Python if a loop produces same output string for each iteration, how to print only once?

Question

a = [['jimmy', '25', 'pancakes'], ['tom', '23', 'brownies'], ['harry', '21', 'cookies']]
for i in range(len(a)):
    if (a[i][1] == '20' or a[i][1] == '26'):
        print 'yes'
    else:
        print 'Not found'

This output's Not found three times. If the output of the every iteration of the if loop is the same, I want it to iterate over the entire list and then print Not found only once.

If I change a[i][1] == '25' and the output becomes:

yes
Not found
Not found

I want to print yes but not Not found.

Ashwini Chaudhary · Accepted Answer · 2012-10-24 16:16:26Z

2

may be you're looking for for-else loop.

and as @Burhan Khalid suggested use for i in a instead if range(len(a)):

a = [['jimmy', '25', 'pancakes'], ['tom', '23', 'brownies'], ['harry', '21', 'cookies']]
for i in a:
    if (i[1] == '25' or i[1] == '26'):
        print 'yes'
else:
    print 'Not found'

output:

yes
Not found

Or may be you're looking for any():

In [200]: if any((i[1]=='25' or i[1]=='26') for i in a):
    print 'yes'
else:    
    print 'not Found'
   .....: 


yes

In [204]: if any((i[1]=='20' or i[1]=='26') for i in a):
    print 'yes'
else:    
    print 'not Found'
   .....: 


not Found

edited Oct 24, 2012 at 16:16

answered Oct 24, 2012 at 15:46

Ashwini Chaudhary

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4 Comments

Burhan Khalid Over a year ago

range(len(a)) can be better setup as for i in a:, then if i[1] == '25 or i[1] == '26':

koogee Over a year ago

Thanks! It solves the first problem. But it still print Not found once for the second problem. How do I get it to print only yes for positive matches and nothing for negative matches?

Ashwini Chaudhary Over a year ago

print nothing for the negative matches , then why do you want to print Not Found?

Ashwini Chaudhary Over a year ago

@koogee see my edited answer may be that is what you were looking for.

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