CODEWITHSUNDEEP
pythonarraysmatrix2doperation
aped
aped

Reputation: 193

Python: One-liner to perform an operation upon elements in a 2d array (list of lists)?

I have a list of lists, each containing a different number of strings. I'd like to (efficiently) convert these all to ints, but am feeling kind of dense, since I can't get it to work out for the life of me. I've been trying: newVals = [int(x) for x in [row for rows in values]]

Where 'values' is the list of lists. It keeps saying that x is a list and can therefore not be the argument if int(). Obviously I'm doing something stupid here, what is it? Is there an accepted idiom for this sort of thing?

Upvotes: 15

Views: 29161

Answers (6)

Aleksey Vlasenko
Aleksey Vlasenko

Reputation: 1098

In order to map list with any number of dimensions you could use numpy.apply_over_axes

import numpy as np
np.apply_over_axes(lambda x,_:x*2, np.array([[1,2,3],[5,2,1]]),[0])
--------------------
array([[ 2,  4,  6],
       [10,  4,  2]])

Unfortunately that doesn't work if you also need to change variable type. Didn't find any library solution for this, so here is the code to do that:

def map_multi_dimensional_list(l, transform):
    if type(l) == list and len(l) > 0:
        if type(l[0]) != list:
            return [transform(v) for v in l]
        else:
            return [map_multi_dimensional_list(v, transform) for v in l]
    else:
        return []
            
map_multi_dimensional_list([[[1,2,3],[5,2,1]],[[10,20,30],[50,20,10]]], lambda x:x*2)
------------------
[[[2, 4, 6], [10, 4, 2]], [[20, 40, 60], [100, 40, 20]]]

Upvotes: 0

Muhammad Younus
Muhammad Younus

Reputation: 1895

Another workaround

a = [[1, 2, 3], [7, 8, 6]]
list(map(lambda i: list(map(lambda j: j - 1, i)), a))
[[0, 1, 2], [6, 7, 5]] #output

Upvotes: 3

Rusty Rob
Rusty Rob

Reputation: 17213

an ugly way is to use evalf:

>>> eval(str(a).replace("'",""))
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

if you don't mind all your numbers in one array you could go:

>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> map(int,sum(a,[]))
[1, 2, 3, 4, 5, 6, 7, 8, 9]

Upvotes: 0

Artsiom Rudzenka
Artsiom Rudzenka

Reputation: 29131

You simply use incorrect order and parenthesis - should be:

inputVals = [['1','2','3'], ['3','3','2','2']]
[int(x) for row in inputVals for x in row]

Or if you need list of list at the output then:

map(lambda row: map(int, row), inputVals)

Upvotes: 1

John La Rooy
John La Rooy

Reputation: 304503

This leaves the ints nested

[map(int, x) for x in values]

If you want them flattened, that's not hard either

for Python3 map() returns an iterator. You could use 

[list(map(int, x)) for x in values]

but you may prefer to use the nested LC's in that case

[[int(y) for y in x] for x in values]

Upvotes: 23

JBernardo
JBernardo

Reputation: 33407

How about:

>>> a = [['1','2','3'],['4','5','6'],['7','8','9']]
>>> [[int(j) for j in i] for i in a]
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Upvotes: 11

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