I have two arrays A = [a1, ..., an] and B = [b1, ..., bn].
I want to get new matrix C that is equal to
[[a1, b1],
[a2, b2],
...
[an, bn]]
How can I do it using numpy.concatenate?
3 Answers
How about this very simple but fastest solution ?
In [73]: a = np.array([0, 1, 2, 3, 4, 5])
In [74]: b = np.array([1, 2, 3, 4, 5, 6])
In [75]: ab = np.array([a, b])
In [76]: c = ab.T
In [77]: c
Out[77]:
array([[0, 1],
[1, 2],
[2, 3],
[3, 4],
[4, 5],
[5, 6]])
But, as Divakar pointed out, using np.column_stack gives the answer directly like:
In [85]: np.column_stack([a, b])
Out[85]:
array([[0, 1],
[1, 2],
[2, 3],
[3, 4],
[4, 5],
[5, 6]])
Efficiency (in descending order)
Interestingly, my simple solution turns out to be the fastest. (little faster than np.concatenate, twice as fast as np.column_stack and thrice as fast as np.vstack)
In [86]: %timeit np.array([a, b]).T
100000 loops, best of 3: 4.44 µs per loop
In [87]: %timeit np.concatenate((a[:,None], b[:,None]), axis=1)
100000 loops, best of 3: 5.6 µs per loop
In [88]: %timeit np.column_stack([a, b])
100000 loops, best of 3: 9.5 µs per loop
In [89]: %timeit np.vstack((a, b)).T
100000 loops, best of 3: 14.7 µs per loop
Comments
You can also use np.vstack then transpose the matrix after
import numpy as np
A = [1, 2, 3]
B = [4, 5, 6]
C = np.vstack((A, B)).T
3 Comments
Michael H.
Since it was asked to use
np.concatenate I'll just add C = np.concatenate((A, B)).reshape(2,3).T here.titipata
oh, that's right! thanks @Michael for updating the solution
Michael H.
Don't get me wrong, I think your solution is better ;-)
In [26]: A=np.arange(5)
In [27]: B=np.arange(10,15)
In [28]: np.concatenate((A[:,None], B[:,None]), axis=1)
Out[28]:
array([[ 0, 10],
[ 1, 11],
[ 2, 12],
[ 3, 13],
[ 4, 14]])
In [29]: _.tolist()
Out[29]: [[0, 10], [1, 11], [2, 12], [3, 13], [4, 14]]
np.column_stack, np.vstack, np.stack all do the same thing, just expanding the dimensions of the arrays in different ways.
np.stack((A,B),-1) expands the arrays like I did, with a newaxis indexing.
np.column_stack((A,B)) uses:
arr = array(arr, copy=False, subok=True, ndmin=2).T
np.vstack((A,B)).T uses:
concatenate([atleast_2d(_m) for _m in tup], 0)
As a matter of curiosity, note this vstack equivalent:
In [38]: np.concatenate((A[None],B[None]))
Out[38]:
array([[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14]])
np.column_stack.np.column_stackis little bit slower please?np.concatenateunder the hoods, so if you are comparing against it, its the function wrapper overhead that we are paying for there.