On 11/19/18 8:15 PM, Asad wrote:
> Hi Avi Gross /All,
>
> Thanks for the reply. Yes you are correct , I would like to to
> open a file and process a line at a time from the file and want to select
> just lines that meet my criteria and print them while ignoring the rest. i
> have created the following code :
>
>
> import re
> import os
>
> f3 = open(r'file1.txt',r)
> f = f3.readlines()
> d = []
> for linenum in range(len(f)):
> if re.search("ERR-1" ,f[linenum])
> print f[linenum]
> break
> if re.search("\d\d\d\d\d\d",f[linenum]) --- > seach for a patch
> number length of six digits for example 123456
> print f[line]
> break
> if re.search("Good Morning",f[linenum])
> print f[line]
> break
> if re.search("Breakfast",f[linenum])
> print f[line]
> break
> ...
> further 5 more hetrogeneus if conditions I have
>
> =======================================================================
> This is beginners approach to print the lines which match the if conditions
> .
>
> How should I make it better may be create a dictionary of search items or a
> list and then iterate over the lines in a file to print the lines matching
> the condition.
We usually suggest using a context manager for file handling, so that
cleanup happens automatically when the context is complete:
with open('file1.txt', 'r') as f3:
# do stuff
# when you get here, f3 is closed
There's no need to do a counting loop, using the count as an index into
an array. That's an idiom from other programing languages; in Python you
may as well just loop directly over the list (array)... lists are iterable.
for line in f:
# search in line
Indeed, there's no real need to read all the lines in with readlines,
you can just loop directly over the file object - the f3 opened above:
for line in f3:
# search in line
There's no need to use a regular expression search if your pattern is a
simple string, you can use the "in" keyword:
if "Breakfast" in line:
print line
Keep your REs for more complex matches.
Do those help?
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