Python way of doing difference of lists containing lists

Question

Consider two lists A and B. I know that list(set(A) - set(B)) will give the difference between A and B. What about the situation whereby elements in both A and B are lists. i.e. A and B are list of list? For e.g.

A = [[1,2], [3,4], [5,6]]
B = [[3,4], [7,8]]

I wish to return the difference A - B as a list of list i.e. [[1,2],[5,6]]

list(set(A) - set(B))
TypeError: unhashable type: 'list'

Possible duplicate of [Get difference between two lists](http://stackoverflow.com/questions/3462143/get-difference-between-two-lists) — GoatsWearHats, Aug 29 '16 at 08:24

score 1 · Answer 1 · answered Aug 29 '16 at 08:24

1

Here's a one-liner you could use:

diff = [x for x in A if x not in B]

Or if you want to use filter:

diff = list(filter(lambda x: x not in B, A))

answered Aug 29 '16 at 08:24

gowrath

2,856
2
16
31

I got this. `TypeError: 'list' object is not callable`. A and B are both list of lists – DougKruger Aug 29 '16 at 08:44
@DougKruger You probably have a variable actually named "list" somewhere in your program which is overriding the builtin list function! Look through your code and if you have a variable by the name "list", change that name to something else. – gowrath Aug 29 '16 at 08:52
@DougKruger Or use the first version. But note there is probably a bug in your program if you have called something in your program "list". – gowrath Aug 29 '16 at 08:55

score 1 · Answer 2 · answered Aug 29 '16 at 08:25

1

>>> [i for i in A if i not in B]
[[1, 2], [5, 6]]

answered Aug 29 '16 at 08:25

Ahsanul Haque

9,865
4
35
51

`TypeError: 'list' object is not callable` A and B are lists of list. – DougKruger Aug 29 '16 at 08:45
Can't reproduce the error in both `python2` and `python3`, works fine . May be the error comes from some other portion of your program. – Ahsanul Haque Aug 29 '16 at 08:55

Israel Unterman · Accepted Answer · 2016-08-29T09:05:10.483

1

The idea is to convert the list of lists to lists of tuples,
which are hashable and are, thus, candidates of making sets themselves:

In [192]: C = set(map(tuple, A)) - set(map(tuple, B))

In [193]: C
Out[193]: {(1, 2), (5, 6)}

And one more touch:

In [196]: [*map(list, C)]
Out[196]: [[1, 2], [5, 6]]

ADDED

In python 2.7 the final touch is simpler:

map(list, C)

edited Aug 29 '16 at 09:05

answered Aug 29 '16 at 08:26

Israel Unterman

12,444
3
26
34

I think, you should reformat your answer in a way that OP needs. A `list of lists` rather than `a set of tuples` may be. – Ahsanul Haque Aug 29 '16 at 08:27
@IsraelUnterman It appears python 2.7 does not support the starred expression `[*map(list, C)]` – DougKruger Aug 29 '16 at 08:36
See my addition for python 2.7 – Israel Unterman Aug 29 '16 at 09:05

score 1 · Answer 4 · answered Aug 29 '16 at 08:52

1

A = [[1, 2], [3, 4], [5, 6]]
B = [[3, 4], [7, 8]]

print[x for x in A if x not in B]

answered Aug 29 '16 at 08:52

BPL

9,532
7
47
105

Python way of doing difference of lists containing lists

4 Answers4