1

This is a continuation from here.

If I have an array a and an array b and a can be smaller or larger than b.Is there a way , having as base array the a, to replace all elements of b (element by element) with elements of a?

For example, this works because a has as many elements as b:

import numpy as np

a = np.array([1,2,3,4,5,6,7,8])
b = np.array([[1,2,3,4],[11,22,33,44]])
print(b[0].shape)

i= 0
for el in b:
    for idx,x in enumerate(el):
        el[idx] = a[i]
        i+= 1

print(b)
[[1 2 3 4]
 [5 6 7 8]]

Now, for example, a can be:

a = np.array([1,2,3,4,5])

and b:

b = np.array([11,22,33,44])

The result I want is b = [1,2,3,4,5] because all the 4 elements of b are replaced by the first 4 elements of a and I add one more element to b (the last of a) in order to be the same size as a(my reference array).

If I have :

a = np.array([1,2,3]) and b = np.array([6,7,8,9]) , then the result b array will be : b = [1,2,3] , so I deleted one element from b (to be the same size as a) and replaced the rest with the a elements.

Now, the whole problem is that:

I will have an array which has number of elements multiples of a size.

So , a = np.array([1,2,3,4,5])

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

b = np.array([[0,1,2,3], [4,5,6,7]])

I want to replace the elements from arr to b in multiples of a size (hence 5).

So , final b = [ [1,2,3,4,5], [6,7,8,9,10])

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6
  • Can you just write -> b = np.array(a) ? Commented Feb 16, 2017 at 8:00
  • bit confused with definition of arr and a here. Can you tell what is arr in your first example? Is it a=np.array([1]) and arr=np.array([1,2,3,4,5])? Commented Feb 16, 2017 at 8:25
  • @Rohanil:In my first example arr was just the elements of b( I changed the name to el now)if you mean that.Generally, I want to replace all b elements by arr values by multiples of a size (so by every 5 elements in my example) Commented Feb 16, 2017 at 8:32
  • When do you want a to add a row? Only when b.size // a.size > 1, b.size % a.size == 0? Commented Feb 16, 2017 at 9:36
  • And what happens when a.size % b.shape[1] /= 0? Commented Feb 16, 2017 at 9:39

2 Answers 2

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3

I think you want

b = arr.reshape(-1, a.size)
b
Out[291]: 
array([[ 1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10]])
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4 Comments

Thanks!It works great!(upvoted because I already accepted)! It gives the same results as the accepted answer (I am answering to your comment below)
Accepted answer has output b=[[1 2 3 4] [5 6 7 8]], which is a different size and value than my answer. Which is correct?
No, it's the same if you run the code.Just the poster have placed wrong output ( I will notify him)
I created another question.
2

Try list slicing assuming b is always either 2d or 1d

import numpy as np

arr = np.array([1,2,3,4,5,6,7,8,9,10])
a = np.array([1,2,3,4,5])
b = np.array([[1,2,3,4],[11,22,33,44]])
print(b[0].shape)

list_arr = arr.tolist()
slice_factor = a.size
j=0

temp = []

for j in range(0,len(list_arr),slice_factor):
    k = min(j + slice_factor,len(list_arr))
    temp.append(list_arr[j:k])

b = np.array(temp)

print(b)
[[1 2 3 4 5]
 [6 7 8 9 10]]
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4 Comments

Thanks!Great!And it works for every size, even if b.size/a.size <1 (upvoted)
That . . .doesn't give the output you asked for.
Please correct your output to be right.The result b is [[ 1, 2, 3, 4, 5], [ 6, 7, 8, 9, 10]] and not [[1 2 3 4] [5 6 7 8]]
I created another question.

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