Replace set of multiple occurring elements in Numpy array with corresponding values

Question

Please consider the following code:

arr = np.array([0, 1, 3, 5, 5, 6, 7, 9, 8, 9, 3, 2, 4, 6])
mapping = np.array([0, 10, 20, 30, 40, 55, 66, 70, 80, 90])
res = np.zeros_like(arr)
min_val = 0
max_val = 10

for val in range(min_val, max_val):
    res[arr == val] = mapping[val]

print(res)

The Numpy array arr can have multiple occurrences of integers from interval [min_val, max_val). mapping array will have mappings for each integer and the size of the mapping array will be max_val. res array is the resultant array.

The for loop replaces multiple occurring elements in arr with the corresponding value in the mapping. For example, 0 value in the arr will be replaced with mapping[0] and 5 in arr with mapping[5].

The result of the above code is as below.

[ 0 10 30 55 55 66 70 90 80 90 30 20 40 66]

Question: How to do this operation using Numpy instead of for loop?

Answer is to use Numpy's fancy indexing

So, mapping[arr]?

Divakar
– Divakar

2019-06-28 10:45:43 +00:00
Commented Jun 28, 2019 at 10:45 — Divakar
– Divakar, Commented Jun 28, 2019 at 10:45
exactly that is doing the job

Albo
– Albo

2019-06-28 10:46:43 +00:00
Commented Jun 28, 2019 at 10:46 — Albo
– Albo, Commented Jun 28, 2019 at 10:46
@Divakar Absolutely, I didn't know that. Thanks.

mrtpk
– mrtpk

2019-06-28 10:55:56 +00:00
Commented Jun 28, 2019 at 10:55 — mrtpk
– mrtpk, Commented Jun 28, 2019 at 10:55

Albo · Accepted Answer · 2019-06-28 12:10:00Z

1

You can simply use arr as an indexing array for mapping:

mapping[arr]

the output is

[ 0 10 30 55 55 66 70 90 80 90 30 20 40 66]

You can read on the official SciPy documentation about the indexing arrays. Example from the documentation:

>>> x = np.arange(10,1,-1)
>>> x
array([10,  9,  8,  7,  6,  5,  4,  3,  2])
>>> x[np.array([3, 3, 1, 8])]
array([7, 7, 9, 2])

The values of the indexing array are used as an index for the source array.

This is also possible with multidimensional arrays:


>>> x = array([[ 0,  1,  2],
...            [ 3,  4,  5],
...            [ 6,  7,  8],
...            [ 9, 10, 11]])
>>> rows = np.array([[0, 0],
...                  [3, 3]], dtype=np.intp)
>>> columns = np.array([[0, 2],
...                     [0, 2]], dtype=np.intp)
>>> x[rows, columns]
array([[ 0,  2],
       [ 9, 11]])

edited Jun 28, 2019 at 12:10

answered Jun 28, 2019 at 10:45

Albo

1,66414 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

mrtpk Over a year ago

Thanks @Albo, I didn't know that.

Jack Bashford · Accepted Answer · 2019-06-28 10:52:48Z

1

Just use mapping[arr] to access the correct element in the new Numpy array:

>>> arr = np.array([0, 1, 3, 5, 5, 6, 7, 9, 8, 9, 3, 2, 4, 6])
>>> mapping = np.array([0, 10, 20, 30, 40, 55, 66, 70, 80, 90])
>>> print(mapping[arr])
array([ 0, 10, 30, 55, 55, 66, 70, 90, 80, 90, 30, 20, 40, 66])

If you want it as a list:

>>> print(list(mapping[arr]))
[0, 10, 30, 55, 55, 66, 70, 90, 80, 90, 30, 20, 40, 66]

answered Jun 28, 2019 at 10:52

Jack Bashford

44.3k11 gold badges56 silver badges84 bronze badges

Collectives™ on Stack Overflow

Replace set of multiple occurring elements in Numpy array with corresponding values

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related