[Tutor] Shorten Code.
From: Wayne Werner Sent: Thursday, November 17, 2011 8:30 PM To: Mic Cc: tutor@python.org Subject: Re: [Tutor] Clock in tkinter? Say that I have a class and I want to make 100 objects. Then it could look like this: snip class Chairs(object): snip code #Create the objects chair1=Chairs(10,20) chair2=Chairs(10,20) chair3=Chairs(10,20) How do I shorten this? I have thought of using a for sling. I have looked in my programming book and on the internet, but I don’t know how to make this shorter. The arguements (“10”, “20”) should be the same for every object, which should make it easier than if they were different each time? If you ever write a line of code more than once, it's a good sign that you have what's called a code smell. This example is very smelly code ;) What you should do instead is have a collection of chairs: chairs = [] for _ in range(100): # the underscore `_` indicates that you don't care about the value chairs.append(Chairs(10,20)) -- Yes, I got that right now. But now that you talked about shortening code, I have a general question. What if I don’t write the same line of code more than once, but I write similiar lines more than once. Is that okay? For example: value=green” value_1=”green” click=-1 click1=-1 click2=-1 I know that I can make this shorter, with a for sling for example, but the problem is that I need to use these variables later in my program, and I don’t know how do to then, to be able to use them later on, in a function for example. Do you have any general tips on how to make your code shorter? I also hope I have learnt to post better now. Thanks! Mic wlEmoticon-smile[1].png___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Shorten Code.
On 18/11/11 08:16, Mic wrote: What if I don’t write the same line of code more than once, but I write similiar lines more than once. Is that okay? Ler For example: value=green” value_1=”green” If you had a lot of these you could shorten it with a loop (BTW the English term in programming terminology is loop not sling ;-) for var in [value,value_1]: var = green But you need to have already created the variables somewhere and unless there is a big list its not usually worth while. One other trick you can use for this specific type of assignment is value = value_1 = green But it gets a bit unreadable for long lists of names. click=-1 click1=-1 click2=-1 Same here, but imagine it had been: click=-1 click1= 2 click2=-1 And here you are changing both name and value. The best abbreviation here is probably tuple expansion: click, click1, click2 = -1,2,-1 Do you have any general tips on how to make your code shorter? Shorter is not necessarily better. Clarity is far more important and you should always consider whether tricks like those above are helping or hindering clarity. Only use them if they make your code easier to read. (Also they all make debugging slightly harder so you should think about that too) HTH, -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] A recursion question
Hi, I am trying to do something which is really stupid :-) I would like to count the number of occurrences of a certain character in a list. This is more of an exercise in recursion rather than the underlying problem. I could have used a *for* loop or simply the list *count* method. Here is the code: class Find_Ex(): count = 0 def slais(self, lst, x): if len(lst) == 0: #if list is empty just give a -1 return -1 elif lst[0] == x: Find_Ex.count += 1 # we have found 'x', increment class attribute 'count' by 1 Find_Ex.slais(self,lst[1:], x)# slice the list and go to the next element else: Find_Ex.slais(self,lst[1:], x)#'x' not found so we move to the next element return Find_Ex.count s = Find_Ex() lst = [4,4,4,5,6,7,4,7,7,4] x = 4 print There are %d occurrences of %d%(s.slais(lst, x),x) It works as advertised but I am convincing myself that it should not! :-) If the list is empty, which is the base case, s.slais([], 4) returns -1. Now using some bush logic, in a non-empty list, in order for the recursion to terminate it has to 'hit' the base case and return -1. Where does this -1 go ? Further, why do not I get a *TypeError* when I use a simple *return* statement in the *if* clause? The reason I am asking that is that I think(wrongly, of course :-)) it should be part of the answer and therefore I should be getting an answer that is off by one or a *TypeError*!! And by the way, the reason I used a *class* was that I could not get a suitable place in the program to initialise my *count* variable otherwise. Thanks... -- Kĩnũthia S 1º 8' 24” E 36º 57' 36” 1522m ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] A recursion question
