merge two numpy.array without a loop

Question

I have a two numpy.arrays, I want to get following result efficiently

1.add the element's of b to a's sub-array

    a=numpy.array([(1,2,3),(1,2,3)])
    b=numpy.array([0,0])
->
    c=[(0,1,2,3),(0,1,2,3)]

code in a loop

a=numpy.array([(1,2,3),(1,2,3)])
b=numpy.array([(0,0)])
c=numpy.zeros(2 , 4)
idx=0
for x in a:
   c[idx]=(a[idx][0],a[idx][1],a[idx][2], b[idx])
   idx = idx+1

and
2. Get an 2-D array with dimension(a.dim*b.dim, 2) from two 1-D arrays

    a=numpy.array([(1,2)])
    b=numpy.array([(3,4)])
->
    c=[(1,3),(1,4),(2,3),(2,4)]

code in a loop

a=numpy.array([(1,2)])
b=numpy.array([(3,4)])
c=numpy.zeros(a.size*b.size , 2)
idx=0
for x in a:
    for y in b:
        c[idx]=(x,y)
        idx = idx+1

Can you show some context? This is possible, and not that hard, but it may not be a good idea. — user2357112
– user2357112, Commented Aug 28, 2013 at 1:53
Okay, now it looks like you're asking two different questions. Can you give a precise definition of the result you're trying to achieve? — user2357112
– user2357112, Commented Aug 28, 2013 at 1:55
I want to these two method to get new array. And I don't want to make this in a loop . I this use loop to make is not efficient — Samuel
– Samuel, Commented Aug 28, 2013 at 1:56
The two things you're trying to achieve look like entirely different things. If you think they're the same, it's not at all clear what you want to do. — user2357112
– user2357112, Commented Aug 28, 2013 at 1:57
Could you write a loop that achieves the effect that you're looking for, so it's clearer what you want? — user2357112
– user2357112, Commented Aug 28, 2013 at 1:58

Community · Accepted Answer · 2017-05-23 12:13:11Z

3

For the first problem, you can define b differently and use numpy.hstack:

a = numpy.array([(1,2,3),(1,2,3)])
b = numpy.array([[0],[0]])
numpy.hstack((b,a))

Regarding the second problem, I would probably use sza's answer and create the numpy array from that result, if necessary. That technique was suggested in an old Stack Overflow question.

edited May 23, 2017 at 12:13

CommunityBot

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answered Aug 28, 2013 at 2:34

Paulo Almeida

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6 Comments

Samuel Over a year ago

Because I have a big array a, and actually b has the same element. Can i don't alloc b the same size of a

Paulo Almeida Over a year ago

I'm not sure I understand what you mean, but you can probably use numpy.empty_like, zeros_like or ones_like. Look into array.fill too.

Samuel Over a year ago

Yes , That's what i mean. But I don't want to use ones_like, This will alloc a big memory. And all the elements of b are the same, So it's a little memory-wasting

Paulo Almeida Over a year ago

I see what you mean now. sza's answer is better for that problem too, then, just use a fixed y.

Samuel Over a year ago

But he also use the loop. I'm afraid it's not efficient. So I should make trade-off on it right?

|

zs2020 · Accepted Answer · 2013-08-28 02:34:31Z

2

For the first one, you can do

>>> a=numpy.array([(1,2,3),(1,2,3)])
>>> b=numpy.array([0,0])
>>> [tuple(numpy.insert(x, 0, y)) for (x,y) in zip(a,b)]
[(0, 1, 2, 3), (0, 1, 2, 3)]

For the 2nd one, you can get the 2-D array like this

>>> a=numpy.array([(1,2)])
>>> b=numpy.array([(3,4)])
>>> import itertools
>>> c = list(itertools.product(a.tolist()[0], b.tolist()[0]))
[(1, 3), (1, 4), (2, 3), (2, 4)]

edited Aug 28, 2013 at 2:34

answered Aug 28, 2013 at 2:21

zs2020

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1 Comment

Samuel Over a year ago

This is list. can I also do this in numpy.array

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