Find element in list n times Python-CodePudding

Given are

a = [1, 4, 2, 5]

b = [[[0, 1]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 6]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 4]],
     [[0, 5]], [[0, 6]], [[0, 4]], [[0, 2]], [[0, 2]], [[0, 4]], [[0, 4]], [[0, 5]], [[0, 5]], [[0, 5]], [[0, 1]],
     [[0, 5]], [[0, 1]], [[0, 1]]]

My goal is to iterate over the list a and to identify (print) those elements in b which have the element of a as element at index 1. The whole thing becomes difficult by the fact that this process may occur exactly three times for each element. After that, the next index in a is to be selected. In concrete terms, the whole thing should look like this at the end:

[[0, 1]]
[[0, 1]]
[[0, 1]]
[[0, 4]]
[[0, 4]]
[[0, 4]]
[[0, 2]]
[[0, 2]]
[[0, 2]]
[[0, 5]]
[[0, 5]]
[[0, 5]]

All elements in b beyond that are to be ignored, even and especially if they occur more than three times.

I have already tried various techniques ( random, while loop, etc.), racked my brain and searched this forum, but I am stuck.

It doesn't matter which elements are selected in b, the main thing is that there are three for each element in a.

CodePudding user response：

If you want to make it easier, you can simply follow the steps below:

a = [1, 4, 2, 5]

b = [[[0, 1]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 6]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 4]],
     [[0, 5]], [[0, 6]], [[0, 4]], [[0, 2]], [[0, 2]], [[0, 4]], [[0, 4]], [[0, 5]], [[0, 5]], [[0, 5]], [[0, 1]],
     [[0, 5]], [[0, 1]], [[0, 1]]]

for val_a in a:
    counter = 0
    for val_b in b:
       if val_a == val_b[0][1]:
           counter  = 1
           if counter < 4:
                print(val_b)

I hope I could help! :D

CodePudding user response：

This will count and filter your b input on a values using a Counter to count the values and appending to the filtered list only is the value count is less than 3.

from collections import Counter

a = [1, 4, 2, 5]

b = [
    [[0, 1]],
    [[0, 2]],
    [[0, 3]],
    [[0, 4]],
    [[0, 5]],
    [[0, 6]],
    [[0, 2]],
    [[0, 3]],
    [[0, 4]],
    [[0, 5]],
    [[0, 4]],
    [[0, 5]],
    [[0, 6]],
    [[0, 4]],
    [[0, 2]],
    [[0, 2]],
    [[0, 4]],
    [[0, 4]],
    [[0, 5]],
    [[0, 5]],
    [[0, 5]],
    [[0, 1]],
    [[0, 5]],
    [[0, 1]],
    [[0, 1]],
]

c = Counter()

b_filtered = []

b.sort(key=lambda x: x[0][-1])

for x in b:
    v = x[0][-1]
    if v in a and c[v] < 3:
        b_filtered.append(x)
    c[v]  = 1

b_filtered contains

[[[0, 1]],
 [[0, 1]],
 [[0, 1]],
 [[0, 2]],
 [[0, 2]],
 [[0, 2]],
 [[0, 4]],
 [[0, 4]],
 [[0, 4]],
 [[0, 5]],
 [[0, 5]],
 [[0, 5]]]

CodePudding user response：

You may follow these steps to achieve your desired output

a = [1, 4, 2, 5]
b = [[[0, 1]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 6]], [[0, 2]], [[0, 3]], [[0, 4]], [[0, 5]], [[0, 4]],
     [[0, 5]], [[0, 6]], [[0, 4]], [[0, 2]], [[0, 2]], [[0, 4]], [[0, 4]], [[0, 5]], [[0, 5]], [[0, 5]], [[0, 1]],
     [[0, 5]], [[0, 1]], [[0, 1]]]

result = []
count = {}
for i in a:
    for j in b:
        if j[0][1] == i:
            if i not in count:
                count[i] = 1
                result.append(j)
            elif count[i] < 3:
                count[i]  = 1
                result.append(j)

for i in result:
    print(i)