On 2011/11/18 03:16 PM, Kĩnũthia Mũchane wrote: Hi, I am trying to do something which is really stupid :-) I would like to count the number of occurrences of a certain character in a list. This is more of an exercise in recursion rather than the underlying problem. I could have used a *for* loop or simply the list *count* method. Here is the code: class Find_Ex(): count = 0 def slais(self, lst, x): if len(lst) == 0: #if list is empty just give a -1 return -1 elif lst[0] == x: Find_Ex.count += 1 # we have found 'x', increment class attribute 'count' by 1 Find_Ex.slais(self,lst[1:], x)# slice the list and go to the next element else: Find_Ex.slais(self,lst[1:], x)#'x' not found so we move to the next element return Find_Ex.count s = Find_Ex() lst = [4,4,4,5,6,7,4,7,7,4] x = 4 print There are %d occurrences of %d%(s.slais(lst, x),x) It works as advertised but I am convincing myself that it should not! :-) If the list is empty, which is the base case, s.slais([], 4) returns -1. Now using some bush logic, in a non-empty list, in order for the recursion to terminate it has to 'hit' the base case and return -1. Where does this -1 go ? Further, why do not I get a *TypeError* when I use a simple *return* statement in the *if* clause? The reason I am asking that is that I think(wrongly, of course :-)) it should be part of the answer and therefore I should be getting an answer that is off by one or a *TypeError*!! And by the way, the reason I used a *class* was that I could not get a suitable place in the program to initialise my *count* variable otherwise. Thanks... If you pop in some print statements you can see what's happening a bit easier. You are creating a stack of functions which each return their values but in a LIFO fashion (Last In, First Out) so you can see the first return is -1 as you expected to happen when the list is exhausted, and then each subsequent return is the count which is why you get the correct return value in the end. Also, why do you think you should get a TypeError when you `return -1` ? class Find_Ex(): count = 0 def slais(self, lst, x): print lst if len(lst) == 0: #if list is empty just give a -1 print 'Returning -1' return -1 elif lst[0] == x: print 'Incrementing Count' Find_Ex.count += 1 # we have found 'x', increment class attribute 'count' by 1 Find_Ex.slais(self,lst[1:], x)# slice the list and go to the next element else: print 'Nothing Found' Find_Ex.slais(self,lst[1:], x)#'x' not found so we move to the next element print 'Returning the count' return Find_Ex.count s = Find_Ex() lst = [4,4,4,5,6,7,4,7,7,4] x = 4 print There are %d occurrences of %d%(s.slais(lst, x),x) [4, 4, 4, 5, 6, 7, 4, 7, 7, 4] Incrementing Count [4, 4, 5, 6, 7, 4, 7, 7, 4] Incrementing Count [4, 5, 6, 7, 4, 7, 7, 4] Incrementing Count [5, 6, 7, 4, 7, 7, 4] Nothing Found [6, 7, 4, 7, 7, 4] Nothing Found [7, 4, 7, 7, 4] Nothing Found [4, 7, 7, 4] Incrementing Count [7, 7, 4] Nothing Found [7, 4] Nothing Found [4] Incrementing Count [] Returning -1 Returning the count Returning the count Returning the count Returning the count Returning the count Returning the count Returning the count Returning the count Returning the count Returning the count There are 5 occurrences of 4 -- Christian Witts Python Developer // ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Encoding
Well, I am assuming that by this you mean converting user input into a
string, and then extracting the numerals (0-9) from it. Next time, please
tell us your version of Python. I'll do my best to help with this. You
might try the following:
the_input = input(Insert string here: ) # change to raw_input in python 2
after =
for char in the_input:
try:
char = int(char)
except:
after += char
If other symbols might be in the string ($, @, etc.), then you might use
the_input = input('Insert string here: ') # change to raw_input in python 2
after = ''
not_allowed = '1234567890-=!@#$%^**()_+,./?`~[]{}\\|'
for char in the_input:
if char in not_allowed:
pass
else:
after += char
This method requires more typing, but it works with a wider variety of
characters. Hopefully this helped.
On Thu, Nov 17, 2011 at 8:45 PM, Nidian Job-Smith nidia...@hotmail.comwrote:
Hi all,
In my programme I am encoding what the user has in-putted.
What the user inputs will in a string, which might a mixture of letters
and numbers.
However I only want the letters to be encoded.
Does any-one how I can only allow the characters to be encoded ??
Big thanks,
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Re: [Tutor] Shorten Code.
