Paul Rubin a écrit :
Bruno Desthuilliers [EMAIL PROTECTED] writes:
(snip)
Python has had functions as first class objects and
(quite-limited-but) anonymous functions, map(), filter() and reduce()
as builtin funcs at least since 1.5.2 (quite some years ago).
True, though no iterators so you
Arnaud Delobelle a écrit :
On Mar 13, 8:53 pm, Bruno Desthuilliers
[EMAIL PROTECTED] wrote:
Paul Rubin a écrit :
[snip]
Iterators like that are a new Python feature
List comps are not that new (2.0 or 2.1 ?):
print \n.join([contact for name, contact in contacts.items() \
All -
I'm currently writing a toy program as I learn python that acts as a
simple address book. I've run across a situation in my search function
where I want to iterate across a filtered list. My code is working
just fine, but I'm wondering if this is the most elegant way to do
this.
Drew [EMAIL PROTECTED] writes:
I'm currently writing a toy program as I learn python that acts as a
simple address book. I've run across a situation in my search function
where I want to iterate across a filtered list. My code is working
just fine, but I'm wondering if this is the most elegant
On Mar 13, 2:42 pm, Paul Rubin http://[EMAIL PROTECTED] wrote:
If I can decipher your Ruby example (I don't know Ruby), I think you
want:
for name,contact in contacts.iteritems():
if re.search('search', name):
print contact
If you just want to filter the dictionary
On Mar 13, 6:04 pm, Drew [EMAIL PROTECTED] wrote:
All -
Hi!
[snip]
http://pastie.caboo.se/46647
There is no need for such a convoluted list comprehension as you
iterate over it immediately! It is clearer to put the filtering logic
in the for loop. Moreover you recalculate the regexp for each
Drew [EMAIL PROTECTED] writes:
You're exactly on the mark. I guess I was just wondering if your first
example (that is, breaking the if statement away from the iteration)
was preferred rather than initially filtering and then iterating.
I think the multiple statement version is more in Python
Arnaud Delobelle [EMAIL PROTECTED] writes:
in the for loop. Moreover you recalculate the regexp for each element
of the list.
The re library caches the compiled regexp, I think.
--
http://mail.python.org/mailman/listinfo/python-list
Paul Rubin a écrit :
Drew [EMAIL PROTECTED] writes:
You're exactly on the mark. I guess I was just wondering if your first
example (that is, breaking the if statement away from the iteration)
was preferred rather than initially filtering and then iterating.
I think the multiple statement
On Mar 13, 7:36 pm, Paul Rubin http://[EMAIL PROTECTED] wrote:
Arnaud Delobelle [EMAIL PROTECTED] writes:
in the for loop. Moreover you recalculate the regexp for each element
of the list.
The re library caches the compiled regexp, I think.
That would surprise me.
How can re.search know
Bruno Desthuilliers [EMAIL PROTECTED] writes:
I don't know if I qualify as a Python traditionalist, but I'm using
Python since the 1.5.2 days, and I usually favor list comps or
generator expressions over old-style loops when it comes to this kind
of operations.
I like genexps when they're
On Mar 13, 8:53 pm, Bruno Desthuilliers
[EMAIL PROTECTED] wrote:
Paul Rubin a écrit :
[snip]
Iterators like that are a new Python feature
List comps are not that new (2.0 or 2.1 ?):
print \n.join([contact for name, contact in contacts.items() \
if search.match(name)])
En Tue, 13 Mar 2007 15:04:50 -0300, Drew [EMAIL PROTECTED] escribió:
I'm currently writing a toy program as I learn python that acts as a
simple address book. I've run across a situation in my search function
where I want to iterate across a filtered list. My code is working
just fine, but
On Mar 13, 8:59 pm, Gabriel Genellina [EMAIL PROTECTED]
wrote:
[snip]
def find(self, search):
search_re = re.compile(search, re.IGNORECASE)
for result in [self.contacts[name] for name in self.contacts if
search_re.match(name)]:
print result
I do not see how
for y in
En Tue, 13 Mar 2007 17:19:53 -0300, Arnaud Delobelle
[EMAIL PROTECTED] escribió:
On Mar 13, 7:36 pm, Paul Rubin http://[EMAIL PROTECTED] wrote:
The re library caches the compiled regexp, I think.
That would surprise me.
How can re.search know that string.lower(search) is the same each
En Tue, 13 Mar 2007 18:16:32 -0300, Arnaud Delobelle
[EMAIL PROTECTED] escribió:
On Mar 13, 8:59 pm, Gabriel Genellina [EMAIL PROTECTED]
wrote:
[snip]
def find(self, search):
search_re = re.compile(search, re.IGNORECASE)
for result in [self.contacts[name] for name in
On Mar 13, 9:31 pm, Gabriel Genellina [EMAIL PROTECTED]
wrote:
En Tue, 13 Mar 2007 17:19:53 -0300, Arnaud Delobelle
[EMAIL PROTECTED] escribió:
On Mar 13, 7:36 pm, Paul Rubin http://[EMAIL PROTECTED] wrote:
The re library caches the compiled regexp, I think.
That would surprise me.
Hi,
On Tuesday 13 of March 2007 22:16:32 Arnaud Delobelle wrote:
for x in L:
if g(x):
do stuff with f(x)
for x in itertools.ifilterfalse(g, L):
do stuff
Maybe this would be even better?
L
--
http://mail.python.org/mailman/listinfo/python-list
En Tue, 13 Mar 2007 19:12:12 -0300, Arnaud Delobelle
[EMAIL PROTECTED] escribió:
py import re
py x = re.compile(ijk)
py y = re.compile(ijk)
py x is y
True
Both, separate calls, returned identical results. You can show the
cache:
OK I didn't realise this. But even so each time there
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