device path from the kernel log with the regular expression.
- Determine the start and end positions of the match (and add /disc or /part1
to the match).
- And use that to mount the usb stick on /mnt/usb - mount -t auto match
/mnt/usb
If anyone can see what i'm doing wrong, please tell me, because I don't
understand it anymore.
Thanks.
Below is the code:
# \d+ = 1 or more digits
# \s = an empty space
import re
def findusbdevice():
''' Returns path of usb device '''
# I did a 'cat /proc/kmsg /log/kmsg' to be able to read the kernel
message.
# Somehow I can't read /proc/kmsg directly.
kmsg = open('/log/kmsg', 'r')
r = re.compile('/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
#r = re.compile('\d+\s/dev/scsi/host\d+/bus\d+/target\d+/lun\d+')
for line in kmsg:
if 'usb-storage' in line and r.match(line):
print 'Success', line
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Re: [Tutor] Regular Expression Misunderstanding
On 14/07/06, Steve Nelson [EMAIL PROTECTED] wrote: What I don't understand is how in the end the RE *does* actually match - which may indicate a serious misunderstanding on my part. re.match(a[bcd]*b, abcbd) _sre.SRE_Match object at 0x186b7b10 I don't see how abcbd matches! It ends with a d and the RE states it should end with a b! What am I missing? It doesn't have to match the _whole_ string. [bcd]* will match, amongst other things, the empty string (ie: 0 repetitions of either a b, a c, or a d). So a[bcd]*b will match ab, which is in the string abcbd. It will also match abcb, which is the longest match, and thus probably the one it found. If you look at the match object returned, you should se that the match starts at position 0 and is four characters long. Now, if you asked for a[bcd]*b$, that would be a different matter! HTH :-) -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
On 7/14/06, John Fouhy [EMAIL PROTECTED] wrote: It doesn't have to match the _whole_ string. Ah right - yes, so it doesn't say that it has to end with a b - as per your comment about ending with $. If you look at the match object returned, you should se that the match starts at position 0 and is four characters long. How does one query a match object in this way? I am learning by fiddling interactively. John. S. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
Steve Nelson wrote: On 7/14/06, John Fouhy [EMAIL PROTECTED] wrote: It doesn't have to match the _whole_ string. Ah right - yes, so it doesn't say that it has to end with a b - as per your comment about ending with $. The matched portion must end with b, but it doesn't have to coincide with the end of the string. The whole regex must be used for it to match; the whole string does not have to be used - the matched portion can be a substring. If you look at the match object returned, you should se that the match starts at position 0 and is four characters long. How does one query a match object in this way? I am learning by fiddling interactively. The docs for match objects are here: http://docs.python.org/lib/match-objects.html match.start() and match.end() will tell you where it matched. You might like to try the regex demo that comes with Python; on Windows it is installed at C:\Python24\Tools\Scripts\redemo.py. It gives you an easy way to experiment with regexes. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
On 14/07/06, Steve Nelson [EMAIL PROTECTED] wrote: How does one query a match object in this way? I am learning by fiddling interactively. If you're fiddling interactively, try the dir() command -- ie: m = re.match(...) dir(m) It will tell you what attributes the match object has. Or you can read the documentation --- a combination of both approaches usually works quite well :-) -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
On 7/14/06, John Fouhy [EMAIL PROTECTED] wrote: m = re.match(...) dir(m) It will tell you what attributes the match object has. Useful - thank you. I am now confuse on this: I have a file full of lines beginning with the letter b. I want a RE that will return the whole line if it begins with b. I find if I do eg: m = re.search(^b, b spam spam spam) m.group() 'b' How do I get it to return the whole line if it begins with a b? S. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
I have a file full of lines beginning with the letter b. I want a RE that will return the whole line if it begins with b. I find if I do eg: m = re.search(^b, b spam spam spam) m.group() 'b' How do I get it to return the whole line if it begins with a b? S. for line in file: if line.strip()[0] == 'b': print line or print [a for a in file if a.strip()[0] == b] if you want to use list comprehension. As for the RE way, I've no idea. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular Expression Misunderstanding
Steve Nelson wrote:
On 7/14/06, John Fouhy [EMAIL PROTECTED] wrote:
m = re.match(...)
dir(m)
It will tell you what attributes the match object has.
Useful - thank you.
I am now confuse on this:
I have a file full of lines beginning with the letter b. I want a
RE that will return the whole line if it begins with b.
I find if I do eg:
m = re.search(^b, b spam spam spam)
m.group()
'b'
How do I get it to return the whole line if it begins with a b?
Use the match object in a test. If the search fails it will return None
which tests false:
for line in lines:
m = re.search(...)
if m:
# do something with line that matches
But for this particular application you might as well use
line.startswith('b') instead of a regex.
Kent
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Re: [Tutor] Regular Expression Misunderstanding
On 7/14/06, Kent Johnson [EMAIL PROTECTED] wrote:
But for this particular application you might as well use
line.startswith('b') instead of a regex.
Ah yes, that makes sense.
Incidentally continuing my reading of the HOWTO I have sat and puzzled
for about 30 mins on the difference the MULTILINE flag makes. I can't
quite see the difference. I *think* it is as follows:
Under normal circumstances, ^ matches the start of a line, only. On a
line by line basis.
With the re.M flag, we get a match after *any* newline?
Similarly with $ - under normal circumstances, $ matches the end of
the string, or that which precedes a newline.
With the MULTILINE flag, $ matches before *any* newline?
Is this correct?
Kent
S.
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Re: [Tutor] Regular Expression Misunderstanding
Steve Nelson wrote:
Incidentally continuing my reading of the HOWTO I have sat and puzzled
for about 30 mins on the difference the MULTILINE flag makes. I can't
quite see the difference. I *think* it is as follows:
Under normal circumstances, ^ matches the start of a line, only. On a
line by line basis.
With the re.M flag, we get a match after *any* newline?
