Assigning values to two dimensional array from two one dimensional ones
Question:
Most probably somebody else already asked this but I couldn’t find it. The question is how can I assign values to a 2D array from two 1D arrays. For example:
import numpy as np
#a is the 2D array. b is the 1D array and should be assigned
#to second coordinate. In this exaple the first coordinate is 1.
a=np.zeros((3,2))
b=np.asarray([1,2,3])
c=np.ones(3)
a=np.vstack((c,b)).T
output:
[[ 1. 1.]
[ 1. 2.]
[ 1. 3.]]
I know the way I am doing it so naive, but I am sure there should be a one line way of doing this.
P.S. In real case that I am dealing with, this is a subarray of an array, and therefore I cannot set the first coordinate from the beginning to one. The whole array’s first coordinate are different, but after applying np.where
they become constant.
Answers:
How about 2 lines?
>>> c = np.ones((3, 2))
>>> c[:, 1] = [1, 2, 3]
And the proof it works:
>>> c
array([[ 1., 1.],
[ 1., 2.],
[ 1., 3.]])
Or, perhaps you want np.column_stack
:
>>> np.column_stack(([1.,1,1],[1,2,3]))
array([[ 1., 1.],
[ 1., 2.],
[ 1., 3.]])
First, there’s absolutely no reason to create the original zeros
array that you stick in a
, never reference, and replace with a completely different array with the same name.
Second, if you want to create an array the same shape and dtype as b
but with all ones, use ones_like
.
So:
b = np.array([1,2,3])
c = np.ones_like(b)
d = np.vstack((c, b).T
You could of course expand b
to a 3×1-array instead of a 3-array, in which case you can use hstack
instead of needing to vstack
then transpose… but I don’t think that’s any simpler:
b = np.array([1,2,3])
b = np.expand_dims(b, 1)
c = np.ones_like(b)
d = np.hstack((c, b))
If you insist on 1 line, use fancy indexing:
>>> a[:,0],a[:,1]=[1,1,1],[1,2,3]
Most probably somebody else already asked this but I couldn’t find it. The question is how can I assign values to a 2D array from two 1D arrays. For example:
import numpy as np
#a is the 2D array. b is the 1D array and should be assigned
#to second coordinate. In this exaple the first coordinate is 1.
a=np.zeros((3,2))
b=np.asarray([1,2,3])
c=np.ones(3)
a=np.vstack((c,b)).T
output:
[[ 1. 1.]
[ 1. 2.]
[ 1. 3.]]
I know the way I am doing it so naive, but I am sure there should be a one line way of doing this.
P.S. In real case that I am dealing with, this is a subarray of an array, and therefore I cannot set the first coordinate from the beginning to one. The whole array’s first coordinate are different, but after applying np.where
they become constant.
How about 2 lines?
>>> c = np.ones((3, 2))
>>> c[:, 1] = [1, 2, 3]
And the proof it works:
>>> c
array([[ 1., 1.],
[ 1., 2.],
[ 1., 3.]])
Or, perhaps you want np.column_stack
:
>>> np.column_stack(([1.,1,1],[1,2,3]))
array([[ 1., 1.],
[ 1., 2.],
[ 1., 3.]])
First, there’s absolutely no reason to create the original zeros
array that you stick in a
, never reference, and replace with a completely different array with the same name.
Second, if you want to create an array the same shape and dtype as b
but with all ones, use ones_like
.
So:
b = np.array([1,2,3])
c = np.ones_like(b)
d = np.vstack((c, b).T
You could of course expand b
to a 3×1-array instead of a 3-array, in which case you can use hstack
instead of needing to vstack
then transpose… but I don’t think that’s any simpler:
b = np.array([1,2,3])
b = np.expand_dims(b, 1)
c = np.ones_like(b)
d = np.hstack((c, b))
If you insist on 1 line, use fancy indexing:
>>> a[:,0],a[:,1]=[1,1,1],[1,2,3]