On 11/18/2011 04:01 AM, Alan Gauld wrote: On 18/11/11 08:16, Mic wrote: What if I don’t write the same line of code more than once, but I write similiar lines more than once. Is that okay? Ler For example: value=green” value_1=”green” If you had a lot of these you could shorten it with a loop (BTW the English term in programming terminology is loop not sling ;-) for var in [value,value_1]: var = green Um, that won't work. You typed that example too quickly. Mic, the problem is not shortening code, but making it more readable, and easier to maintain. If you have a series of variables that hold similar or identical values, or which are treated in consistent ways, then you should probably make a list out of them. And that will naturally shorten code like this. student0 = 4 student1 = 3 student2 = 12 student3 = 11 Replace with students = list((4,3,12,11)) Then if you want to deal with a particular student, you might do student[2] = 5 But if you want to deal with the ith student, you could do student[i] = and if you want to do something with all of them: for index, student in enumerate(students): students[indes] += 65 There are plenty of more advanced methods that would make even that simpler, but I'm trying to keep it simple. -- DaveA ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] A recursion question
On 11/18/2011 08:16 AM, Kĩnũthia Mũchane wrote: Hi, I am trying to do something which is really stupid :-) I would like to count the number of occurrences of a certain character in a list. This is more of an exercise in recursion rather than the underlying problem. I could have used a *for* loop or simply the list *count* method. Here is the code: class Find_Ex(): count = 0 def slais(self, lst, x): if len(lst) == 0: #if list is empty just give a -1 return -1 elif lst[0] == x: Find_Ex.count += 1 # we have found 'x', increment class attribute 'count' by 1 Find_Ex.slais(self,lst[1:], x)# slice the list and go to the next element else: Find_Ex.slais(self,lst[1:], x)#'x' not found so we move to the next element return Find_Ex.count s = Find_Ex() lst = [4,4,4,5,6,7,4,7,7,4] x = 4 print There are %d occurrences of %d%(s.slais(lst, x),x) It works as advertised but I am convincing myself that it should not! :-) Well, it doesn't count the number of occurrences correctly if the list is empty. It should get zero, and it gets -1. But for any non-empty list, nothing ever looks at the -1 value, so it doesn't matter what you put there. This example diverges from traditional recursion in many ways, but the chief reason it's not traditional recursion is that it never uses the return value. within the function. If the list is empty, which is the base case, s.slais([], 4) returns -1. Now using some bush logic, in a non-empty list, in order for the recursion to terminate it has to 'hit' the base case and return -1. Where does this -1 go ? Nowhere. You don't use it, so it gets discarded. if you were going to use it, your internal calls to the method would look something like: b = self.slais(lst[1:], x) and then you'd do something with b. Further, why do not I get a *TypeError* when I use a simple *return* statement in the *if* clause? You would if you actually used the value for arithmetic. But the return itself would be perfectly legal. it's just a shortcut for 'return None' The reason I am asking that is that I think(wrongly, of course :-)) it should be part of the answer and therefore I should be getting an answer that is off by one or a *TypeError*!! And by the way, the reason I used a *class* was that I could not get a suitable place in the program to initialise my *count* variable otherwise. Using a class is fine. But you're abusing it. What happens if you try to sum a second list? (Hint, it gives you a higher number) Thanks... If you really want to recursively count, you need a structure which uses no objects of global lifetime. The intermediate values should be stored in local variables to that method. def counter(mylist, val): if len(mylist == 0): return 0 prev = counter(mylist[1:], val) #this is actually using the recursive return value if mylist[0] == val: return prev + 1 else: return prev Totally untested. And there are certainly many other variants possible. But the key is you have to do something with the value returned to you by the lower level function. -- DaveA ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] FW: urgent help!! THANKS EVERYONE!
Forwarding to the list since I wasn't the only person who helped ;)
Ramit
Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
712 Main Street | Houston, TX 77002
work phone: 713 - 216 - 5423
--
From: ADRIAN KELLY [mailto:kellyadr...@hotmail.com]
Sent: Thursday, November 17, 2011 6:08 PM
To: Prasad, Ramit
Subject: RE: [Tutor] urgent help!! THANKS EVERYONE!
Thanks for your help i just got it going my way below - but your way looks
easier and better!
thanks for all your help everyone. feel free to comment on my method -
its awkward but it works..
adrian
def exchange(cash_in):
euro=1
dollar=1.35
base=50
if cash_inbase:
totalreturn=cash_in*dollar
else:
totalreturn=0
return totalreturn
def main():
amount=0
amount = float(raw_input('how much do you want to change:'))
while amount50:
print 'enter an amount over 50'
amount = float(raw_input('how much do you want to change:'))
else:
total=exchange(amount)
print 'Your exchange comes to: ',total
main()
From: ramit.pra...@jpmorgan.com
To: kellyadr...@hotmail.com; waynejwer...@gmail.com
CC: tutor@python.org
Subject: RE: [Tutor] urgent help!!!
