«

how to get rid of 'for' operator in the code?


import os, os.path

def _test():
src = 'C:\\Documents and Settings\\åÇÏÒ\\My Documents\\My Music\\'

for i in [x for x in os.listdir(src) if os.path.isfile(os.path.join(src,
x)) and len(x.split('.')) > 1 and x.split('.')[-1].lower() == 'm3u']:
os.remove(os.path.join(src, i))

if __name__ == '__main__':
_test()

On Thu, 13 Jan 2005 15:55:10 +1000, Egor Bolonev <> wrote:
> how to get rid of 'for' operator in the code?
>
> import os, os.path
>
> def _test():
> src = 'C:\\Documents and Settings\\Егор\\My Documents\\My Music\\'
>
> for i in [x for x in os.listdir(src) if os.path.isfile(os.path.join(src,
> x)) and len(x.split('.')) > 1 and x.split('.')[-1].lower() == 'm3u']:
> os.remove(os.path.join(src, i))
>
> if __name__ == '__main__':
> _test()

import glob
for x in glob.glob("*.m3u"):
os.remove(x)

Regards,
Stephen Thorne

"Stephen Thorne" <> �ообщил/�ообщила в ново�т�х
�ледующее: news:mailman.611.1105598828.22381.python-...
On Thu, 13 Jan 2005 15:55:10 +1000, Egor Bolonev <> wrote:
> how to get rid of 'for' operator in the code?
>
> import os, os.path
>
> def _test():
> src = 'C:\\Documents and Settings\\Егор\\My Documents\\My Music\\'
>
> for i in [x for x in os.listdir(src) if
> os.path.isfile(os.path.join(src,
> x)) and len(x.split('.')) > 1 and x.split('.')[-1].lower() == 'm3u']:
> os.remove(os.path.join(src, i))
>
> if __name__ == '__main__':
> _test()

import glob
for x in glob.glob("*.m3u"):
os.remove(x)

i want to wipe out 'for x in []: f(x)' using map, lambda, reduce, filter and
List
Comprehensions [x for x in []]
just don't get how to add string to all elements of list

On Thu, 13 Jan 2005 17:05:39 +1000, Egor Bolonev <> wrote:
>
> "Stephen Thorne" <> �ообщил/�ообщила в ново�т�х
> �ледующее: news:mailman.611.1105598828.22381.python-...
> On Thu, 13 Jan 2005 15:55:10 +1000, Egor Bolonev <> wrote:
> > how to get rid of 'for' operator in the code?
> >
> > import os, os.path
> >
> > def _test():
> > src = 'C:\\Documents and Settings\\Егор\\My Documents\\My Music\\'
> >
> > for i in [x for x in os.listdir(src) if
> > os.path.isfile(os.path.join(src,
> > x)) and len(x.split('.')) > 1 and x.split('.')[-1].lower() == 'm3u']:
> > os.remove(os.path.join(src, i))
> >
> > if __name__ == '__main__':
> > _test()
>
> import glob
> for x in glob.glob("*.m3u"):
> os.remove(x)
>
> i want to wipe out 'for x in []: f(x)' using map, lambda, reduce, filter and
> List
> Comprehensions [x for x in []]
> just don't get how to add string to all elements of list

Here's a few ways,

map(os.remove, glob.glob("*.m3u"))
[os.remove(x) for x in glob.glob("*.m3u")]

[os.remove(x) for x in os.listdir(src) if
os.path.isfile(os.path.join(src, x))
and len(x.split('.')) > 1
and x.split('.')[-1].lower() == 'm3u']

def myfilter(x):
return os.path.isfile(os.path.join(src, x)) and len(x.split('.')) >
1 and x.split('.')[-1].lower() == 'm3u'
map(os.remove, filter(myfilter, os.listdir(src)))

Regards,
Stephen Thorne.

I understand you want to do it in an applicative programming style?
Not recommended in general. But here goes:

..# c.l.p. question:
..# "using map, lambda, reduce, filter and List Comprehensions [x for
..# x in []] just don't get how to add string to all elements of list"
..
..##
..# A function that returns a function, taking a function as argument.
..def allapply(fn):
.. def f(seq): return map(fn, seq)
.. return f
..
..def add_string_to_element(stringval):
.. def f(x): return x + stringval
.. return f
..
..add_string_to_all = allapply(add_string_to_element('mystring'))
..
..print add_string_to_all(['d:', 'c:\windows\\','something/'])

this outputs:
['d:mystring', 'c:\\windows\\mystring', 'something/mystring']
--
look Ma, no for and no lambda!
-- Will Stuyvesant

Egor Bolonev wrote:

