如何在列表列表中获取相同值的所有出现的索引列表？

Question

Having a list of lists,

mylist = [[1, 3, 4], [3, 6, 7], [8, 0, -1, 3]]

I need a method to get list containing indexes of all elements of some certain value, '3' for example. So for value of '3' such list of indexes should be

[[0, 1], [1, 0], [2, 3]]

Thanks in advance.

Update : we can have several instances of sought-for value ('3' in our case) in same sublist, like

my_list = [[1, 3, 4], [3, 6, 7], [8, 0, -1, 3, 3]]

So desired output will be [[0, 1], [1, 0], [2, 3], [2, 4]] .

I suppose that my solution is slightly naive, here it is:

my_list = [[1, 3, 4], [3, 6, 7], [8, 0, -1, 3, 3]]
value = 3
list_of_indexes = []

for i in range(len(my_list)):
        for j in range(len(my_list[i])):
            if my_list[i][j] == value:
                index = i, j
                list_of_indexes.append(index)

print list_of_indexes

>>[(0, 1), (1, 0), (2, 3), (2, 4)]]

It will be great to see more compact solution

Answer 1

Assuming the value appears once in each sublist, you could use the following function, which makes use of the built in enumerate function and a list comprehension:

my_list = [[1, 3, 4], [3, 6, 7], [8, 0, -1, 3]]

def func(my_list, x):
    return [[idx, sublist.index(x)] for idx, sublist in enumerate(my_list)]

answer = func(my_list, 3)
print(answer)

Output

[[0, 1], [1, 0], [2, 3]]

Answer 2

Easy way,

def get_val(mylist, val):
    for ix, item in enumerate(mylist):
        yield [ix, item.index(val)]

mylist = [[1, 3, 4], [3, 6, 7], [8, 0, -1, 3]]

val = list(get_val(mylist, 3))
print(val)

Output:

[[0, 1], [1, 0], [2, 3]]