Date: Thu, 17 Nov 2011 23:43:10 +
def exchange(cash_in):
euro=1
dollar=1.35
base=50
if cash_inbase:
totalreturn=cash_in*dollar
else:
totalreturn=0
return totalreturn
amount=0
# this would be better placed inside the main function.
def main():
while amount50:
amount = raw_input(float('how much do you want to change:'))
# This should be
# amount = float( raw_input('how much do you want to change:' ) )
if amount50:
total=0
print 'enter an amount over 50: '
else:
total=exchange(amount)
print 'Your exchange comes to: ',total
Ramit
Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
712 Main Street | Houston, TX 77002
work phone: 713 - 216 - 5423
--
This email is confidential and subject to important disclaimers and
conditions including on offers for the purchase or sale of
securities, accuracy and completeness of information, viruses,
confidentiality, legal privilege, and legal entity disclaimers,
available at http://www.jpmorgan.com/pages/disclosures/email.
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Re: [Tutor] FW: urgent help!! THANKS EVERYONE!
On Fri, Nov 18, 2011 at 9:26 AM, Prasad, Ramit ramit.pra...@jpmorgan.comwrote:
Forwarding to the list since I wasn't the only person who helped ;)
From: ADRIAN KELLY [mailto:kellyadr...@hotmail.com]
Sent: Thursday, November 17, 2011 6:08 PM
To: Prasad, Ramit
Subject: RE: [Tutor] urgent help!! THANKS EVERYONE!
Thanks for your help i just got it going my way below - but your way looks
easier and better!
thanks for all your help everyone. feel free to comment on my method -
its awkward but it works.. snip
def main():
amount=0
amount = float(raw_input('how much do you want to change:'))
while amount50:
print 'enter an amount over 50'
amount = float(raw_input('how much do you want to change:'))
else:
total=exchange(amount)
print 'Your exchange comes to: ',total
In these sorts of cases I actually prefer the recursive solution. And from
a usability standpoint it's much nicer to warn the user ahead of time:
def amount_over_50():
amount = float(raw_input(How much do you want to change (minimum $50)?
Amount: $))
if amount 50.0:
print 'Please enter an amount greater than $50'
return amount_over_50()
return amount
Then you can just do this:
def main():
print 'Your total comes to', exchange(amount_over_50())
The chances of hitting the maximum recursion depth is pretty slim, unless
your user manages to type in a wrong number about 1000 times ;)
HTH,
Wayne
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[Tutor] In a pickle over pickles….Python 3
Hi All,Sorry to trouble you all again with more nooby problems! Only been
programming a week so still all a haze especially since self taught…..
I am opening a url and saving it to a file on my computer, then I am trying to
pickle it. I have written the url file to a file on my computer then opened it
and assigned the contents to a var 'filecontents' and tried to pickle it it is
giving the error:
Traceback (most recent call last): File
/Users/joebatt/Desktop/python/pickling puzzle 5.py, line 39, in module
geturlfile(urlfile) File /Users/joebatt/Desktop/python/pickling puzzle 5.py,
line 28, in geturlfilepickle.dump(filecontents,pickledfile)TypeError: must
be str, not bytes
Yet I tested the variable filecontents using print (type(file contents)) and
it is saying that it is a str yet the error seems to indicate it is a byte, can
anyone point me in the right direction please?
(I know my coding is very untidy and verbose, sorry I am very new and at the
moment I have to think in baby steps with little knowledge!)
Joe
Pickle an
object- ###
import pickleimport urllib
def geturlfile(urlfile):#gets the user URL from main opens it and saves it to
var fileurlfrom urllib.request import urlopen
fileurl=urlopen(urlfile)#opens the file on computer and writes the var fileurl
to ithtml_file=open('///Users/joebatt/Desktop/python/puzzle5.txt','w')
for line in fileurl.readlines():linestring=line.decode(utf-8)
#ensures written in string not bytehtml_file.write(linestring)
html_file.close() #closes the puzzle file#just prints file to check its correct
html_file=open('///Users/joebatt/Desktop/python/puzzle5.txt','r')
filecontents=html_file.read()html_file.close()print (filecontents)
print (type(filecontents))
#pickles the file filecontents containing the url file
pickledfile=open('///Users/joebatt/Desktop/python/pickledpuzzle5.txt','w')
pickle.dump(filecontents,pickledfile)pickledfile.close()
return ()
# Main programurlfile=input('Please input the URL ')geturlfile(urlfile)
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[Tutor] Saving read-only or encoded text files?
