Python Return List from Function

Python Return List from Function

I have a function that parses a file into a list. I'm trying to return that list so I can use it in other functions.

def splitNet():
    network = []
    for line in open("/home/tom/Dropbox/CN/Python/CW2/network.txt","r").readlines():
        line = line.replace("\r\n", "")
        line = string.split(line, ',')
        line = map(int, line)
        network.append(line)
    return network

When I try to print the list outside of the function (for debugging) I get this error:

NameError: name 'network' is not defined

Is there something simple I am doing wrong or is there a better way to pass variables between functions without using globals?

5

8 Answers

Variables cannot be accessed outside the scope of a function they were defined in.

Simply do this:

network = splitNet()
print network
3

I assume you are not assigning the returned value to a variable in scope.

ie. you can't do

splitNet()
print network

instead you would

network = splitNet()
print network

or for that matter

my_returned_network_in_scope = splitNet()
print my_returned_network_in_scope

otherwise you could declare network outside of the splitNet function, and make it global, but that is not the recommended approach.

2

The names of variables in a function are not visible outside, so you need to call your function like this:

networks = splitNet()
print(networks)

A couple of other notes:

  • You may want to convert your function to an iterator, using yield.
  • You don't need to call readlines; the function itself is an iterator.
  • Your function may be leaking the file handle. Use the with statement.
  • You can use str.split, which is more readable and easier to understand than string.split.
  • Your file looks to be a CSV file. Use the csv module.

In summary, this is how your code should look like:

import csv
def splitNet():
    with open("/home/tom/Dropbox/CN/Python/CW2/network.txt") as nf:
        for line in csv.reader(nf, delimiter=','):
            yield map(int, line)
network = list(splitNet())
print (network)

Your function is returning a list so you have to assign it to a variable and than try to print it.

network = splitNet()
print network

For example

>>> def mylist():
...    myl = []
...    myl.append('1')
...    return myl
...
>>> my_list = mylist()
>>> my_list
['1']
>>>

Have you actually called the function yet? This works fine (in the Python interpreter)

 >>> def f():
 ...   network = []
 ...   network.append(1)
 ...   network.append(2)
 ...   network.append(3)
 ...   return network
 ...
 >>> network = f()
 >>> print network
 [1, 2, 3]

You may declare the name of the variable assigned to the list as global, like this:

def get_list():
    global destination_list
    destination_list = []
    destination_list.extend(('1','2','3'))
    return destination_list

get_list()
print(destination_list)
1
L=[1, 2, 3]

def rl(l): 
    return l

[*ll] = rl(L) # ll is in a list

ll
# >>> [1, 2, 3]

*t, = rl(L)   # ll is in a tuple

t
# >>> [1, 2, 3]

If you want to return an item or list from a definition, you could define it before hand and use it as a variable during the initial writing of said definition. Unless it has to be defined within the definition. In this case you won't need to write in a return command at the end.

network = []

def splitNet(network):
    for line in open("/home/tom/Dropbox/CN/Python/CW2/network.txt","r").readlines():
        line = line.replace("\r\n", "")
        line = string.split(line, ',')
        line = map(int, line)
        network.append(line)

print network # Will print the list you've appended. But it is now a usable object. 

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Robert Thorne
Author

Robert Thorne

Robert Thorne covers electric vehicle innovations, autonomous driving systems, global mobility trends, and automotive engineering developments.