Python从函数返回列表

Question

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?

Answer 1

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

Simply do this:

network = splitNet()
print network

Answer 2

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.

Answer 3

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)

Answer 4

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']
>>>

Answer 5

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]

Answer 6

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]

Answer 7

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)

Answer 8

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. 

Answer 9

L=[1,2,3]
def rl(l): return l

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

Out[45]: [1, 2, 3]

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

Out[47]: [1, 2, 3]