Similarly with $ - under normal circumstances, $ matches the end of
the string, or that which precedes a newline.
With the MULTILINE flag, $ matches before *any* newline?
Is this correct?
I'm not sure, I think you are a little confused. MULTILINE only matters
if the string you are matching contains newlines. Without MULTILINE, ^
will match only at the beginning of the string. With it, ^ will match
after any newline. For example,
In [1]: import re
A string containing two lines:
In [2]: s='one\ntwo'
The first line matches without MULTILINE:
In [3]: re.search('^one', s)
Out[3]: _sre.SRE_Match object at 0x00C3E640
The second one does not (result of the search is None so nothing prints):
In [4]: re.search('^two', s)
With MULTILINE ^two will match:
In [5]: re.search('^two', s, re.MULTILINE)
Out[5]: _sre.SRE_Match object at 0x00E901E0
Kent
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Re: [Tutor] regular expression matching a dot?
Hi [Christian|List] This post is not regarding your special problem (which anyway has been solved by now), but I'd like to share some general tip on working with regular expressions. There are some nice regex-debuggers out there that can help clearify what went wrong when a regex doesn't match when it should or vice versa. Kodos http://kodos.sourceforge.net/ is one of them, but there are many others that can make your life easier ; at least in terms of regex-debugging ;) Probably most of you (especially all regex-gurus) know about this already, but i thought it was worth the post as a hint for all beginners hth Frank On Wed, 2005-10-19 at 09:45 +0200, Christian Meesters wrote: Hi I've got the problem that I need to find a certain group of file names within a lot of different file names. Those I want to match with a regular expression are a bit peculiar since they all look like: ... ...(regex-problem) ... Any ideas what I could do else? TIA Christian PS Hope that I described the problem well enough ... ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression matching a dot?
I personally fancy Kiki, that comes with the WxPython installer... It has very nice coloring for grouping in Regexes. Hugo Kent Johnson wrote: Frank Bloeink wrote: There are some nice regex-debuggers out there that can help clearify what went wrong when a regex doesn't match when it should or vice versa. Kodos http://kodos.sourceforge.net/ is one of them, but there are many others that can make your life easier ; at least in terms of regex-debugging ;) Yes, these programs can be very helpful. There is even one that ships with Python - see C:\Python24\Tools\Scripts\redemo.py Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression matching a dot?
Christian Meesters wrote: Hi I've got the problem that I need to find a certain group of file names within a lot of different file names. Those I want to match with a regular expression are a bit peculiar since they all look like: 07SS.INF , 10SE.INF, 13SS.INF, 02BS.INF, 05SS.INF. Unfortunately there are similar file names that shouldn't be matched, like: 01BE.INF, 02BS.INF Any other extension than 'INF' should also be skipped. (There are names like 07SS.E00, wich I don't want to see matched.) So I tried the following pattern (using re): \d+[SS|SE]\.INF - as there should be at least one digit, the group 'SE' or 'SS' followed by a dot and the extension 'INF'. Use parentheses () for grouping. Brackets [] define a group of characters. Your re says to match \d+ one or more digits [SS|SE] exactly one of the characters S, |, E \.INF literal .INF Since there are TWO characters S or E, nothing matches. Change the [] to () and it works. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression matching a dot?
[ Workaround ]
What about using the glob module?
http://docs.python.org/lib/module-glob.html
you can use something like
glob.glob('./[0-9][0-9]S[E|S].INF')
(Not tested)
Misto
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Re: [Tutor] regular expression matching a dot?
Actually, your answer did help to open my eyes. The expression is
\d+S[S|E]\.INF: Ouch!
Thanks a lot,
Christian
On 19 Oct 2005, at 12:11, Misto . wrote:
[ Workaround ]
What about using the glob module?
http://docs.python.org/lib/module-glob.html
you can use something like
glob.glob('./[0-9][0-9]S[E|S].INF')
(Not tested)
Misto
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Re: [Tutor] regular expression matching a dot?
Christian Meesters wrote: Actually, your answer did help to open my eyes. The expression is \d+S[S|E]\.INF: Ouch! That will work, but what you really mean is one of these: \d+S[SE]\.INF \d+S(S|E)\.INF Your regex will match 0S|.INF Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression matching a dot?
