Break statement in Python

Question

I am trying to break out of a for loop, but for some reason the following doesn't work as expected:

for out in dbOutPut:
    case_id = out['case_id']
    string = out['subject']
    vectorspace = create_vector_space_model(case_id, string, tfidf_dict)
    vectorspace_list.append(vectorspace)
    case_id_list.append(case_id)

    print len(case_id_list)

    if len(case_id_list) >= kcount:
        print "true"
        break

It just keeps iterating untill the end of dbOutput. What am I doing wrong?

What is kcount? Please show all relevant parts of your code. — Greg Hewgill
– Greg Hewgill, Commented Jan 27, 2012 at 20:56
hi kcount is a variable i am inputting..but its way less than the size of that list..kcount = 20...and total length of dboutput is like 100000 — frazman
– frazman, Commented Jan 27, 2012 at 20:57
Have you tried adding print kcount to verify condition is actually met? — Li0liQ
– Li0liQ, Commented Jan 27, 2012 at 20:57
Does it ever print true? If not then your if statement isn't being triggered, so your break is never happening. — Herms
– Herms, Commented Jan 27, 2012 at 20:57
@Fraz: I'm suggesting that your kcount is not an integer. It might be a string (and so Python is not doing what you expect when you compare an integer with a string). Please show the code that initialises kcount. — Greg Hewgill
– Greg Hewgill, Commented Jan 27, 2012 at 20:59

Community · Accepted Answer · 2017-05-23 09:59:11Z

6

I'm guessing, based on your previous question, that kcount is a string, not an int. Note that when you compare an int with a string, (in CPython version 2) the int is always less than the string because 'int' comes before 'str' in alphabetic order:

In [12]: 100 >= '2'
Out[12]: False

If kcount is a string, then the solution is add a type to the argparse argument:

import argparse
parser=argparse.ArgumentParser()
parser.add_argument('-k', type = int, help = 'number of clusters')
args=parser.parse_args()
print(type(args.k))   
print(args.k)

running

% test.py -k 2

yields

<type 'int'>
2

This confusing error would not arise in Python3. There, comparing an int and a str raises a TypeError.

edited May 23, 2017 at 9:59

CommunityBot

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answered Jan 27, 2012 at 20:58

unutbu

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Community · Accepted Answer · 2017-05-23 12:34:51Z

4

Could it happen that kcount is actually a string, not an integer and, therefore, could never become less than any integer?
See string to int comparison in python question for more details.

edited May 23, 2017 at 12:34

CommunityBot

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answered Jan 27, 2012 at 21:00

Li0liQ

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