Returning elements from a loop, one at a time

Question

I have a simple function to parse a .csv as follows

def grabber3(datafile):

with open(datafile, 'rb') as f:
    r =csv.DictReader(f)
    for line in r:
        del line['thisthing']
        print line

This print out each dict line correctly which is about 50 lines such as

{'NetworkStatus': '1', 'ID': '1469'}
{'NetworkStatus': '1', 'ID': '1470'}
etc etc

However I i want to return this when the function is called so I change the print statement to return as

def grabber3(datafile):

with open(datafile, 'rb') as f:
    r =csv.DictReader(f)
    for line in r:
        del line['thisthing']
    return line

but it only returns the first line

{'NetworkStatus': '1', 'ID': '1469'}

How do I return each line of the loop/all of the dict lines instead of just the first one?

first, you need to fix your tabs. the only way that's happening is if `return ` is part of the `for line...` block. — acushner, Aug 26 '14 at 19:46
Correct the indentation in you function definitions to make it readable and understandable. It seems that in your second function definition you should append lines to some list and return a list of dictionaries. — Radosław Roszkowiak, Aug 26 '14 at 19:46
or, what you really want, is a map of `ID` to `NetworkStatus` — acushner, Aug 26 '14 at 19:48
In your current edit, the code would return the last line and not the first. I think your `del` and `return` are the same indentation level when you observe the first line being returned. — shaktimaan, Aug 26 '14 at 19:50

score 7 · Accepted Answer · edited May 23 '17 at 10:33

7

An efficient way of doing this is to use yield:

def grabber3(datafile):
    with open(datafile, 'rb') as f:
        r =csv.DictReader(f)
        for line in r:
            del line['thisthing']
            yield line

And then in the code that calls this function, you can do:

dict_generator = grabber3(a_file)

And then iterate through this dict_generator as:

for a_dict in dict_generator:
    print a_dict

Returning elements from a loop, one at a time

1 Answers1