Thanks, corrected. I was happy now - and then too fast ;-). Cheers Christian On 19 Oct 2005, at 13:50, Kent Johnson wrote: Christian Meesters wrote: Actually, your answer did help to open my eyes. The expression is \d+S[S|E]\.INF: Ouch! That will work, but what you really mean is one of these: \d+S[SE]\.INF \d+S(S|E)\.INF Your regex will match 0S|.INF Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression error
Bernard Lebel wrote: Hello, I'm using regular expressions to check if arguments of a function are valid. When I run the code below, I get an error. Basically, what the regular expression should expect is either an integer, or an integer followed by a letter (string). I convert the tested argument to a string to make sure. Below is the code. # Create regular expression to match oPattern = re.compile( r(\d+|\d+\D), re.IGNORECASE ) This will match a string of digits followed by any non-digit, is that what you want? If you want to restrict it to digits followed by a letter you should use r(\d+|\d+[a-z]) Also this will match something like 123A456B, if you want to disallow anything after the letter you need to match the end of the string: r(\d+|\d+[a-z])$ # Iterate provided arguments for oArg in aArgs: # Attempt to match the argument to the regular expression oMatch = re.match( str( oArg ), 0 ) The problem is you are calling the module (re) match, not the instance (oPattern) match. re.match() expects the second argument to be a string. Just use oMatch = oPattern.match( str( oArg ), 0 ) The hint in the error is expected string or buffer. So you are not giving the expected argument types which should send you to the docs to check... Kent The error I get is this: #ERROR : Traceback (most recent call last): # File Script Block , line 208, in BuildProjectPaths_Execute #aShotInfos = handleArguments( args ) # File Script Block , line 123, in handleArguments #oMatch = re.match( str( oArg ), 0 ) # File D:\Python24\Lib\sre.py, line 129, in match #return _compile(pattern, flags).match(string) #TypeError: expected string or buffer # - [line 122] ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
D Elliott wrote: I wonder if anyone can help me with an RE. I also wonder if there is an RE mailing list anywhere - I haven't managed to find one. I'm trying to use this regular expression to delete particular strings from a file before tokenising it. I want to delete all strings that have a full stop (period) when it is not at the beginning or end of a word, and also when it is not followed by a closing bracket. I want to delete file names (eg. fileX.doc), and websites (when www/http not given) but not file extensions (eg. this is in .jpg format). I also don't want to delete the last word of each sentence just because it precedes a fullstop, or if there's a fullstop followed by a closing bracket. fullstopRe = re.compile (r'\S+\.[^)}]]+') There are two problems with this is: - The ] inside the [] group must be escaped like this: [^)}\]] - [^)}\]] matches any whitespace so it will match on the ends of words It's not clear from your description if the closing bracket must immediately follow the full stop or if it can be anywhere after it. If you want it to follow immediately then use \S+\.[^)}\]\s]\S* If you want to allow the bracket anywhere after the stop you must force the match to go to a word boundary otherwise you will match foo.bar when the word is foo.bar]. I think this works: (\S+\.[^)}\]\s]+)(\s) but you have to include the second group in your substitution string. BTW C:\Python23\pythonw.exe C:\Python24\Tools\Scripts\redemo.py is very helpful with questions like this... Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
But I only want to ignore B if A is a match. If A is not a match,
I'd like it to advance on to B.
On Mar 9, 2005, at 12:07 PM, Marcos Mendonça wrote:
Hi
Not and regexp expert. But it seems to me that if you want to ignora
B then it should be
(A) | (^B)
Hope it helps!
On Wed, 9 Mar 2005 11:11:57 -0800, Mike Hall
[EMAIL PROTECTED] wrote:
I'm having some strange results using the or operator. In every
test
I do I'm matching both sides of the | metacharacter, not one or the
other as all documentation says it should be (the parser supposedly
scans left to right, using the first match it finds and ignoring the
rest). It should only go beyond the | if there was no match found
before it, no?
Correct me if I'm wrong, but your regex is saying match dog, unless
it's followed by cat. if it is followed by cat there is no match on
this side of the | at which point we advance past it and look at the
alternative expression which says to match in front of cat.
However, if I run a .sub using your regex on a string contain both dog
and cat, both will be replaced.
A simple example will show what I mean:
import re
x = re.compile(r(A) | (B))
s = X R A Y B E
r = x.sub(13, s)
print r
X R 13Y13 E
...so unless I'm understanding it wrong, B is supposed to be ignored
if A is matched, yet I get both matched. I get the same result if I
put A and B within the same group.
On Mar 8, 2005, at 6:47 PM, Danny Yoo wrote:
Regular expressions are a little evil at times; here's what I think
you're
thinking of:
###
import re
pattern = re.compile(rdog(?!cat)
...| (?=dogcat), re.VERBOSE)
pattern.match('dogman').start()
0
pattern.search('dogcatcher').start()
Hi Mike,
Gaaah, bad copy-and-paste. The example with 'dogcatcher' actually
does
come up with a result:
###
pattern.search('dogcatcher').start()
6
###
Sorry about that!
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Re: [Tutor] regular expression question
Actually, you should get that anyway...
|
Alternation, or the ``or'' operator. If A and B are regular
expressions, A|B will match any string that matches either A or B.
| has very low precedence in order to make it work reasonably when
you're alternating multi-character strings. Crow|Servo will match
either Crow or Servo, not Cro, a w or an S, and ervo.
So, for each letter in that string, it's checking to see if any letter
matches 'A' or 'B' ...
the engine steps through one character at a time.
sorta like -
for letter in s:
if letter == 'A':
#Do some string stuff
elif letter == 'B':
#do some string stuff
i.e.
k = ['A','B', 'C', 'B']
for i in range(len(k)):
if k[i] == 'A' or k[i]=='B':
k[i]==13
print k
[13, 13, 'C', 13]
You can limit substitutions using an optional argument, but yeah, it
seems you're expecting it to examine the string as a whole.
Check out the example here -
http://www.amk.ca/python/howto/regex/regex.html#SECTION00032
Also
http://www.regular-expressions.info/alternation.html
Regards,
Liam Clarke
On Thu, 10 Mar 2005 09:09:13 +1300, Liam Clarke [EMAIL PROTECTED] wrote:
Hi Mike,
Do you get the same results for a search pattern of 'A|B'?
On Wed, 9 Mar 2005 11:11:57 -0800, Mike Hall
[EMAIL PROTECTED] wrote:
I'm having some strange results using the or operator. In every test
I do I'm matching both sides of the | metacharacter, not one or the
other as all documentation says it should be (the parser supposedly
scans left to right, using the first match it finds and ignoring the
rest). It should only go beyond the | if there was no match found
before it, no?
Correct me if I'm wrong, but your regex is saying match dog, unless
it's followed by cat. if it is followed by cat there is no match on
this side of the | at which point we advance past it and look at the
alternative expression which says to match in front of cat.
However, if I run a .sub using your regex on a string contain both dog
and cat, both will be replaced.
A simple example will show what I mean:
import re
x = re.compile(r(A) | (B))
s = X R A Y B E
r = x.sub(13, s)
print r
X R 13Y13 E
...so unless I'm understanding it wrong, B is supposed to be ignored
if A is matched, yet I get both matched. I get the same result if I
put A and B within the same group.
