On Monday, April 20, 2015 at 5:00:15 AM UTC-7, subhabrat...@gmail.com wrote:
> Dear Group,
>
> I am trying to open multiple files at one time.
> I am trying to do it as,
>
> for item in [ "one", "two", "three" ]:
> f = open (item + "world.txt", "w")
> f.close()
>
> This is fine. But I was looking if I do not know the number of
> text files I would create beforehand, so not trying xrange option
> also.
>
> And if in every run of the code if the name of the text files have
> to created on its own.
>
> Is there a solution for this?
>
> If anybody may please suggest.
>
> Regards,
> Subhabrata Banerjee.
Create a list and append each file object to the list. Something like:
fileList = []
for item in ['one', 'two', 'three']:
fileList.append(open(item + 'world.txt', 'w')
fileList will then have a list of open file objects. Of course, later you'll
have to close them all. That would be as easy as:
for f in fileList:
f.close()
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