How to return values without breaking the loop?

Question

I want to know how to return values without breaking a loop in Python.

Here is an example:

def myfunction():
    list = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
    print(list)
    total = 0
    for i in list:
        if total < 6:
            return i  #get first element then it breaks
            total += 1
        else:
            break

myfunction()

return will only get the first answer then leave the loop, I don’t want that, I want to return multiple elements till the end of that loop.

How can resolve this, is there any solution?

Asked By: Hiroyuki Nuri

||

Source

Answer 1

You can create a generator for that, so you could yield values from your generator (your function would become a generator after using the yield statement).

See the topics below to get a better idea of how to work with it:

An example of using a generator:

def simple_generator(n):
    i = 0
    while i < n:
        yield i
        i += 1

my_simple_gen = simple_generator(3) # Create a generator
first_return = next(my_simple_gen) # 0
second_return = next(my_simple_gen) # 1

Also you could create a list before the loop starts and append items to that list, then return that list, so this list could be treated as list of results "returned" inside the loop.

Example of using list to return values:

def get_results_list(n):
    results = []
    i = 0
    while i < n:
        results.append(i)
        i += 1
    return results


first_return, second_return, third_return = get_results_list(3)

NOTE: In the approach with list you have to know how many values your function would return in results list to avoid too many values to unpack error

Answered By: Andriy Ivaneyko

Answer 2

Using a generator is a probable way:

def myfunction():
    l = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
    total = 0
    for i in l:
        if total < 6:
            yield i  #yields an item and saves function state
            total += 1
        else:
            break

g = myfunction()

Now you can access all elements returned with yield i by calling next() on it:

>>> val = next(g)
>>> print(v)
a
>>> v = next(g)
>>> print(v)
b

Or, in a for loop by doing:

>>> for returned_val in myfunction():
...    print(returned_val)
a
b
c
d
e
f

Answered By: Dimitris Fasarakis Hilliard

Answer 3

What you want is most easily expressed with list slicing:

>>> l = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
>>> l[:6]
# ['a', 'b', 'c', 'd', 'e', 'f']

Alternatively create another list which you will return at the end of the function.

def myfunction():
    l = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
    ret = []

    total = 0
    for i in l:
        if total < 6:
            total += 1
            ret.append(i)

    return ret

myfunction()

# ['a', 'b', 'c', 'd', 'e', 'f']

Answered By: jumbopap

Answer 4

A yield statement to create a generator is what you want.

What does the "yield" keyword do in Python?

Then use the next method to get each value returned from the loop.

var = my_func().next()

Answered By: BenJ

How to return values without breaking the loop?

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