On Mar 8, 2005, at 6:47 PM, Danny Yoo wrote:
Regular expressions are a little evil at times; here's what I think
you're
thinking of:
###
import re
pattern = re.compile(rdog(?!cat)
...| (?=dogcat), re.VERBOSE)
pattern.match('dogman').start()
0
pattern.search('dogcatcher').start()
Hi Mike,
Gaaah, bad copy-and-paste. The example with 'dogcatcher' actually does
come up with a result:
###
pattern.search('dogcatcher').start()
6
###
Sorry about that!
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Re: [Tutor] regular expression question
but yeah, it
seems you're expecting it to examine the string as a whole.
I guess I was, good point.
On Mar 9, 2005, at 12:28 PM, Liam Clarke wrote:
Actually, you should get that anyway...
|
Alternation, or the ``or'' operator. If A and B are regular
expressions, A|B will match any string that matches either A or B.
| has very low precedence in order to make it work reasonably when
you're alternating multi-character strings. Crow|Servo will match
either Crow or Servo, not Cro, a w or an S, and ervo.
So, for each letter in that string, it's checking to see if any letter
matches 'A' or 'B' ...
the engine steps through one character at a time.
sorta like -
for letter in s:
if letter == 'A':
#Do some string stuff
elif letter == 'B':
#do some string stuff
i.e.
k = ['A','B', 'C', 'B']
for i in range(len(k)):
if k[i] == 'A' or k[i]=='B':
k[i]==13
print k
[13, 13, 'C', 13]
You can limit substitutions using an optional argument, but yeah, it
seems you're expecting it to examine the string as a whole.
Check out the example here -
http://www.amk.ca/python/howto/regex/
regex.html#SECTION00032
Also
http://www.regular-expressions.info/alternation.html
Regards,
Liam Clarke
On Thu, 10 Mar 2005 09:09:13 +1300, Liam Clarke [EMAIL PROTECTED]
wrote:
Hi Mike,
Do you get the same results for a search pattern of 'A|B'?
On Wed, 9 Mar 2005 11:11:57 -0800, Mike Hall
[EMAIL PROTECTED] wrote:
I'm having some strange results using the or operator. In every
test
I do I'm matching both sides of the | metacharacter, not one or the
other as all documentation says it should be (the parser supposedly
scans left to right, using the first match it finds and ignoring the
rest). It should only go beyond the | if there was no match found
before it, no?
Correct me if I'm wrong, but your regex is saying match dog, unless
it's followed by cat. if it is followed by cat there is no match on
this side of the | at which point we advance past it and look at
the
alternative expression which says to match in front of cat.
However, if I run a .sub using your regex on a string contain both
dog
and cat, both will be replaced.
A simple example will show what I mean:
import re
x = re.compile(r(A) | (B))
s = X R A Y B E
r = x.sub(13, s)
print r
X R 13Y13 E
...so unless I'm understanding it wrong, B is supposed to be
ignored
if A is matched, yet I get both matched. I get the same result if
I
put A and B within the same group.
On Mar 8, 2005, at 6:47 PM, Danny Yoo wrote:
Regular expressions are a little evil at times; here's what I think
you're
thinking of:
###
import re
pattern = re.compile(rdog(?!cat)
...| (?=dogcat), re.VERBOSE)
pattern.match('dogman').start()
0
pattern.search('dogcatcher').start()
Hi Mike,
Gaaah, bad copy-and-paste. The example with 'dogcatcher' actually
does
come up with a result:
###
pattern.search('dogcatcher').start()
6
###
Sorry about that!
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'There is only one basic human right, and that is to do as you damn
well please.
And with it comes the only basic human duty, to take the consequences.
--
'There is only one basic human right, and that is to do as you damn
well please.
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Re: [Tutor] regular expression question
Mike Hall wrote: A simple example will show what I mean: import re x = re.compile(r(A) | (B)) s = X R A Y B E r = x.sub(13, s) print r X R 13Y13 E ...so unless I'm understanding it wrong, B is supposed to be ignored if A is matched, yet I get both matched. I get the same result if I put A and B within the same group. The problem is with your use of sub(), not with |. By default, re.sub() substitutes *all* matches. If you just want to substitute the first match, include the optional count parameter: import re s = X R A Y B E re.sub(r(A) | (B), '13', s) 'X R 13Y13 E' re.sub(r(A) | (B), '13', s, 1) 'X R 13Y B E' BTW, there is a very handy interactive regex tester that comes with Python. On Windows, it is installed at C:\Python23\Tools\Scripts\redemo.py Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
First, thanks for the response. Using your re: my_re = re.compile(r'(dog)(cat)?') ...I seem to simply be matching the pattern Dog. Example: str1 = The dog chased the car str2 = The dog cat parade was under way x1 = re.compile(r'(dog)(cat)?') rep1 = x1.sub(REPLACE, str1) rep2 = x2.sub(REPLACE, str2) print rep1 The REPLACE chased the car print rep2 The REPLACE cat parade was under way ...what I'm looking for is a match for the position in front of Cat, should it exist. On Mar 8, 2005, at 5:54 PM, Sean Perry wrote: Mike Hall wrote: I'd like to get a match for a position in a string preceded by a specified word (let's call it Dog), unless that spot in the string (after Dog) is directly followed by a specific word(let's say Cat), in which case I want my match to occur directly after Cat, and not Dog. I can easily get the spot after Dog, and I can also get it to ignore this spot if Dog is followed by Cat. But what I'm having trouble with is how to match the spot after Cat if this word does indeed exist in the string. . import re . my_re = re.compile(r'(dog)(cat)?') # the ? means find one or zero of these, in other words cat is optional. . m = my_re.search(This is a nice dog is it not?) . dir(m) ['__copy__', '__deepcopy__', 'end', 'expand', 'group', 'groupdict', 'groups', 'span', 'start'] . m.span() (15, 18) . m = my_re.search(This is a nice dogcat is it not?) . m.span() (15, 21) If m is None then no match was found. span returns the locations in the string where the match occured. So in the dogcat sentence the last char is 21. . This is a nice dogcat is it not?[21:] ' is it not?' Hope that helps. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
Mike Hall wrote: First, thanks for the response. Using your re: my_re = re.compile(r'(dog)(cat)?') ...I seem to simply be matching the pattern Dog. Example: str1 = The dog chased the car str2 = The dog cat parade was under way x1 = re.compile(r'(dog)(cat)?') rep1 = x1.sub(REPLACE, str1) rep2 = x2.sub(REPLACE, str2) print rep1 The REPLACE chased the car print rep2 The REPLACE cat parade was under way ...what I'm looking for is a match for the position in front of Cat, should it exist. Because my regex says 'look for the word dog and remember where you found it. If you also find the word cat, remember that too'. Nowhere does it say watch out for whitespace. r'(dog)\s*(cat)?' says match 'dog' followed by zero or more whitespace (spaces, tabs, etc.) and maybe 'cat'. There is a wonderful O'Reilly book called Mastering Regular Expressions or as Danny points out the AMK howto is good. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] regular expression question
On Tue, 8 Mar 2005, Mike Hall wrote:
Yes, my existing regex is using a look behind assertion:
(?=dog)
...it's also checking the existence of Cat:
(?!Cat)
...what I'm stuck on is how to essentially use a lookbehind on Cat,
but only if it exists.
