Replacing an element in a list of array?

Question

I have a list like this:

x = [array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]),
     array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.])]

and then I declare another list:

y = [2, 8]

How can I replace an element in the first list if the value equals to 2 (or in index of y[0]) with 0? And do the same thing on the second list, which I should replace the value 8 with 0?

So the desired output should be like this:

x = [array([ 0.,  1.,  0.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]),
     array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  0.,  9., 10.])]

Ch3steR · Accepted Answer · 2020-04-01 09:33:20Z

Here's a vectorized solution. When both x and y are NumPy arrays

Make your x and y NumPy arrays. Then use this approach.

x[x==y[:,None]]=0

Complete example:

x=np.array([[ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.],
            [ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  0.,  9., 10.]])
y=np.array([2,8])

x[x==y[:,None]]=0

#array([[ 0.,  1.,  0.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.],
#       [ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  0.,  9., 10.]])

Timeit analysis:

In [50]: timeit x[x==y[:,None]]=0 #My solution
2.2 µs ± 34.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [53]: timeit [np.where(a==v, 0, a)  for v, a in zip(y, x)] #kederrac's solution
10.6 µs ± 309 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [54]: %%timeit
    ...: x = [np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]),
    ...:      np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.])]
    ...: y = [2, 8]
    ...: for i in range(len(x)):
    ...:   arr = x[i]
    ...:   arr[arr == y[i]] = 0
    ...:
6.61 µs ± 310 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) #Sam's solution

incarnadine · Accepted Answer · 2020-04-01 09:06:12Z

1

The following code will replace all values of the each array in x with the corresponding index of the values in y.

import numpy as np

x = [np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]),
     np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.])]

y = [2, 8]

for i in range(len(x)):
  arr = x[i]
  arr[arr == y[i]] = 0

For future note, you are using both numpy arrays and python lists, yet you are calling them both lists. Try not to do this as they are different things and it might get confusing when you get into more complicated projects

answered Apr 1, 2020 at 9:06

incarnadine

7009 silver badges22 bronze badges

1 Comment

Ch3steR Over a year ago

Nice answer. I added your timeit analysis to my answer.

kederrac · Accepted Answer · 2020-04-01 09:56:15Z

1

you can use a list comprehension with np.where and the built-in function zip:

x = [np.where(a==v, 0, a)  for v, a in zip(y, x)]
x

output:

[array([ 0.,  1.,  0.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]),
 array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  0.,  9., 10.])]

to be a bit faster since np.where will return a new array you could filter base on y and change the value without creating a new array:

for a, v in zip(x, y):
    x[x==v] = 0

edited Apr 1, 2020 at 9:56

answered Apr 1, 2020 at 9:02

kederrac

17.4k6 gold badges36 silver badges58 bronze badges

3 Comments

Ch3steR Over a year ago

Without casting this can be used. Nice answer.+1

Ch3steR Over a year ago

I timeit Sam's answer it faster than list comprehension may be the culprit is np.where. I'll time your answer using == instead of np.where.

kederrac Over a year ago

@Ch3steR sam is doing in-place replacements, np.where is returning a new array

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

Solution:

x = [[ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.],
     [ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9., 10.]]
     
y = [2, 8]

for i, array in enumerate(x):
    for j, number in enumerate(array):
        if number == y[i]:  # this might fail for floats
            array[j] = 0.0
    print(array)

Output:

[0.0, 1.0, 0.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0]
[0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 0.0, 9.0, 10.0]

Notes:

Note that comparing floats can become tricky. If you know you will only compare integers, best to cast them as such at least, or use some package/module to properly compare them (either by comparing their difference to a very small delta, or using decimal.Decimal field instead, for example)

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Apr 1, 2020 at 8:57

Victor

2,9191 gold badge16 silver badges20 bronze badges

Comments

Erik McKelvey · Accepted Answer · 2020-04-01 09:03:03Z

0

Assuming the lists are of the same size, which you can check like this:

if len(x) == len(y):
    print("Lists are the same size")

You can loop through and check for the values:

for i in range(len(x)):        # For each array in x
    for j in range(len(x[i])): # For each item in the array
        if x[i][j] == y[i]:    # If the item is equal to the value you are searching
            x[i][j] = 0

answered Apr 1, 2020 at 9:03

Erik McKelvey

1,6372 gold badges17 silver badges25 bronze badges

Collectives™ on Stack Overflow

Replacing an element in a list of array?

5 Answers 5

Comments

1 Comment

3 Comments

Solution:

Output:

Notes:

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

1 Comment

3 Comments

Solution:

Output:

Notes:

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related