Python – Comparing two lists to make a third list

Question

I have two lists, and I have a specific logic task that I’m not sure how to implement. Here are the lists:

list_1 = [826, 829, 830, 834, 837, 838, 842, 845, 846, 850]

list_2 = [826, 834, 842, 850]

list_2 is a subset of list_1.

The task I need is to make a list_3 using only the values in list_1 that precede the values in list_2. So it would be the list_1 items preceding 834, 842, and 850 (skipping 826 since nothing precedes it):

# Thus, we would end up with: 
list_3 = [830, 838, 846]

This is a bit of a specific logic question if anyone has the chops to help me out.
Much appreciated!

Asked By: Samsonite Manly

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Source

Answer 1

You could start with for item in list_2 to work with every elements form list_2 separately, and use list_1.index(item) to get its position on other list, and put on list_3 item list_1[position-1] if position-1 >= 0 (or if position > 0)

list_1 = [826, 829, 830, 834, 837, 838, 842, 845, 846, 850]

list_2 = [826, 834, 842, 850]

list_3 = []

for item in list_2:
    position = list_1.index(item)
    if position > 0:  # if position-1 >= 0:
        predecessor = list_1[position-1]
        list_3.append(predecessor)
        print(predecessor)

Answered By: furas

Answer 2

lists in python can be converted to sets as well, so you don’t have to worry about n^2 duration.

list_3 = [list_1[i - 1] for i in range(1, len(list_1)) if list_1[i] in list_2 and list_1[i - 1] not in set(list_2)]

or

list_3 = [list_1[i - 1] for i in range(1, len(list_1)) if list_1[i] in set(list_2)]

if you dont care that the previous item could be in list_2 also.

Answered By: yitzchak24

Answer 3

if they’re in the same order, you can solve it in one pass using two pointers, like

j = 0
list_3 = []
for i in range(len(list_1)):
    if list_1[i]==list_2[j]:
        if i>0:
            list_3.append(list_1[i-1])
        j += 1

Answered By: Michael Pruglo

Answer 4

If there is no repeted elements in list_1:

list_1 = [826, 829, 830, 834, 837, 838, 842, 845, 846, 850]

list_2 = [826, 834, 842, 850]

list_idx = [list_1.index(x) - 1 for x in list_2]
list_idx = list(filter(lambda x: x >= 0, list_idx))

list_3 = [list_1[x] for x in list_idx]

If list_1 can have repeted elements:

list_idx = [
    idx - 1 for idx, value in enumerate(list_1) for item2 in list_2 if value == item2
]
list_idx = list(filter(lambda x: x >= 0, list_idx))

list_3 = [list_1[x] for x in list_idx]

Answered By: itogaston

Answer 5

Assuming that you have unique items in list_1, you can use a dictionary to map the previous values. The complexity is O(n):

d = dict(zip(list_1[1:], list_1))
list_3 = [d[x] for x in list_2 if x in d]

Output: [830, 838, 846]

Intermediate d:

{829: 826, 830: 829, 834: 830, 837: 834, 838: 837, 842: 838, 845: 842, 846: 845, 850: 846}

NB. If you can have duplicates in list_1, then it is possible to generalize by using a dictionary of lists, but the expected behavior should be defined.

Answered By: mozway

Python – Comparing two lists to make a third list

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