CODEWITHSUNDEEP
pythonlist
EasilyBaffled
EasilyBaffled

Reputation: 3882

Getting the different between two 2D lists

I have a 2D list of lists, I am doing some stuff to it and getting, as a result a slightly modified 2d lists of lists. I cannot track what changes are being made until after I get the new list back. I want to get a list of all the items that have been changed such that [[1,2,3], [4,5,6], [7,8,9]] becomes [[1,None,3], [4,None,6], [7,None, None]] and I would get a list [(0,1), (1,1), (2, 1), (2,2)] I know you can normally do list(set(a)-set(b)) but when I tried it I got TypeError: unhashable type: 'list' So what is the most efficient way of doing this?

Upvotes: 2

Views: 1723

Answers (3)

Francesco Montesano
Francesco Montesano

Reputation: 8668

If the lists have a regular structure, that is each of the sub-lists has the same length, and you don't mind using external packages, numpy can help.

import numpy as np
a = np.array([[1,2,3], [4,5,6], [7,8,9]])
b = np.array([[1,None,3], [4,None,6], [7,None, None]])

print(np.where(a!=b))

>>>(array([0, 1, 2, 2]), array([1, 1, 1, 2]))

Upvotes: 0

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 251096

Using zip, enumerate and a generator function:

def diff(lis1, lis2):
    for i, (x, y) in enumerate(zip(lis1, lis2)):
        for j, (x1, y1) in enumerate(zip(x, y)):
            if x1 != y1:
                yield i, j
...                 
>>> lis1 = [[1,2,3], [4,5,6], [7,8,9]]
>>> lis2 = [[1,None,3], [4,None,6], [7,None, None]]
>>> list(diff(lis1, lis2))
[(0, 1), (1, 1), (2, 1), (2, 2)]

Upvotes: 4

falsetru
falsetru

Reputation: 369314

Using list comprehension:

>>> a = [[1,2,3], [4,5,6], [7,8,9]]
>>> b = [[1,None,3], [4,None,6], [7,None, None]]
>>> [(i,j) for i, row in enumerate(a) for j, x in enumerate(row) if b[i][j] != x]
[(0, 1), (1, 1), (2, 1), (2, 2)]

Upvotes: 3

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