Using regex (re.search) in a python list

Question

I have the following code

import re

pattern = ['A-minor Type I AGC', 'A-minor Type I AGC', 'A-minor Type I AGC', 'A-minor Type I AUA', 'A-minor Type I AUA', 'A-minor Type II AGC', 'A-minor Type II AGC']

n = len(pattern)
print pattern
pattern_str = ', '.join(pattern)
print pattern_str
for x in range(0, n):
    if re.search(r'\bType I\b', pattern_str):
        print "Hello Type I"
    elif re.search(r'\bType II\b', pattern_str):
        print "Hello Type II"
    else:
        print "An error has occured"

The desired output should be:

Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type II
Hello Type II

But I'm not getting the desired output. My current output is:

Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type I

Can someone point out the problem? I suspect it has to be something to do with the list to str conversion. I have managed to solve the problem using the following code:

for x in pattern:
    if re.search(r'\bType I\b', x):
        print "Hello Type I"
    elif re.search(r'\bType II\b', x):
        print "Hello Type II"
    else:
        print "An error has occured"

But I would like to know why my first code didn't work and how can I make it work. Any help is appreciated

Well - you initialized the pattern_str only once in the list and that's why you are getting the result. You could do - what you've done (more correct) or following change for x in range(0,n): pattern_str = pattern[x] — gabhijit
– gabhijit, Commented May 21, 2015 at 3:14
@gabhijit Thank you very much for your answer. Your advice to change pattern_str = pattern[x] made the first code to work. Thanks again :) — TheEmperor
– TheEmperor, Commented May 21, 2015 at 3:26

Eric Renouf · Accepted Answer · 2015-05-21 02:55:13Z

2

You're joining the whole list into a single string, then testing that a whole bunch of times. Instead, what you want is to test each string in the list like

for pattern_str in pattern:
    if re.search(r'\bType I\b', pattern_str):
        print "Hello Type I"
    elif re.search(r'\bType II\b', pattern_str):
        print "Hello Type II"
    else:
        print "An error has occured"

so you are searching each pattern, one at a time

answered May 21, 2015 at 2:55

Eric Renouf

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Comments

Lim H. · Accepted Answer · 2015-05-21 02:55:30Z

1

What you want: search through each string in your list.

What your code does

re.search(r'\bType I\b', pattern_str)

It searches through pattern_str in every iteration of the loop. What is pattern_str:

pattern_str = ', '.join(pattern)

Thus, in every iteration, it searches through the same string, which is the concatanation of the whole list, which always matches Type I in A-minor Type I AGC.

Changing in to search through each x in pattern does the trick

answered May 21, 2015 at 2:55

Lim H.

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Comments

Martin Konecny · Accepted Answer · 2015-05-21 02:54:36Z

Your pattern is the same for each iteration. You need to loop through your patterns in your for-loop.

Use for p in pattern

This will mean p is 'A-minor Type I AGC' on the first iteration, 'A-minor Type I AGC' on the second iteration, etc.

import re

pattern = ['A-minor Type I AGC', 'A-minor Type I AGC', 'A-minor Type I AGC', 'A-minor Type I AUA', 'A-minor Type I AUA', 'A-minor Type II AGC', 'A-minor Type II AGC']

for p in pattern:
    if re.search(r'\bType I\b', p):
        print "Hello Type I"
    elif re.search(r'\bType II\b', p):
        print "Hello Type II"
    else:
        print "An error has occured"

Output:

Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type I
Hello Type II
Hello Type II

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