A function that creates a lists with input from another function

Question

I have a function that will run a couple of times (in a recursive function, so with a 'while' condition). Everytime it runs, it will return a certain integer. So for example, the first time it runs it returns a 3, the second time a 5 and the third time a 9.

Now I need to save these returns in a list. So I thought to create a separate function that would take these values and store them. So the endstate I'm looking for is to have a list = [3,5,8].

B = [3,6,5,7,8,10]
def function_1(A):
    for i in range(len(A)/2):
         factor = A[2*i]
         list_of_diagonals(factor)
    return factor`

def list_of_diagonals(d):
    factor_list = []
    factor_list = factor_list.append(d)
    return factor_list`

Now I would expect that print function_1(B) would produces [3,5,8] but instead it just produces 8.

I think it has something to do with the fact that I define factor_list=[] right at the start of the function, but how could I work around that?

You are just returning factor from function_1 which holds the last iterated value of A[2*i] — user459872
– user459872, Commented Jan 17, 2019 at 6:35
I don't think "recursive" is the same as a "while" condition. You also don't have a while statement in your code anywhere. Just a for loop. — gilch
– gilch, Commented Jan 17, 2019 at 6:37
I'm not familiar with yield. But I return factor in the first function, just to make sure I could use it in the second. I return factor_list in the second, so I could use it later, right? — Mathbeginner
– Mathbeginner, Commented Jan 17, 2019 at 6:37
@SpghttCd The function has to be defined before it's called, not before it appears in another definition. Python just looks up the name in the globals dict at runtime. I don't see any problem with the order of the functions. — gilch
– gilch, Commented Jan 17, 2019 at 6:41

sahasrara62 · Accepted Answer · 2019-01-17 06:44:25Z

3

using generator you can do this this way , better , readable and more pythonic

B = [3,6,5,7,8,10]
def function_1(A):
    for i in range(len(A)//2):
         factor = A[2*i]
         yield factor

result = list(function_1(B))
# output [3, 5, 8]

answered Jan 17, 2019 at 6:44

sahasrara62

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1 Comment

gilch Over a year ago

@Mathbeginner that's the yield I was talking about.

Amit Nanaware · Accepted Answer · 2019-01-17 06:37:15Z

1

you are creating empty factor_list every time whenever list of diagonals call happens. Actually you don't need another function to store result in list , try below code:

B = [3,6,5,7,8,10]
def function_1(A):
    l1 = []
    for i in range(len(A)/2):
         factor = A[2*i]
         l1.append(factor)
    return l1
print function_1(B)

answered Jan 17, 2019 at 6:37

Amit Nanaware

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6 Comments

Vineeth Sai Over a year ago

have you checked before posting. This doesn't seem to work.

Amit Nanaware Over a year ago

@VineethSai what is not working here can you explain? Yes.. it is working I am getting desired output [3, 5, 8]

Vineeth Sai Over a year ago

Maybe it's the issue with range being a generator in python3.x

Vineeth Sai Over a year ago

TypeError: 'float' object cannot be interpreted as an integer

Amit Nanaware Over a year ago

Strange.. At my end its working fine.. I am using python ... Are you using same B list?

|

iGian · Accepted Answer · 2019-01-17 09:05:14Z

1

Other users already pointed the option to use a list directly in the first method, but I'd suggest one way to refactor your code to make it work.

def function_1(array):
    list_of_diagonals_ = [] # <-- initialize the recipient
    for i in range(len(array)//2): # <-- # // to return integer
         factor = array[2*i]
         list_of_diagonals_ = list_of_diagonals(factor, list_of_diagonals_) # <-- call the function and store to recipient
    return list_of_diagonals_ # <-- return the recipient

def list_of_diagonals(element, factor_list = None): # <-- need an argument to memorize past results
    if factor_list == None: factor_list = [] # <-- just in case
    factor_list.append(element)
    return factor_list

B = [3,6,5,7,8,10]
print (function_1(B))
#=> [3, 5, 8]

answered Jan 17, 2019 at 9:05

iGian

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1 Comment

Mathbeginner Over a year ago

NIce one, I appreciate it and it's definitely useful for my understanding.

Menglong Li · Accepted Answer · 2019-01-17 06:47:03Z

0

The following codes is under Python2.7

using islice in itertools can handle your case

from itertools import islice

B = [3, 6, 5, 7, 8, 10]


def function_1(A):
    return islice(A, 0, None, 2)

# [3, 5, 8]
print(list(function_1(B)))

answered Jan 17, 2019 at 6:47

Menglong Li

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