获取包含值的第一个子列表的索引的最快方法

Question

I have a list of lists in python of the form

A=[[1,2,3,4],
   [5,6,7,8],
   [9,10,11,12]]

I need to get a fast way to get the row index of an element in that structure.

method(2) = 0

method(8) = 1

method(12) = 2

and so on. As always, the fastest the method the better, as my actual list of lists is quite large.

Answer 1

In this state, the data structure (list of lists) is not quite convenient and efficient for the queries you want to make on it . Restructure it to have it in a form:

item -> list of sublist indexes  # assuming items can be present in multiple sublists

This way the lookups would be instant, by key - O(1) . Let's use defaultdict(list) :

>>> from collections import defaultdict
>>>
>>> d = defaultdict(list)
>>> for index, sublist in enumerate(A):
...     for item in sublist:
...         d[item].append(index)
... 
>>> d[2]
[0]
>>> d[8]
[1]
>>> d[12]
[2]

Answer 2

It is very simple using next() with a generator expression:

def method(lists, value):
    return next(i for i, v in enumerate(lists) if value in v)

The problem with that is that it will have an error if value does not occur. With a slightly longer function call, you can make a default of -1:

def method(lists, value):
    return next((i for i,v in enumerate(lists) if value in v), -1)

Answer 3

Here is another way using numpy

import numpy

A = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

my_array = numpy.array(A)

numpy.where(my_array==2) ## will return both the list and the index within the list
numpy.where(my_array==12)

## As a follow up if we want only the index we can always do :
numpy.where(my_array==12)[0][0] # will return 2 , index of list
numpy.where(my_array==12)[1][0] # will return 3 , index within list

Answer 4

find operation in list is linear. Following is simple code in python to find an element in list of lists.

A=[[1,2,3,4],
   [5,6,7,8],
   [9,10,11,12]]

def method(value):
    for idx, list in enumerate(A):
        if value in list:
            return idx
    return -1

print (method(12))