Compare each item of two lists in Python

Question

For example, suppose list_1 = [a,b,c] and list_2 = [m,n,o]. I want to compare each item from one list to the other, for example, create an output list such that

[a == m, a == n, a == o, b == m, b == n, b == o...]

For simplicity I'm using the == operation but this could also be a summation, e.g.

[a + m, a + n, a + o, b + m...]

I know how I could do this with two loops, but I'm wondering a lambda function with map() or list comprehensions could be used? I've searched online but only found one-to-one comparisons of list items, e.g. map(lambda x,y: x + y, list_1, list_2).

This doesn't work as I expect

Hamman Samuel
– Hamman Samuel

2015-07-03 12:12:59 +00:00
Commented Jul 3, 2015 at 12:12 — Hamman Samuel
– Hamman Samuel, Commented Jul 3, 2015 at 12:12

Roel Schroeven · Accepted Answer · 2015-07-03 12:03:21Z

11

itertools.product() can generate all combinations for you:

import itertools
list_1 = [1,5,4]
list_2 = [2,3,4]
# using list comprehensions
comparisons = [a == b for (a, b) in itertools.product(list_1, list_2)]
sums = [a + b for (a, b) in itertools.product(list_1, list_2)]
# using map and lambda
comparisons = map(lambda (a, b): a == b, itertools.product(list_1, list_2))
sums = map(lambda (a, b): a + b, itertools.product(list_1, list_2))

answered Jul 3, 2015 at 12:03

Roel Schroeven

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Hamman Samuel Over a year ago

Thanks to everyone for the the excellent answers, I chose this as the best because of it's conciseness and use of examples

Community · Accepted Answer · 2017-05-23 11:59:23Z

8

To get all the permutations of elements with a list comprehension:

[a == b for a in list_1 for b in list_2]

Functionality is the same as the nested for loops:

list_3 = []
for a in list_1:
    for b in list_2:
        list_3.append(a == b)  # Or a + b, etc.

Functional implementation is a bit more confusing:

list_3 = map(lambda x: map(lambda y: y == x, list_2), list_1)

This creates a list of lists, however, so you'd want to flatten it with any of the techniques described here

sum(list_3, [])

Or use itertools.product as suggested by @bereal.

edited May 23, 2017 at 11:59

CommunityBot

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answered Jul 3, 2015 at 11:52

jayelm

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2 Comments

bereal Over a year ago

itertools.product is nice, too.

jayelm Over a year ago

@bereal didn't know about that one, nice suggestion.

Anand S Kumar · Accepted Answer · 2015-07-03 12:08:49Z

4

For python 3.x - Yes you can do this with map function and itertools.product function and lambda expression -

>>> lst1 = [1,5,4]
>>> lst2 = [2,3,4]
>>> lst3 = list(map(lambda x: x[0] == x[1] , itertools.product(lst1,lst2)))
>>> lst3
[False, False, False, False, False, False, False, False, True]

For Python 2.x you can use same expression, just map function in Python 2.x returns a list so the list(...) is not required.

edited Jul 3, 2015 at 12:08

answered Jul 3, 2015 at 11:51

Anand S Kumar

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6 Comments

SuperBiasedMan Over a year ago

Out of curiosity, what does map return in Python 3 that you have to add the explicit casting?

jayelm Over a year ago

Question seems to be asking for a list with 9 elements (all possible combinations), not three?

bereal Over a year ago

The OP wants to compare each element of lst1 with each element of lst2, so should be 9.

Anand S Kumar Over a year ago

@JesseMu Sorry read the question wrongly, have fixed it to use itertools.product instead of zip

Anand S Kumar Over a year ago

@SuperBiasedMan zip returns an iterator in Python 3 , not a list.

|

René van den Berg · Accepted Answer · 2015-07-03 12:09:31Z

2

You want to apply an operator on pairs from the cartesian product of two lists. Let's say op defines the operation you want to apply to the two items, e.g:

op = lambda x, y: x == y

and have two lists

a = [1, 2, 3]
b = [2, 3, 4]

You can apply op on all pairs as a list comprehension as follows:

c = [op(x, y) for y in b for x in a]

To use the map function you first need to create the cartesian product using itertools.product. This effectively creates a double loop over the elements of both list just like the list comprehension. You will need to adjust the op definition slightly for this since it will only receive one argument consisting of the tuple (x, y). For example:

op2 = lambda t: t[0] == t[1]
d = map(op2, itertools.product(a, b))

edited Jul 3, 2015 at 12:09

answered Jul 3, 2015 at 12:03

René van den Berg

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1 Comment

Hamman Samuel Over a year ago

Thanks for pointing out the correct terminology, this is a product operation

Mrudula Balachander · Accepted Answer · 2015-07-06 06:14:23Z

1

You can do this:

list_1 = [1,2,3]
list_2 = [1,4,6]
is_it_equal_function = lambda x:x in list_1
is_it_equal_list = map(is_it_equal_function, list_2)

It will return:

[True, False, False]

edited Jul 6, 2015 at 6:14

answered Jul 3, 2015 at 16:10

Mrudula Balachander

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3 Comments

celticminstrel Over a year ago

I suggest making sure your code block is indented by exactly 4 spaces, as any more will cause the Python interpreter to complain about unexpected indent if the code is copy-pasted into it (which is the case right now).

Hardik Raval Over a year ago

To me, it's returning this results don't know why: [True, False, False, False, False, False, False, False, False]

Hardik Raval Over a year ago

import itertools lst1 = [ 1,2,3 ] lst2 = [ 1,4,6 ] lst3 = list(map(lambda x: x[0] == x[1] , itertools.product(lst1,lst2))) print(lst3)

Jiangtang Hu · Accepted Answer · 2018-09-19 14:32:06Z

-2

If we like the idea of vectorization, we could try numpy:

import numpy as np
list_1 = ['a','b','c']
list_2 = ['m','n','o']
l1 = np.array(list_1)
l2 = np.array(list_2)
l1 == l2

answered Sep 19, 2018 at 14:32

Jiangtang Hu

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