Hi Mike,
[Note: Please do a reply-to-all next time, so that everyone can help you.]
Regular expressions are a little evil at times; here's what I think you're
thinking of:
###
import re
pattern = re.compile(rdog(?!cat)
...| (?=dogcat), re.VERBOSE)
pattern.match('dogman').start()
0
pattern.search('dogcatcher').start()
pattern.search('dogman').start()
0
pattern.search('catwoman')
###
but I can't be sure without seeing some of the examples you'd like the
regular expression to match against.
Best of wishes to you!
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Re: [Tutor] regular expression question
This will match the position in front of dog: (?=dog) This will match the position in front of cat: (?=cat) This will not match in front of dog if dog is followed by cat: (?=dog)\b (?!cat) Now my question is how to get this: (?=cat) ...but ONLY if cat is following dog. If dog does not have cat following it, then I simply want this: (?=dog) ...if that makes sense :) thanks. On Mar 8, 2005, at 6:05 PM, Danny Yoo wrote: On Tue, 8 Mar 2005, Mike Hall wrote: I'd like to get a match for a position in a string preceded by a specified word (let's call it Dog), unless that spot in the string (after Dog) is directly followed by a specific word(let's say Cat), in which case I want my match to occur directly after Cat, and not Dog. Hi Mike, You may want to look at lookahead assertions. These are patterns of the form '(?=...)' or '(?!...). The documentation mentions them here: http://www.python.org/doc/lib/re-syntax.html and AMK's excellent Regular Expression HOWTO covers how one might use them: http://www.amk.ca/python/howto/regex/ regex.html#SECTION00054 ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression re.search() object . Please help
On Thu, 13 Jan 2005, kumar s wrote:
My list looks like this: List name = probe_pairs
Name=AFFX-BioB-5_at
Cell1=96 369 N control AFFX-BioB-5_at
Cell2=96 370 N control AFFX-BioB-5_at
Cell3=441 3 N control AFFX-BioB-5_at
Cell4=441 4 N control AFFX-BioB-5_at
Name=223473_at
Cell1=307 87 N control 223473_at
Cell2=307 88 N control 223473_at
Cell3=367 84 N control 223473_at
My Script:
name1 = '[N][a][m][e][=]'
Hi Kumar,
The regular expression above can be simplified to:
'Name='
The character-class operator that you're using, with the brackets '[]', is
useful when we want to allow different kind of characters. Since the code
appears to be looking at a particular string, the regex can be greatly
simplified by not using character classes.
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
key
_sre.SRE_Match object at 0x00E37A68
_sre.SRE_Match object at 0x00E37AD8
_sre.SRE_Match object at 0x00E37A68
_sre.SRE_Match object at 0x00E37AD8
_sre.SRE_Match object at 0x00E37A68
. (cont. 10K
lines)
Here it prints a bunch of reg.match objects. However when I say group()
it prints only one object why?
Is it possible that the edited code may have done something like this?
###
for i in range(len(probe_pairs)):
key = re.match(name1, probe_pairs[i])
print key
###
Without seeing what the literal code looks like, we're doomed to use our
imaginations and make up a reasonable story. *grin*
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
key.group()
Ok, I think I see what you're trying to do. You're using the interactive
interpreter, which tries to be nice when we use it as a calculator. The
interactive interpreter has a special feature that prints out the result
of expressions, even though we have not explicitely put in a print
statement.
When we using a loop, like:
###
for i in range(10):
... i, i*2, i*3
...
(0, 0, 0)
(1, 2, 3)
(2, 4, 6)
(3, 6, 9)
(4, 8, 12)
(5, 10, 15)
(6, 12, 18)
(7, 14, 21)
(8, 16, 24)
(9, 18, 27)
###
If the body of the loop contains a single expression, then Python's
interactive interpreter will try to be nice and print that expression
through each iteration.
The automatic expression-printing feature of the interactive interpreter
is only for our convenience. If we're not running in interactive mode,
Python will not automatically print out the values of expressions!
So in a real program, it is much better to explicity write out the command
statement to 'print' the expression to screen, if that's what you want:
###
for i in range(10):
... print (i, i*2, i*3)
...
(0, 0, 0)
(1, 2, 3)
(2, 4, 6)
(3, 6, 9)
(4, 8, 12)
(5, 10, 15)
(6, 12, 18)
(7, 14, 21)
(8, 16, 24)
(9, 18, 27)
###
After I get the reg.match object, I tried to remove
that match object like this:
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
del key
print probe_pairs[i]
The match object has a separate existance from the string
'probe_pairs[i]'. Your code does drop the 'match' object, but this has no
effect in making a string change in probe_pairs[i].