Hi. I've been using a lot of text files recently, and I'm starting to worry about a user hacking some element by editing the text files. I know that I can pickle my data instead, creating less easily editable (try saying that five times fast) .dat files, but I'd rather store individual variables rather than lists of objects. Is there a way to make my text files either read-only or saved in some way that they can't be opened, or at least not so easily as double-clicking on them? I just want some slightly more secure code, though it's not too important. I just thought I'd ask. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Saving read-only or encoded text files?
Hi. I've been using a lot of text files recently, and I'm starting to worry about a user hacking some element by editing the text files. I know that I can pickle my data instead, creating less easily editable (try saying that five times fast) .dat files, but I'd rather store individual variables rather than lists of objects. Is there a way to make my text files either read-only or saved in some way that they can't be opened, or at least not so easily as double-clicking on them? I just want some slightly more secure code, though it's not too important. I just thought I'd ask. == Any file will eventually be able to be reverse engineered, but it matters how much effort you care to obfuscate it. The way you can do it will vary based on your OS. For Windows, you can change the file extension to something that is not read by most text editors '.zxy'. It will still be able to be read if they try and open it with a text editor, but double clicking will not work by default. You can also try setting the file attribute directly: http://code.activestate.com/recipes/303343-changing-file-attributes-on-windows/ For *nix/OS X, you can prepend the file with . as those files are hidden by default on most *nix systems I have used. You can also try to use os.chmod(0###, 'filename'). Keep in mind that all of these solutions are probably user reversible since the application will have the permissions of the user account it is run as; in most cases this is the same as the logged in user. Ramit Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology 712 Main Street | Houston, TX 77002 work phone: 713 - 216 - 5423 -- This email is confidential and subject to important disclaimers and conditions including on offers for the purchase or sale of securities, accuracy and completeness of information, viruses, confidentiality, legal privilege, and legal entity disclaimers, available at http://www.jpmorgan.com/pages/disclosures/email. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Saving read-only or encoded text files?
Thank you. This will work perfectly. On Nov 18, 2011, at 11:58 AM, Prasad, Ramit wrote: Hi. I've been using a lot of text files recently, and I'm starting to worry about a user hacking some element by editing the text files. I know that I can pickle my data instead, creating less easily editable (try saying that five times fast) .dat files, but I'd rather store individual variables rather than lists of objects. Is there a way to make my text files either read-only or saved in some way that they can't be opened, or at least not so easily as double-clicking on them? I just want some slightly more secure code, though it's not too important. I just thought I'd ask. == Any file will eventually be able to be reverse engineered, but it matters how much effort you care to obfuscate it. The way you can do it will vary based on your OS. For Windows, you can change the file extension to something that is not read by most text editors '.zxy'. It will still be able to be read if they try and open it with a text editor, but double clicking will not work by default. You can also try setting the file attribute directly: http://code.activestate.com/recipes/303343-changing-file-attributes-on-windows/ For *nix/OS X, you can prepend the file with . as those files are hidden by default on most *nix systems I have used. You can also try to use os.chmod(0###, 'filename'). Keep in mind that all of these solutions are probably user reversible since the application will have the permissions of the user account it is run as; in most cases this is the same as the logged in user. Ramit Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology 712 Main Street | Houston, TX 77002 work phone: 713 - 216 - 5423 -- This email is confidential and subject to important disclaimers and conditions including on offers for the purchase or sale of securities, accuracy and completeness of information, viruses, confidentiality, legal privilege, and legal entity disclaimers, available at http://www.jpmorgan.com/pages/disclosures/email. ___ 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] In a pickle over pickles….Python 3
On 11/18/2011 11:12 AM, Joe Batt wrote:
Hi All,Sorry to trouble you all again with more nooby problems! Only been
programming a week so still all a haze especially since self taught…..
