Is this the best way to test every item in a list?
def alltrue(f,l):
return reduce(bool.__and__,map(f,l))
def onetrue(f,l):
return reduce(bool.__or__,map(f,l))
>>> alltrue(lambda x:x>1,[1,2,3])
False
>>>
>>> alltrue(lambda x:x>=1,[1,2,3])
True
>>>
Thanks
--
http://mail.python.org/mai
Thanks for the answers, I suspected something like any(), all()
existed. Also got me thinking about generator expressions, my code is
full of list comprehensions for lists I don't keep.
--
http://mail.python.org/mailman/listinfo/python-list
Had the same issue. What you want is: reload()
--
http://mail.python.org/mailman/listinfo/python-list
if you require a SOAP client, then I would recommend SUDS
https://fedorahosted.org/suds/
Other options are not in active development, so is difficult to get
support.
--
http://mail.python.org/mailman/listinfo/python-list
I installed from here:
http://bazaar-vcs.org/WindowsInstall
first pycrypto-2.0.1.win32-py2.5.zip
then paramiko-1.7.1-ctypes.win32.exe
--
http://mail.python.org/mailman/listinfo/python-list
There is http://www.codeplex.com/IronPythonStudio
--
http://mail.python.org/mailman/listinfo/python-list
On Jul 23, 1:27 am, Clay Hobbs <[EMAIL PROTECTED]> wrote:
> I am making a program that (with urllib) that downloads two jpeg files
> and, if they are different, displays the new one. I need to find a way
> to compare two files in Python. How is this done?
>
> -- Ratfink
import hashlib
file = op
I would like to say something like:
for x in l if :
e.g.
for filename in os.listdir(DIR) if filename[-4:] == '.xml':
instead of having to say:
for filename in os.listdir(DIR):
if filename[-4:] == '.xml':
or
for filename in (f for f in os.listdir(DIR) if f[-4] == '.xml