How to make a copy of a 2D array in Python?

Question

X is a 2D array. I want to have a new variable Y that which has the same value as the array X. Moreover, any further manipulations with Y should not influence the value of the X.

It seems to me so natural to use y = x. But it does not work with arrays. If I do it this way and then changes y, the x will be changed too. I found out that the problem can be solved like that: y = x[:]

But it does not work with 2D array. For example:

x = [[1,2],[3,4]]
y = x[:]
y[0][0]= 1000
print x

returns [ [1000, 2], [3, 4] ]. It also does not help if I replace y=x[:] by y = x[:][:].

Does anybody know what is a proper and simple way to do it?

score 130 · Answer 1 · answered Jun 30 '11 at 09:59

130

Using deepcopy() or copy() is a good solution. For a simple 2D-array case

y = [row[:] for row in x]

answered Jun 30 '11 at 09:59

Ryan Ye

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+1 for the list comprehension version. It works fine for a 2d array and is way faster that `deepcopy(x)`. – Eugene Yarmash Dec 26 '13 at 21:02

score 87 · Accepted Answer · answered Jun 30 '11 at 09:46

87

Try this:

from copy import copy, deepcopy
y = deepcopy(x)

I'm not sure, maybe copy() is sufficient.

answered Jun 30 '11 at 09:46

Constantinius

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for 2D array of floats copy do not work, but deepcopy does work, thanks! – Evalds Urtans Feb 28 '17 at 13:08
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copy() is not sufficient, it is a shallow copy so the result will be a copy of references to the original row lists. – Richard Whitehead Jan 13 '22 at 16:33

Анатолий Панин · Answer 3 · 2017-12-19T18:23:02.647

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For 2D arrays it's possible use map function:

old_array = [[2, 3], [4, 5]]
# python2.*
new_array = map(list, old_array)
# python3.*
new_array = list(map(list, old_array))

edited Dec 19 '17 at 18:23

answered Jan 20 '14 at 18:30

Анатолий Панин

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I think you need to "cast" that, like new_array = list(map(list, old_array)) – Fran Marzoa Jun 06 '17 at 16:26
distance = list(map(lambda i: list(map(lambda j: j, i)), graph)). I've also seen this.... – Linus Fernandes Mar 12 '23 at 08:04

Artsiom Rudzenka · Answer 4 · 2011-06-30T10:05:37.933

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In your case(since you use list of lists) you have to use deepcopy, because 'The difference between shallow and deep copying is only relevant for compound objects (objects that contain other objects, like lists or class instances): A shallow copy constructs a new compound object and then (to the extent possible) inserts references into it to the objects found in the original. A deep copy constructs a new compound object and then, recursively, inserts copies into it of the objects found in the original.'

Note that sample below is simply intended to show you an example(don't beat me to much) how deepcopy could be implemented for 1d and 2d arrays:

arr = [[1,2],[3,4]]

deepcopy1d2d = lambda lVals: [x if not isinstance(x, list) else x[:] for x in lVals]

dst = deepcopy1d2d(arr)

dst[1][1]=150
print dst
print arr

edited Jun 30 '11 at 10:05

answered Jun 30 '11 at 09:59

Artsiom Rudzenka

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"you have to use deepcopy" Not really. There's always a more complicated, but (maybe) faster way. See Ryan Ye's answer. Yours is fine, but this statement is not perfect. – palsch Jan 02 '16 at 20:02
@palsch as i can see there is no difference between our answers. Mine deepcopy1d2d doing the same and even more)))) – Artsiom Rudzenka Jan 04 '16 at 04:42

score 0 · Answer 5 · answered Aug 01 '15 at 08:27

I think np.tile also might be useful

>>> a = np.array([0, 1, 2])
>>> np.tile(a, 2)
array([0, 1, 2, 0, 1, 2])
>>> np.tile(a, (2, 2))
array([[0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2]])
>>> np.tile(a, (2, 1, 2))
array([[[0, 1, 2, 0, 1, 2]],
       [[0, 1, 2, 0, 1, 2]]])

How to make a copy of a 2D array in Python?

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