获取列表中给定元素的列表号

Question

Let's say we have N elements from 0 to N - 1 ( N is known) scattered into k lists as follows:

X = [[0, 1, 6], [2, 4], [3, 5]] for N = 7, k = 3

Given the number how can I determine the number of the group it is in? For example 1 is in first group and 5 is int the third group.

I can nest loops and check it manually (for each given number check each group where it can be) but how do I do this with more clean and functional-like style in Python? It'd more useful especially when I have list of numbers [1, 3, 5] and want to transform it into list of groups of each of elements ie. [1, 3, 3]

EDIT:

I was asked to present the way I am doing it now so here it comes:

results = []
numbers = [1, 3, 5]
for number in numbers:
    for i in range(len(X)):
        if number in X[i]:
            results.append(i)

Answer 1

Probably the easiest way to do this without re-computing for every new number you're looking for is to make a dictionary:

X = [[0, 1, 6], [2, 4], [3, 5]]
X_map = {element: group for group, elts in enumerate(X) for element in elts}
# {0: 0, 1: 0, 2: 1, 3: 2, 4: 1, 5: 2, 6: 0}

Now for any given number, just look it up:

X_map.get(5, None)  # 2
X_map.get(7, None)  # None

Answer 2

 [(index, row.index(your_number)) for index, row in enumerate(X) if your_number in row]

output:

your_number = 1
[(0, 1)]

means: array number 0, index 1