I am opening a url and saving it to a file on my computer, then I am trying to
pickle it. I have written the url file to a file on my computer then opened it
and assigned the contents to a var 'filecontents' and tried to pickle it it is
giving the error:
Traceback (most recent call last): File /Users/joebatt/Desktop/python/pickling puzzle 5.py,
line 39, inmodule geturlfile(urlfile) File /Users/joebatt/Desktop/python/pickling
puzzle 5.py, line 28, in geturlfilepickle.dump(filecontents,pickledfile)TypeError: must be str,
not bytes
Yet I tested the variable filecontents using print (type(file contents)) and
it is saying that it is a str yet the error seems to indicate it is a byte, can
anyone point me in the right direction please?
(I know my coding is very untidy and verbose, sorry I am very new and at the
moment I have to think in baby steps with little knowledge!)
Joe
Pickle an
object- ###
import pickleimport urllib
def geturlfile(urlfile):#gets the user URL from main opens it and saves it to var fileurl
from urllib.request import urlopenfileurl=urlopen(urlfile)#opens the file on
computer and writes the var fileurl to it
html_file=open('///Users/joebatt/Desktop/python/puzzle5.txt','w')for line in
fileurl.readlines():linestring=line.decode(utf-8) #ensures written in
string not bytehtml_file.write(linestring)html_file.close() #closes the
puzzle file#just prints file to check its correct
html_file=open('///Users/joebatt/Desktop/python/puzzle5.txt','r')
filecontents=html_file.read()html_file.close()print (filecontents)print
(type(filecontents))
#pickles the file filecontents containing the url file
pickledfile=open('///Users/joebatt/Desktop/python/pickledpuzzle5.txt','w')
pickle.dump(filecontents,pickledfile)pickledfile.close()
return ()
# Main programurlfile=input('Please input the URL ')geturlfile(urlfile)
You forgot to post in text mode, so everything ran together.
--
DaveA
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Re: [Tutor] Saving read-only or encoded text files?
On Fri, Nov 18, 2011 at 11:17 AM, Max gmail maxskywalk...@gmail.com wrote: Thank you. This will work perfectly. On Nov 18, 2011, at 11:58 AM, Prasad, Ramit wrote: snip Any file will eventually be able to be reverse engineered, but it matters how much effort you care to obfuscate it. The way you can do it will vary based on your OS. For Windows, you can change the file extension to something that is not read by most text editors '.zxy'. It will still be able to be read if they try and open it with a text editor, but double clicking will not work by default. You can also try setting the file attribute directly: http://code.activestate.com/recipes/303343-changing-file-attributes-on-windows/ For *nix/OS X, you can prepend the file with . as those files are hidden by default on most *nix systems I have used. You can also try to use os.chmod(0###, 'filename'). Keep in mind that all of these solutions are probably user reversible since the application will have the permissions of the user account it is run as; in most cases this is the same as the logged in user. snip As an addition, you can also create the file using the zipfile module ( http://docs.python.org/library/zipfile.html). This adds another layer of obfuscation and has the added benefit of making your files smaller. Of course, the savvy user won't have any issue getting past this (Office files are just archive files). On Ubuntu at least, if you just remove the extension then it will attempt to discover what the filetype, and can usually guess archive types. It's security through obscurity, of course, so it all depends on who you're worried about accessing this data. HTH, Wayne ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Encoding
On 11/17/2011 8:45 PM, Nidian Job-Smith wrote: Hi all, In my programme I am encoding what the user has in-putted. What the user inputs will in a string, which might a mixture of letters and numbers. However I only want the letters to be encoded. I am assuming that you meant only accept characters and not actual text encoding. The following example is untested and is limited. It will not really work with non-ASCII letters (i.e. Unicode). import string input_string = raw_input( 'Enter something' ) #use input in Python3 final_input = [] # append to a list instead of concatenating a string # because it is faster to ''.join( list ) for char in input_string: if char in string.letters: final_input.append( char ) input_string = ''.join( final_input ) Ramit Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology 712 Main Street | Houston, TX 77002 work phone: 713 - 216 - 5423 -- This email is confidential and subject to important disclaimers and conditions including on offers for the purchase or sale of securities, accuracy and completeness of information, viruses, confidentiality, legal privilege, and legal entity disclaimers, available at http://www.jpmorgan.com/pages/disclosures/email. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Shorten Code.