The code above, removing those two lines that play with the 'key', reduces
down back to:
###
for i in range(len(probe_pairs)):
print probe_pairs[i]
###
which is why you're not seeing any particular change in the output.
I'm not exactly sure you really need to do regular expression stuff here.
Would the following work for you?
###
for probe_pair in probe_pairs:
if not probe_pair.startswith('Name='):
print probe_pair
###
Name=AFFX-BioB-5_at
Cell1=96 369 N control AFFX-BioB-5_at
Cell2=96 370 N control AFFX-BioB-5_at
Cell3=441 3 N control AFFX-BioB-5_at
Result shows that that Name** line has not been deleted.
What do you want to see? Do you want to see:
###
AFFX-BioB-5_at
Cell1=96369 N control AFFX-BioB-5_at
Cell2=96370 N control AFFX-BioB-5_at
Cell3=441 3 N control AFFX-BioB-5_at
###
or do you want to see this instead?
###
Cell1=96369 N control AFFX-BioB-5_at
Cell2=96370 N control AFFX-BioB-5_at
Cell3=441 3 N control AFFX-BioB-5_at
###
Good luck to you!
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Re: [Tutor] Regular expression re.search() object . Please help
...as do I. openFile=file(probe_pairs.txt,r) probe_pairs=openFile.readlines() openFile.close() indexesToRemove=[] for lineIndex in range(len(probe_pairs)): if probe_pairs[lineIndex].startswith(Name=): indexesToRemove.append(lineIndex) for index in indexesToRemove: probe_pairs[index]=' Could just be openFile=file(probe_pairs.txt,r) probe_pairs=openFile.readlines() openFile.close() indexesToRemove=[] for lineIndex in range(len(probe_pairs)): if probe_pairs[lineIndex].startswith(Name=): probe_pairs[lineIndex]='' On Fri, 14 Jan 2005 09:38:17 +1300, Liam Clarke [EMAIL PROTECTED] wrote: name1 = '[N][a][m][e][=]' for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) key _sre.SRE_Match object at 0x00E37A68 _sre.SRE_Match object at 0x00E37AD8 _sre.SRE_Match object at 0x00E37A68 _sre.SRE_Match object at 0x00E37AD8 _sre.SRE_Match object at 0x00E37A68 You are overwriting key each time you iterate. key.group() gives the matched characters in that object, not a group of objects!!! You want name1 = '[N][a][m][e][=]' keys=[] for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) keys.append[key] print keys 'Name=' 1. My aim: To remove those Name= lines from my probe_pairs list Why are you deleting the object key? for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) del key print probe_pairs[i] Here's the easy way. Assuming that probe_pairs is stored in a file callde probe_pairs.txt openFile=file(probe_pairs.txt,r) probe_pairs=openFile.readlines() openFile.close() indexesToRemove=[] for lineIndex in range(len(probe_pairs)): if probe_pairs[lineIndex].startswith(Name=): indexesToRemove.append(lineIndex) for index in indexesToRemove: probe_pairs[index]=' Try that. Argh, my head. You do some strange things to Python. Liam Clarke On Thu, 13 Jan 2005 10:56:00 -0800 (PST), kumar s [EMAIL PROTECTED] wrote: Dear group: My list looks like this: List name = probe_pairs Name=AFFX-BioB-5_at Cell1=96369 N control AFFX-BioB-5_at Cell2=96370 N control AFFX-BioB-5_at Cell3=441 3 N control AFFX-BioB-5_at Cell4=441 4 N control AFFX-BioB-5_at Name=223473_at Cell1=307 87 N control 223473_at Cell2=307 88 N control 223473_at Cell3=367 84 N control 223473_at My Script: name1 = '[N][a][m][e][=]' for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) key _sre.SRE_Match object at 0x00E37A68 _sre.SRE_Match object at 0x00E37AD8 _sre.SRE_Match object at 0x00E37A68 _sre.SRE_Match object at 0x00E37AD8 _sre.SRE_Match object at 0x00E37A68 . (cont. 10K lines) Here it prints a bunch of reg.match objects. However when I say group() it prints only one object why? Alternatively: for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) key.group() 'Name=' 1. My aim: To remove those Name= lines from my probe_pairs list with name1 as the pattern, I asked using re.match() method to identify the lines and then remove by using re.sub(pat,'',string) method. I want to substitute Name=*** line by an empty string. After I get the reg.match object, I tried to remove that match object like this: for i in range(len(probe_pairs)): key = re.match(name1,probe_pairs[i]) del key print probe_pairs[i] Name=AFFX-BioB-5_at Cell1=96369 N control AFFX-BioB-5_at Cell2=96370 N control AFFX-BioB-5_at Cell3=441 3 N control AFFX-BioB-5_at Result shows that that Name** line has not been deleted. Is the way I am doing a good one. Could you please suggest a good simple method. Thanks in advance K __ Do you Yahoo!? Yahoo! Mail - Easier than ever with enhanced search. Learn more. http://info.mail.yahoo.com/mail_250 ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor -- 'There is only one basic human right, and that is to do as you damn well please. And with it comes the only basic human duty, to take the consequences. -- 'There is only one basic human right, and that is to do as you damn well please. And with it comes the only basic human duty, to take the consequences. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression re.search() object . Please help
Liam Clarke wrote:
openFile=file(probe_pairs.txt,r)
probe_pairs=openFile.readlines()
openFile.close()
indexesToRemove=[]
for lineIndex in range(len(probe_pairs)):
if probe_pairs[lineIndex].startswith(Name=):
probe_pairs[lineIndex]=''
If the intent is simply to remove all lines that begin with Name=,
and setting those lines to an empty string is just shorthand for that,
it'd make more sense to do this with a filtering list comprehension:
openfile = open(probe_pairs.txt,r)
probe_pairs = openfile.readlines()
openfile.close()
probe_pairs = [line for line in probe_pairs \
if not line.startswith('Name=')]
(The '\' line continuation isn't strictly necessary, because the open
list-comp will do the same thing, but I'm including it for
readability's sake.)
