如何跨函数传递变量？

Question

I want to pass values (as variables) between different functions.

For example, I assign values to a list in one function, then I want to use that list in another function:

list = []

def defineAList():
    list = ['1','2','3']
    print "For checking purposes: in defineAList, list is",list
    return list

def useTheList(list):
    print "For checking purposes: in useTheList, list is",list

def main():
    defineAList()
    useTheList(list)

main()

I would expect this to do as follows:

1. Initialize 'list' as an empty list; call main (this, at least, I know I've got right...)
2. Within defineAList(), assign certain values into the list; then pass the new list back into main()
3. Within main(), call useTheList(list)
4. Since 'list' is included in the parameters of the useTheList function, I would expect that useTheList would now use the list as defined by defineAList(), NOT the empty list defined before calling main.

However, my output is:

For checking purposes: in defineAList, list is ['1', '2', '3']
For checking purposes: in useTheList, list is []

"return" does not do what I expected. What does it actually do? How do I take the list from defineAList() and use it within useTheList()?

I don't want to use global variables.

Answer 1

This is what is actually happening:

global_list = []

def defineAList():
    local_list = ['1','2','3']
    print "For checking purposes: in defineAList, list is", local_list 
    return local_list 

def useTheList(passed_list):
    print "For checking purposes: in useTheList, list is", passed_list

def main():
    # returned list is ignored
    returned_list = defineAList()   

    # passed_list inside useTheList is set to global_list
    useTheList(global_list) 

main()

This is what you want:

def defineAList():
    local_list = ['1','2','3']
    print "For checking purposes: in defineAList, list is", local_list 
    return local_list 

def useTheList(passed_list):
    print "For checking purposes: in useTheList, list is", passed_list

def main():
    # returned list is ignored
    returned_list = defineAList()   

    # passed_list inside useTheList is set to what is returned from defineAList
    useTheList(returned_list) 

main()

You can even skip the temporary returned_list and pass the returned value directly to useTheList :

def main():
    # passed_list inside useTheList is set to what is returned from defineAList
    useTheList(defineAList()) 

Answer 2

You're just missing one critical step. You have to explicitly pass the return value in to the second function.

def main():
    l = defineAList()
    useTheList(l)

Alternatively:

def main():
    useTheList(defineAList())

Or (though you shouldn't do this! It might seem nice at first, but globals just cause you grief in the long run.):

l = []

def defineAList():
    global l
    l.extend(['1','2','3'])

def main():
    global l
    defineAList()
    useTheList(l)

The function returns a value, but it doesn't create the symbol in any sort of global namespace as your code assumes. You have to actually capture the return value in the calling scope and then use it for subsequent operations.

Answer 3

return returns a value. It doesn't matter what name you gave to that value. Returning it just "passes it out" so that something else can use it. If you want to use it, you have to grab it from outside:

lst = defineAList()
useTheList(lst)

Returning list from inside defineAList doesn't mean "make it so the whole rest of the program can use that variable". It means "pass the value of this variable out and give the rest of the program one chance to grab it and use it". You need to assign that value to something outside the function in order to make use of it. Also, because of this, there is no need to define your list ahead of time with list = [] . Inside defineAList , you create a new list and return it; this list has no relationship to the one you defined with list = [] at the beginning.

Incidentally, I changed your variable name from list to lst . It's not a good idea to use list as a variable name because that is already the name of a built-in Python type. If you make your own variable called list , you won't be able to access the builtin one anymore.

Answer 4

你的回报是没有用的，如果你不分配它

list=defineAList()

Answer 5

Read up the concept of a name space. When you assign a variable in a function, you only assign it in the namespace of this function. But clearly you want to use it between all functions.

def defineAList():
    #list = ['1','2','3'] this creates a new list, named list in the current namespace.
    #same name, different list!

    list.extend['1', '2', '3', '4'] #this uses a method of the existing list, which is in an outer namespace
    print "For checking purposes: in defineAList, list is",list
    return list

Alternatively, you can pass it around:

def main():
    new_list = defineAList()
    useTheList(new_list)

Answer 6

passing variable from one function as argument to other functions can be done like this

define functions like this

def function1():
    global a
    a=input("Enter any number\t")
 
def function2(argument):
    print ("this is the entered number - ",argument)

call the functions like this

function1()
function2(a)

Answer 7

Try do the following:

def funa():
    name=("Bob")
    age=("42")
    return name, age

def funb():
    name, age=funa()
    print(name)
    print(age)

funb()

Happens that you gotta call it via funa() instead of age .

Answer 8

Just return it.

Example:

def initiate_excel_corvert():
 
    df = pd.ExcelFile('empty_book.xlsx').parse('Sheet')
    x=[]
    x.append(df['Numbers'])
    x.clear()
    x.extend(df['Numbers'])
    print(len(x))
    print('\n')
    len_of_column_0 =+1 + len(x)
    print("+ add-on ::")
    print(len_of_column_0)
    print('\n')
    
    return render_template('index.html'), insert_into_excel(len_of_column_0)
    
def insert_into_excel(len_of_column_0):
    print ("len_of_column_0 = ", len_of_column_0)
    print('\n')