for var in [value,value_1]: var = green Um, that won't work. You typed that example too quickly. Oops! Yes. You'd need to enumerate and access the variables via an index. yuk. Don't do it folks! :-) Excuse: It was early morning and I hadn't had any coffee... Alan G.___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] In a pickle over pickles….Python 3
On Fri, Nov 18, 2011 at 11:12 AM, Joe Batt joeb...@hotmail.co.uk wrote:
pickledfile=open('///Users/joebatt/Desktop/python/pickledpuzzle5.txt','w')
pickle.dump(filecontents,pickledfile)
pickledfile.close()
A pickle is a binary object, so you'll need to open your picklefile in
binary mode:
pickledfile=open('///Users/joebatt/Desktop/python/pickledpuzzle5.txt','wb')
--
Jerry
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Re: [Tutor] A recursion question
On 11/18/2011 06:04 PM, Dave Angel wrote: On 11/18/2011 08:16 AM, Kĩnũthia Mũchane wrote: snip Well, it doesn't count the number of occurrences correctly if the list is empty. It should get zero, and it gets -1. But for any non-empty list, nothing ever looks at the -1 value, so it doesn't matter what you put there. This example diverges from traditional recursion in many ways, but the chief reason it's not traditional recursion is that it never uses the return value. within the function. If the list is empty, which is the base case, s.slais([], 4) returns -1. Now using some bush logic, in a non-empty list, in order for the recursion to terminate it has to 'hit' the base case and return -1. Where does this -1 go ? Nowhere. You don't use it, so it gets discarded. if you were going to use it, your internal calls to the method would look something like: b = self.slais(lst[1:], x) and then you'd do something with b. Further, why do not I get a *TypeError* when I use a simple *return* statement in the *if* clause? You would if you actually used the value for arithmetic. But the return itself would be perfectly legal. it's just a shortcut for 'return None' The reason I am asking that is that I think(wrongly, of course :-)) it should be part of the answer and therefore I should be getting an answer that is off by one or a *TypeError*!! And by the way, the reason I used a *class* was that I could not get a suitable place in the program to initialise my *count* variable otherwise. Using a class is fine. But you're abusing it. What happens if you try to sum a second list? (Hint, it gives you a higher number) That is very true, I tried it with a second list and the value was higher. Thanks... If you really want to recursively count, you need a structure which uses no objects of global lifetime. The intermediate values should be stored in local variables to that method. def counter(mylist, val): if len(mylist == 0): return 0 prev = counter(mylist[1:], val) #this is actually using the recursive return value if mylist[0] == val: return prev + 1 else: return prev I was trying to come up with something like this but I was totally stumped. Now I understand, perfectly. My heartfelt thanks to you, Dave, and Christian ! ;-) Totally untested. And there are certainly many other variants possible. But the key is you have to do something with the value returned to you by the lower level function. -- Kĩnũthia S 1º 8' 24” E 36º 57' 36” 1522m ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] binary file query
No *need* but, often convenient, regardless of whether the data was written by a C struct or not.(*) It is a way to interpret binary data in terms of fundamental data types such as ints, floats, strings etc. Trying to convert raw bytes representing floats into a floating point number without using struct would be an interesting exercise. Using struct it's trivial. Interesting, I will keep it in mind. Thanks! Ramit Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology 712 Main Street | Houston, TX 77002 work phone: 713 - 216 - 5423 -- This email is confidential and subject to important disclaimers and conditions including on offers for the purchase or sale of securities, accuracy and completeness of information, viruses, confidentiality, legal privilege, and legal entity disclaimers, available at http://www.jpmorgan.com/pages/disclosures/email. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Shorten Code.
On Nov 18, 2011, at 13:15, ALAN GAULD alan.ga...@btinternet.com wrote: for var in [value,value_1]: var = green Um, that won't work. You typed that example too quickly. Oops! Yes. You'd need to enumerate and access the variables via an index. yuk. Don't do it folks! :-) Excuse: It was early morning and I hadn't had any coffee... Alan G. In your defense Alan, after you typed that code in your response you mentioned the necessity of defining the variables: But you need to have already created the variables somewhere and unless there is a big list its not usually worth while. Alexander Etter ___ 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] urgent help!!!!!!!!!!!