>
> "Stephen Thorne" <> �ообщил/�ообщила в ново�т�х
> �ледующее: news:mailman.611.1105598828.22381.python-...
> On Thu, 13 Jan 2005 15:55:10 +1000, Egor Bolonev <> wrote:
>
>> how to get rid of 'for' operator in the code?
>>
>> import os, os.path
>>
>> def _test():
>> src = 'C:\\Documents and Settings\\Егор\\My Documents\\My Music\\'
>>
>> for i in [x for x in os.listdir(src) if
>> os.path.isfile(os.path.join(src,
>> x)) and len(x.split('.')) > 1 and x.split('.')[-1].lower() == 'm3u']:
>> os.remove(os.path.join(src, i))
>>
>> if __name__ == '__main__':
>> _test()
>
>
> import glob
> for x in glob.glob("*.m3u"):
> os.remove(x)
>
> i want to wipe out 'for x in []: f(x)' using map, lambda, reduce, filter
> and List
> Comprehensions [x for x in []]
> just don't get how to add string to all elements of list
>

Any statement of the form

for i in [x for x in something]:

can be rewritten as

for i in something:

Note that this doesn't mean you never want to iterate over a list
comprehension. It's the easiest way, for example, to iterate over the
first item of each list in a list of lists:

for i in [x[0] for x in something]:

regards
Steve
--
Steve Holden http://www.holdenweb.com/
Python Web Programming http://pydish.holdenweb.com/
Holden Web LLC +1 703 861 4237 +1 800 494 3119

On Thu, 13 Jan 2005 09:16:40 -0500, Steve Holden <> wrote:
[...]
>
>Any statement of the form
>
> for i in [x for x in something]:
>
>can be rewritten as
>
> for i in something:
>
>Note that this doesn't mean you never want to iterate over a list
>comprehension. It's the easiest way, for example, to iterate over the
>first item of each list in a list of lists:
>
> for i in [x[0] for x in something]:
>
As I'm sure you know, with 2.4's generator expressions you
don't have to build the temporary list.
Which could be important if 'something'
is (or generates) a huge sequence.

for i in (x[0] for x in something):

>>> something = ([x] for x in xrange(10,20))
>>> something
<generator object at 0x02EF176C>
>>> list(something)
[[10], [11], [12], [13], [14], [15], [16], [17], [18], [19]]
>>> for i in (x[0] for x in something): print i,
...

oops, that list() used it up

>>> something = [[x] for x in xrange(10,20)]
>>> for i in (x[0] for x in something): print i,
...
10 11 12 13 14 15 16 17 18 19

Really nice.

Regards,
Bengt Richter

On Fri, 14 Jan 2005 00:08:09 GMT, rumours say that (Bengt
Richter) might have written:

>As I'm sure you know, with 2.4's generator expressions you
>don't have to build the temporary list.
>Which could be important if 'something'
>is (or generates) a huge sequence.
>
> for i in (x[0] for x in something):

and for some functional fun:

from itertools import imap
from operator import itemgetter
for i in imap(itemgetter(0), something):
--
TZOTZIOY, I speak England very best.
"Be strict when sending and tolerant when receiving." (from RFC195
I really should keep that in mind when talking with people, actually...

Bengt Richter wrote:

> On Thu, 13 Jan 2005 09:16:40 -0500, Steve Holden <> wrote:
> [...]
>
>>Any statement of the form
>>
>> for i in [x for x in something]:
>>
>>can be rewritten as
>>
>> for i in something:
>>
>>Note that this doesn't mean you never want to iterate over a list
>>comprehension. It's the easiest way, for example, to iterate over the
>>first item of each list in a list of lists:
>>
>> for i in [x[0] for x in something]:
>>
>
> As I'm sure you know, with 2.4's generator expressions you
> don't have to build the temporary list.
> Which could be important if 'something'
> is (or generates) a huge sequence.
>
> for i in (x[0] for x in something):
>
Yes. While I haven't yet done any more than play with generator
sequences I do really feel that more of "the best of Icon" has arrived
in Python with this new addition.

> >>> something = ([x] for x in xrange(10,20))
> >>> something
> <generator object at 0x02EF176C>
> >>> list(something)
> [[10], [11], [12], [13], [14], [15], [16], [17], [18], [19]]
> >>> for i in (x[0] for x in something): print i,
> ...
>
> oops, that list() used it up
>
> >>> something = [[x] for x in xrange(10,20)]
> >>> for i in (x[0] for x in something): print i,
> ...
> 10 11 12 13 14 15 16 17 18 19
>
> Really nice.
>
I quite agree. It's particularly useful for infinite sequences

regards
Steve
--
Steve Holden http://www.holdenweb.com/
Python Web Programming http://pydish.holdenweb.com/
Holden Web LLC +1 703 861 4237 +1 800 494 3119