Python - Comparing two lists to make a third list-CodePudding

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!

CodePudding user response：

lists in python can be searched as 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, list_1) if list_1[i] in list_2 and list_1[i - 1] not in list_2]

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

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

CodePudding user response：

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

CodePudding user response：

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)

CodePudding user response：

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]

CodePudding user response：

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.