If one wants to avoid list comprehensions, you could instead do:
openfile = open(probe_pairs.txt,r)
probe_pairs = []
for line in openfile.readlines():
if not line.startswith('Name='):
probe_pairs.append(line)
openfile.close()
Either way, lines that start with 'Name=' get thrown away, and all
other lines get kept.
Jeff Shannon
Technician/Programmer
Credit International
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Re: [Tutor] Regular expression re.search() object . Please help
Quoting Jeff Shannon [EMAIL PROTECTED]:
If the intent is simply to remove all lines that begin with Name=,
and setting those lines to an empty string is just shorthand for that,
it'd make more sense to do this with a filtering list comprehension:
[...]
If one wants to avoid list comprehensions, you could instead do:
[...]
Or, since we're filtering, we could use the filter() function!
probe_pairs = filter(lambda x: not x.startswith('Name='), probe_pairs)
(hmm, I wonder which is the faster option...)
--
John.
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Re: [Tutor] Regular expression re.search() object . Please help
Quoting Jeff Shannon [EMAIL PROTECTED]: [ snip ] (hmm, I wonder which is the faster option...) Propably grep -v ^Name= filename -- John. 0.2 EUR, Jo! -- Wir sind jetzt ein Imperium und wir schaffen uns unsere eigene Realität. Wir sind die Akteure der Geschichte, und Ihnen, Ihnen allen bleibt nichts, als die Realität zu studieren, die wir geschaffen haben. -- Karl Rove zu Ron Suskind (NYT) +++ GMX - die erste Adresse für Mail, Message, More +++ 1 GB Mailbox bereits in GMX FreeMail http://www.gmx.net/de/go/mail ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression re.search() object . Please help
Hello group:
thank you for the suggestions. It worked for me using
if not line.startswith('Name='): expression.
I have been practising regular expression problems. I
tumle over one simple thing always. After obtaining
either a search object or a match object, I am unable
to apply certain methods on these objects to get
stuff.
I have looked into many books including my favs(
Larning python and Alan Gaulds Learn to program using
python) I did not find the basic question, how can I
get what I intend to do with returned reg.ex match
object (search(), match()).
For example:
I have a simple list like the following:
seq
['probe:HG-U133B:20_s_at:164:623;
Interrogation_Position=6649; Antisense;',
'TCATGGCTGACAACCCATCTTGGGA']
Now I intend to extract particular pattern and write
to another list say: desired[]
What I want to extract:
I want to extract 164:623:
Which always comes after _at: and ends with ;
2. The second pattern/number I want to extract is
6649:
This always comes after position=.
How I want to put to desired[]:
desired
['164:623|6649', 'TCATGGCTGACAACCCATCTTGGGA']
I write a pattern:
pat = '[0-9]*[:][0-9]*'
pat1 = '[_Position][=][0-9]*'
for line in seq:
pat = '[0-9]*[:][0-9]*'
pat1 = '[_Position][=][0-9]*'
print (re.search(pat,line) and re.search(pat1,line))
_sre.SRE_Match object at 0x163CAF00
None
Now I know that I have a hit in the seq list evident
by _sre.SRE_Match object at 0x163CAF00.
Here is the black box:
What kind of operations can I do on this to get those
two matches:
164:623 and 6649.
I read
http://www.python.org/doc/2.2.3/lib/re-objects.html
This did not help me to progress further. May I
request tutors to give a small note explaining things.
In Alan Gauld's book, most of the explanation stopped
at
_sre.SRE_Match object at 0x163CAF00 this level.
After that there is no example where he did some
operations on these objects. If I am wrong, I might
have skipped/missed to read it. Aplogies for that.
Thank you very much in advance.
K
--- Liam Clarke [EMAIL PROTECTED] wrote:
...as do I.
openFile=file(probe_pairs.txt,r)
probe_pairs=openFile.readlines()
openFile.close()
indexesToRemove=[]
for lineIndex in range(len(probe_pairs)):
if
probe_pairs[lineIndex].startswith(Name=):
indexesToRemove.append(lineIndex)
for index in indexesToRemove:
probe_pairs[index]='
Could just be
openFile=file(probe_pairs.txt,r)
probe_pairs=openFile.readlines()
openFile.close()
indexesToRemove=[]
for lineIndex in range(len(probe_pairs)):
if
probe_pairs[lineIndex].startswith(Name=):
probe_pairs[lineIndex]=''
On Fri, 14 Jan 2005 09:38:17 +1300, Liam Clarke
[EMAIL PROTECTED] wrote:
name1 = '[N][a][m][e][=]'
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
key
_sre.SRE_Match object at 0x00E37A68
_sre.SRE_Match object at 0x00E37AD8
_sre.SRE_Match object at 0x00E37A68
_sre.SRE_Match object at 0x00E37AD8
_sre.SRE_Match object at 0x00E37A68
You are overwriting key each time you iterate.
key.group() gives the
matched characters in that object, not a group of
objects!!!
You want
name1 = '[N][a][m][e][=]'
keys=[]
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
keys.append[key]
print keys
'Name='
1. My aim:
To remove those Name= lines from my
probe_pairs
list
Why are you deleting the object key?
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
del key
print probe_pairs[i]
Here's the easy way. Assuming that probe_pairs is
stored in a file callde
probe_pairs.txt
openFile=file(probe_pairs.txt,r)
probe_pairs=openFile.readlines()
openFile.close()
indexesToRemove=[]
for lineIndex in range(len(probe_pairs)):
if
probe_pairs[lineIndex].startswith(Name=):
indexesToRemove.append(lineIndex)
for index in indexesToRemove:
probe_pairs[index]='
Try that.
Argh, my head. You do some strange things to
Python.
