Python list for different number of strings per line

Question

I have output file like below:

junk
abc 123 xyz test
123 abc test 123
bob ani kepu maro
exist

What i am trying with this file is I am searching kepu is present in column 3 in the file.

I am trying below code:

   name = "kepu"
   with open("listfile.txt") as f:
      for line in f:
         line = line.split(" ")[2]
         line = line.lower()
         if line == name:
            print "Present"
         else:
            print "Not Present"

Since in file does not have equal number of strings, In first line "junk" and last line "exist" contains only one string and trying

line = line.split(" ")[2]

This is the reason i am getting below error:

 IndexError: list index out of range

I tried removing first and last line that is junk and exist keyword it works. but my output file generation is random.

Please help me what is the best way to find the string is exit in the third column in file.

yeniv · Accepted Answer · 2017-05-11 03:04:50Z

2

Add a check to ensure that the line has at least 3 words:

wordsInLine = line.split(" ")
if len(wordsInLine) > 2:
    if wordsInLine[2].lower() == name:
        print "Present"
     else:
        print "Not Present"

answered May 11, 2017 at 3:04

yeniv

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Comments

Phung Duy Phong · Accepted Answer · 2017-05-11 03:05:23Z

1

Or you can throw catch an exception:

name = "kepu"
with open("listfile.txt") as f:
for line in f:
    try:
        line = line.split(" ")[2]
    except IndexError:
        continue
    line = line.lower()
    if line == name:
        print "Present"
    else:
        print "Not Present"

answered May 11, 2017 at 3:05

Phung Duy Phong

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Comments

A.J. Uppal · Accepted Answer · 2017-05-11 03:02:16Z

0

Just do a simple if statement:

name = "kepu"
with open("listfile.txt") as f:
  for line in f:
     if len(line.split(" ")) >= 3:
         line = line.split(" ")[2]
         line = line.lower()
         if line == name:
            print "Present"
         else:
            print "Not Present"

answered May 11, 2017 at 3:02

A.J. Uppal

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Comments

fan · Accepted Answer · 2017-05-11 03:25:08Z

0

May be you understand the "for line in f" is not clear: you cant true this code ,may be you can find why your problem happend:

 with open("listfile.txt") as f:
...:     for line in f:
...:        print 'line=%s'%line
...:
line=junk
line=abc 123 xyz test
line=123 abc test 123
line=bob ani kepu maro
line=exist

answered May 11, 2017 at 3:25

fan

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Comments

Jan Christoph Terasa · Accepted Answer · 2017-05-11 05:21:33Z

0

Use regular expressions:

import re

re_kepu = re.compile('^(\w+ ){2}kepu')

with open('listfile.txt', 'r') as f:
    for line in f:
        if re_kepu.match(line):
            print('Present')
            break
    else:
        print('Not present')

answered May 11, 2017 at 5:21

Jan Christoph Terasa

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