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Python Python Collections (2016, retired 2019) Tuples Combo

Benjamin Guyton
Benjamin Guyton
6,858 Points
Posted by Benjamin Guyton
Benjamin Guyton
6,858 Points

combo.py

Where am I going wrong? I'm not sure I understand how to return different values inside the same tuple while looping over it

combo.py 
# combo([1, 2, 3], 'abc')
# Output:
# [(1, 'a'), (2, 'b'), (3, 'c')]
def combo(my_list, my_string):
    # create a list that will have tuples appended to it
    tuples = []
    count = 0
    # create tuples containing the corresponding index to my_list and my_string
    # loop through both arguments of combo()
    for list_item in my_list: 
        for string_item in my_string:
            add_tuples[0] = (my_list[0], my_string[0])
            tuples.append(add_tuples)
            count += 1
    #return list of created tuples
    return tuples

2 Answers

Stuart McIntosh
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Stuart McIntosh
Python Web Development Techdegree Graduate 22,874 Points
Stuart McIntosh
.a{fill-rule:evenodd;}techdegree seal-36
Stuart McIntosh
Python Web Development Techdegree Graduate 22,874 Points

Hi there, you are very close, however, a couple of suggestions

You only need one for loop. You have a counter so you can just directly access the second iterable via the count for instance:

for list_item in my_list: 
       tuples.append((list_item, my_string(count))

You can also remove the count by using enumerate()

for count, list_item in enumerate(my_list):

Hopefully, this is helpful

Ryan Cross
Ryan Cross
5,742 Points
Ryan Cross
Ryan Cross
5,742 Points

looks like for every itteration of the first for block you will itterate all of the 2nd for block?

Goekalp Cicek
Goekalp Cicek
9,322 Points
Goekalp Cicek
Goekalp Cicek
9,322 Points 
def combo(list1,list2):
    d={}
    ind=0
    l=[]
    for i in list1:
        d[i]=list2[ind]
        ind+=1
    for i in d.items():
        l.append(i)
    return l