在迭代中将列表与其他列表进行比较，并在 Python 中的嵌套列表中返回差异

Question

I have a list like this:

my_list = [['a', 'b', 'c', 'd'], ['a', 'e', 'f', 'd'], ['a', 'b', 'c', 'g'], ['d', 'e', 'f', 'd']]

and I want to compare its items with another list:

main_list = ["a", "b", "f", "d"]

And I want to return the indexes that they differ. My code so far looks like this:

differences = []
my_list = [['a', 'b', 'c', 'd'], ['a', 'e', 'f', 'd'], ['a', 'b', 'c', 'g'], ['d', 'e', 'f', 'd']]
main_list = ["a", "b", "f", "d"]

for i in range(len(my_list)):
    for index, (first, second) in enumerate(zip(my_list[i], main_list), start=1):
        if first != second:
            differences.append(index)
print(differences)

With the above code I get this output:

[3, 2, 3, 4, 1, 2]

Which is exactly the indices the main list differs with the original list. However I would like to get this list as an output, which gives me a nested list of which each index is the indices the main list differs with my_list[0], then with my_list[1] and so on:

[[3], [2], [3, 4], [1, 2]]

I would appreciate some help on modifying the code to get the ideal output. Thanks!

Answer 1

my_list = [['a', 'b', 'c', 'd'], ['a', 'e', 'f', 'd'], ['a', 'b', 'c', 'g'], ['d', 'e', 'f', 'd']]
main_list = ["a", "b", "f", "d"]

out = []
for subl in my_list:
    out.append([i for i, (a, b) in enumerate(zip(subl, main_list), 1) if a != b])

print(out)

Prints:

[[3], [2], [3, 4], [1, 2]]

---

Or one liner:

out = [[i for i, (a, b) in enumerate(zip(subl, main_list), 1) if a != b] for subl in my_list]
print(out)