Liam Clarke
On Thu, 13 Jan 2005 10:56:00 -0800 (PST), kumar s
[EMAIL PROTECTED] wrote:
Dear group:
My list looks like this: List name = probe_pairs
Name=AFFX-BioB-5_at
Cell1=96369 N control
AFFX-BioB-5_at
Cell2=96370 N control
AFFX-BioB-5_at
Cell3=441 3 N control
AFFX-BioB-5_at
Cell4=441 4 N control
AFFX-BioB-5_at
Name=223473_at
Cell1=307 87 N control
223473_at
Cell2=307 88 N control
223473_at
Cell3=367 84 N control
223473_at
My Script:
name1 = '[N][a][m][e][=]'
for i in range(len(probe_pairs)):
key =
Re: [Tutor] Regular expression re.search() object . Please help
My list looks like this: List name = probe_pairs
Name=AFFX-BioB-5_at
Cell1=96 369 N control AFFX-BioB-5_at
Cell2=96 370 N control AFFX-BioB-5_at
Cell3=441 3 N control AFFX-BioB-5_at
Cell4=441 4 N control AFFX-BioB-5_at
...
My Script:
name1 = '[N][a][m][e][=]'
Why not just: 'Name=' - the result is the same.
for i in range(len(probe_pairs)):
andwhy not just
for line in probe_pairs:
key = re.match(name1,line)
Although I suspect it would be easier still to use
line.startswith('Name=')
especially combined with the fileinput module.
It is really quite good for line by line
matching/processing of files, and I assume this
data comes from a file originally?.
key = re.match(name1,probe_pairs[i])
key
_sre.SRE_Match object at 0x00E37A68
One per line that matches.
when I say group() it prints only one object why?
Because the group is the string you are looking for.
But I may be missing something since there is no
indentation in the post, its hard to tell whats
inside and whats outside the loop.
1. My aim:
To remove those Name= lines from my probe_pairs
list
for i in range(len(probe_pairs)):
key = re.match(name1,probe_pairs[i])
del key
That will remove the match object, not the line from
the list!
To filter the list I'd have thought you'd be better using
a list comprehension:
filtered = [line for line in probe_pairs if not
line.startswith('Name=')]
HTH,
Alan G.
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Re: [Tutor] Regular expression re.search() object . Please help
I have looked into many books including my favs( Larning python and Alan Gaulds Learn to program using Yes this is pushing regex a bit further than I show in my book. What I want to extract: I want to extract 164:623: Which always comes after _at: and ends with ; You should be able to use the group() method to extract the matching string out of the match object. 2. The second pattern/number I want to extract is 6649: This always comes after position=. How I want to put to desired[]: desired ['164:623|6649', 'TCATGGCTGACAACCCATCTTGGGA'] I write a pattern: pat = '[0-9]*[:][0-9]*' pat1 = '[_Position][=][0-9]*' for line in seq: pat = '[0-9]*[:][0-9]*' pat1 = '[_Position][=][0-9]*' pat1 = [_Position] will match any *one* of the characters in _Position, is that really what you want? I suspect its: '_Position=[0-9]*' Which is the fixed string followed by any number(including zero) of digits. print (re.search(pat,line) and re.search(pat1,line)) This is asking print to print the boolean value of your expression which if the first search fails will be that failure and if it succeeeds will be the result of the second search. Check the section on Functional Programming in my tutor to see why. _sre.SRE_Match object at 0x163CAF00 None Looks like your searches worked but of course you don't have the match objects stored so you can't use them for anything. But I'm suspicious of where the None is coming from... Again without any indentation showing its not totally clear what your code looks like. What kind of operations can I do on this to get those two matches: 164:623 and 6649. I think that if you keep the match objects you can use group() to extract the thing that matched which should be close to what you want. In Alan Gauld's book, most of the explanation stopped at _sre.SRE_Match object at 0x163CAF00 this level. Yep, personally I use regex to find the line then extract the data I need from the line manually. Messing around with match objects is something I try to avoid and so left it out of the tutor. :-) Alan G. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Regular expression re.search() object . Please help
I assume that both you and Liam are using previous-er versions of python?
Now files are iterators by line and you can do this.
openFile = open(probe_pairs.txt,r)
indexesToRemove = []
for line in openFile:
if line.startswith(Name=):
line = '' ## Ooops, this won't work because it just changes what
line references or points to or whatever, it doesn't
##actually change the object
openFile.close()
I think this is a great flaw with using for element in list instead of for
index in range(len(list))
Of course you later put IMHO better solutions (list comprehensions,etc.) --
I just wanted to point out that files have been
iterators for a few versions now and that is being used more and more. It
also saves the memory problem of using big files
and reading them all at once with readlines.
Jacob
Liam Clarke wrote:
openFile=file(probe_pairs.txt,r)
probe_pairs=openFile.readlines()
openFile.close()
indexesToRemove=[]
for lineIndex in range(len(probe_pairs)):
if probe_pairs[lineIndex].startswith(Name=):
probe_pairs[lineIndex]=''
If the intent is simply to remove all lines that begin with Name=,
and setting those lines to an empty string is just shorthand for that,
it'd make more sense to do this with a filtering list comprehension:
openfile = open(probe_pairs.txt,r)
probe_pairs = openfile.readlines()
openfile.close()
probe_pairs = [line for line in probe_pairs \
if not line.startswith('Name=')]
(The '\' line continuation isn't strictly necessary, because the open
list-comp will do the same thing, but I'm including it for
readability's sake.)
If one wants to avoid list comprehensions, you could instead do:
openfile = open(probe_pairs.txt,r)
probe_pairs = []
for line in openfile.readlines():
if not line.startswith('Name='):
probe_pairs.append(line)
openfile.close()
Either way, lines that start with 'Name=' get thrown away, and all
other lines get kept.
Jeff Shannon
Technician/Programmer
Credit International
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