Python: return statement doesnt work

Question

I am trying to print a list of names using the following program

def search():
    with open('business_ten.json') as f:
        data=f.read()
        jsondata=json.loads(data)


    for row in jsondata['rows']:
        #print row['text']
        a=str(row['name'])

        #print a    
        return a


search()

If I put the return statement , it doesnt print , but if i put the print statement it works I want the return statement to work. I want the return statement instead of the print statement

In Python indentation is part of syntax. Please format your code correctly. — Łukasz Rogalski, Nov 24 '14 at 11:47
Return ends the method and returns the value of `a` on the first iteration. Have a look at [this lesson](http://learnpythonthehardway.org/book/ex21.html). — Jørgen R, Nov 24 '14 at 11:47
Also, you probably don't want `return` in there, but `yield` instead. — Rusty, Nov 24 '14 at 11:49
Instead of `print` you can append in a list and return that list after `for` loop ends. — Ashwani, Nov 24 '14 at 11:53

Vishnu Upadhyay · Answer 1 · 2014-11-24T12:00:23.033

0

This is due to indentation problem so that your return statement is not working. Put for loop with in search

def search():
    with open('business_ten.json') as f:
        data=f.read()
        jsondata=json.loads(data)
        for row in jsondata['rows']:
            a=str(row['name'])  
            yield a

data = search()
print list(data)

or

print list(search())

edited Nov 24 '14 at 12:00

answered Nov 24 '14 at 11:48

Vishnu Upadhyay

5,043
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I tried it , its not working . it doesnt print anything – dipit Nov 24 '14 at 11:51
@dipit check the updated code. you need to catch the return statement in some variable then `print` it. – Vishnu Upadhyay Nov 24 '14 at 11:52
It will only return the first row. Instead store return data in list and return that list after `for` loop ends. – Ashwani Nov 24 '14 at 11:54
This still won't work, after calling return the loop is going to end, so you will only get one result from the potentially many generated by jsondata['rows'] – Benjamin James Drury Nov 24 '14 at 11:54
@AshwaniDausodia then generator will help. i updated. – Vishnu Upadhyay Nov 24 '14 at 11:56
@Ashwani Dausodia could you please write it down – dipit Nov 24 '14 at 11:58

score 0 · Answer 2 · answered Nov 24 '14 at 11:49

You cannot return from inside a loop, as the method will just finish after the first part of the loop. It seems like you need to print the a, append a to a list of your data, then after the loop has finished, return said list.

Also your indentation is very... interesting. I'm not sure if that's an issue with how you have copied over your code, or if that is actually what your code looks like, however I would imagine the former otherwise it would have never run. Could you possibly edit the question to correct the issue?

score 0 · Answer 3 · answered Nov 24 '14 at 11:55

I guess this is what you want, but you could have explained it better...

def search():
    with open('business_ten.json') as f:
        data=f.read()
        jsondata=json.loads(data)
        for row in jsondata['rows']:
            a=str(row['name'])  
            yield a

print list(search())

score 0 · Answer 4 · answered Nov 24 '14 at 12:02

Create a list of all strings you were printing earlier and return that list.

def search():
    with open('business_ten.json') as f:
        data=f.read()
        jsondata=json.loads(data)
    return_data = []
    for row in jsondata['rows']:
        #print row['text']
        a=str(row['name'])

        #print a
        return_data.append(a)

    return return_data


print(search())

Python: return statement doesnt work

4 Answers4