On 11/17/2011 3:26 PM ADRIAN KELLY said... i know i'm stupid but i have tried everything to get one line of text working, i have written out pseudo and read every website. now i am getting this error Traceback (most recent call last): File F:\VTOS ATHLONE\PYTHON_VTOS\foreign exchange\f_ex4 - Copy.py, line 24, in module main() File F:\VTOS ATHLONE\PYTHON_VTOS\foreign exchange\f_ex4 - Copy.py, line 14, in main while amount50: UnboundLocalError: local variable 'amount' referenced before assignment You've seen the fixes, but perhaps more explanation helps too. Python organizes variables into namespaces. The normal namespace resolution sequence is local, global, builtin. Assigning to a name within a function creates a local variable, otherwise accessing a name in a function will use the global or builtin namespaces. Python decides these issues in part during initial loading of the module where it sees that you assign to the variable amount within the function, thus creating a local variable. Later, during execution, you first test the value of amount (in the while statement), but amount isn't assigned to until after the test, ergo, UnboundLocalError. HTH, Emile ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] python telnet
I installed the telnet client but still the same error.
Traceback (most recent call last):
File C:\Users\Rayon\Documents\projects1\super_hia\main.py, line 6, in
module
child = winspawn('telnet 192.168.0.55:210')
File
C:\Python26\lib\site-packages\winpexpect-1.5-py2.6.egg\winpexpect.py, line
346, in __init__
logfile=logfile, cwd=cwd, env=env)
File C:\Python26\lib\site-packages\winpexpect-1.5-py2.6.egg\pexpect.py,
line 429, in __init__
self._spawn (command, args)
File
C:\Python26\lib\site-packages\winpexpect-1.5-py2.6.egg\winpexpect.py, line
369, in _spawn
raise ExceptionPexpect, 'Command not found: %s' % self.command
ExceptionPexpect: Command not found: telnet
-Original Message-
From: tutor-bounces+evosweet=hotmail@python.org
[mailto:tutor-bounces+evosweet=hotmail@python.org] On Behalf Of Steven
D'Aprano
Sent: 17 November 2011 19:52
To: tutor@python.org
Subject: Re: [Tutor] python telnet
Rayon wrote:
I am trying to use winpexpect to connect a telnet session.
I keep getting this error.
raise ExceptionPexpect, 'Command not found: %s' % self.command
ExceptionPexpect: Command not found: telnet
Please copy and paste the entire traceback, not just the last couple of
lines. But judging by just the small bit you show, it looks like telnet is
not a command that winpexpect understands, so it raises an error command
not found.
Perhaps it is a bug in winpexpect. Are you using the latest version?
Do you actually have telnet available on your system? If not, then you can't
expect winpexpect to use something which isn't there.
--
Steven
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Re: [Tutor] python telnet
Thank you I think I understand now, thank you very much. -Original Message- From: tutor-bounces+evosweet=hotmail@python.org [mailto:tutor-bounces+evosweet=hotmail@python.org] On Behalf Of Steven D'Aprano Sent: 17 November 2011 19:52 To: tutor@python.org Subject: Re: [Tutor] python telnet Rayon wrote: I am trying to use winpexpect to connect a telnet session. I keep getting this error. raise ExceptionPexpect, 'Command not found: %s' % self.command ExceptionPexpect: Command not found: telnet Please copy and paste the entire traceback, not just the last couple of lines. But judging by just the small bit you show, it looks like telnet is not a command that winpexpect understands, so it raises an error command not found. Perhaps it is a bug in winpexpect. Are you using the latest version? Do you actually have telnet available on your system? If not, then you can't expect winpexpect to use something which isn't there. -- Steven ___ 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] A recursion question
Another way to do that is to avoid any intermediate variables altogether That may be easier to understand YMMV def counter(mylist, val): if len(mylist == 0): return 0 if mylist[0] == val: return 1 + counter(mylist[1:], val) else: return counter(mylist[1:]) Asokan Pichai ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] FW: urgent help!! THANKS EVERYONE!
Wayne Werner wrote: The chances of hitting the maximum recursion depth is pretty slim, unless your user manages to type in a wrong number about 1000 times ;) I've had to provide desktop support to people who have done that. Only-slightly-exaggerating-ly 'yrs, -- Steven ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] A recursion question
On 11/19/2011 06:03 AM, Asokan Pichai wrote: Another way to do that is to avoid any intermediate variables altogether That may be easier to understand YMMV def counter(mylist, val): if len(mylist == 0): return 0 if mylist[0] == val: return 1 + counter(mylist[1:], val) else: return counter(mylist[1:]) The intermediate variable explanation by Dave actually clinched it for me. Actually, the one I wrote is suspiciously similar to yours ;-). Anyway, thanks Asokan! Asokan Pichai ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor -- Kĩnũthia S 1º 8' 24” E 36º 57' 36” 1522m ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
