iterating through files

Steve Holden steve at holdenweb.com
Thu Feb 19 23:56:30 CET 2009


Chris Rebert wrote:
> On Thu, Feb 19, 2009 at 2:28 PM, Mike Driscoll <kyosohma at gmail.com> wrote:
>> On Feb 19, 4:22 pm, Mike Driscoll <kyoso... at gmail.com> wrote:
>>> On Feb 19, 3:56 pm, oamram <oam... at gmail.com> wrote:
>>>
>>>> Hi Pythonist,
>>>> new to python. i have a directory with about 50 text file and i need to
>>>> iterate through them and get
>>>> line 7 to 11 from each file and write those lines into another file(one file
>>>> that will contain all lines).
>>>> Cheers, Omer.
>>>> --
>>>> View this message in context:http://www.nabble.com/iterating-through-files-tp22048070p22048070.html
>>>> Sent from the Python - python-list mailing list archive at Nabble.com.
>>> I would recommend using the glob module to grab a list of the files
>>> you want or you can just create your own list. Then use a loop to grab
>>> the lines you want. Something like this:
>>>
>>> f = open(textFile)
>>> newFile = open(newFileName, "a")
>>> x = 1
>>> for line in f.readlines():
>>>     if x >=7 and x <=11:
>>>          newFile.write(line + "\n")
>>>
>>> You could even put the above inside a loop that loops over the list of
>>> files. Anyway, that's one approach. I'm sure there are many others.
>>>
>>> Mike
>> Oops...I forgot to iterate the counter. The code should look like
>> this:
>>
>> <code>
>>
>> f = open(textFile)
>> newFile = open(newFileName, "a")
>> x = 1
>> for line in f.readlines():
>>    if x >=7 and x <=11:
>>         newFile.write(line + "\n")
>>    x +=1
>>
>> </code>
> 
> Or you could use enumerate(); also, readlines() isn't necessary:
> 
> f = open(textFile)
> newFile = open(newFileName, "a")
> for x, line in enumerate(f):
>     if x >=7 and x <=11:
>         newFile.write(line + "\n")
> 
> Sounds a bit like homework to me though...
> 
But all these solutions read the whole file.  You should really read six
lines throwing them away then read five lines to keep.

Since this might be homework I'll not write the code, but it will
involve f.next() ...

regards
  Steve
-- 
Steve Holden        +1 571 484 6266   +1 800 494 3119
Holden Web LLC              http://www.holdenweb.com/




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