I have one 2D array and one 1D array. I would like to zip them together.
import numpy as np
arr2D = [[5.88964708e-02, -2.38142395e-01, -4.95821417e-01, -7.07269274e-01],
[0.53363666, 0.1654723 , -0.16439857, -0.44880487]]
arr2D = np.asarray(arr2D)
arr1D = np.arange(7, 8.5+0.5, 0.5)
arr1D = np.asarray(arr1D)
res = np.array(list(zip(arr1D, arr2D)))
print(res)
which results in:
[[7.0 array([ 0.05889647, -0.2381424 , -0.49582142, -0.70726927])]
[7.5 array([ 0.53363666, 0.1654723 , -0.16439857, -0.44880487])]]
But I am trying to get:
[[(7.0, 0.05889647), (7.5, -0.2381424), (8.0, -0.49582142), (8.5, -0.70726927)]]
[[(7.0, 0.53363666), (7.5, 0.1654723),(8.0, -0.16439857), (8.5, -0.44880487)]]
How can I do this?
In [382]: arr2D = [[5.88964708e-02, -2.38142395e-01, -4.95821417e-01, -7.07269274e-01],
...: [0.53363666, 0.1654723 , -0.16439857, -0.44880487]]
...: arr2D = np.asarray(arr2D)
...: arr1D = np.arange(7, 8.5+0.5, 0.5) # already an array
In [384]: arr2D.shape
Out[384]: (2, 4)
In [385]: arr1D.shape
Out[385]: (4,)
zip
iterates on the first dimension of the arguments, and stops with the shortest:
In [387]: [[i,j[0:2]] for i,j in zip(arr1D, arr2D)]
Out[387]:
[[7.0, array([ 0.05889647, -0.2381424 ])],
[7.5, array([0.53363666, 0.1654723 ])]]
If we transpose the 2d, so it is now (4,2), we get a four element list:
In [389]: [[i,j] for i,j in zip(arr1D, arr2D.T)]
Out[389]:
[[7.0, array([0.05889647, 0.53363666])],
[7.5, array([-0.2381424, 0.1654723])],
[8.0, array([-0.49582142, -0.16439857])],
[8.5, array([-0.70726927, -0.44880487])]]
We could add another level of iteration to get the desired pairs:
In [390]: [[(i,k) for k in j] for i,j in zip(arr1D, arr2D.T)]
Out[390]:
[[(7.0, 0.0588964708), (7.0, 0.53363666)],
[(7.5, -0.238142395), (7.5, 0.1654723)],
[(8.0, -0.495821417), (8.0, -0.16439857)],
[(8.5, -0.707269274), (8.5, -0.44880487)]]
and with list transpose idiom:
In [391]: list(zip(*_))
Out[391]:
[((7.0, 0.0588964708), (7.5, -0.238142395), (8.0, -0.495821417), (8.5, -0.707269274)),
((7.0, 0.53363666), (7.5, 0.1654723), (8.0, -0.16439857), (8.5, -0.44880487))]
Or we can get that result directly by moving the zip
into an inner loop:
[[(i,k) for i,k in zip(arr1D, row)] for row in arr2D]
In other words, you are pairing the elements of arr1D
with the elements of each row of 2D, rather than with the whole row.
Since you already have arrays, one of the array solutions might be better, but I'm trying to clarify what is happening with zip
.
There are various ways of building a numpy array from these arrays. Since you want to repeat the arr1D
values:
This repeat
makes a (4,2) array that matchs arr2D
(tile
also works):
In [400]: arr1D[None,:].repeat(2,0)
Out[400]:
array([[7. , 7.5, 8. , 8.5],
[7. , 7.5, 8. , 8.5]])
In [401]: arr2D
Out[401]:
array([[ 0.05889647, -0.2381424 , -0.49582142, -0.70726927],
[ 0.53363666, 0.1654723 , -0.16439857, -0.44880487]])
which can then be joined on a new trailing axis:
In [402]: np.stack((_400, arr2D), axis=2)
Out[402]:
array([[[ 7. , 0.05889647],
[ 7.5 , -0.2381424 ],
[ 8. , -0.49582142],
[ 8.5 , -0.70726927]],
[[ 7. , 0.53363666],
[ 7.5 , 0.1654723 ],
[ 8. , -0.16439857],
[ 8.5 , -0.44880487]]])
Or a structured array with tuple-like display:
In [406]: arr = np.zeros((2,4), dtype='f,f')
In [407]: arr
Out[407]:
array([[(0., 0.), (0., 0.), (0., 0.), (0., 0.)],
[(0., 0.), (0., 0.), (0., 0.), (0., 0.)]],
dtype=[('f0', '<f4'), ('f1', '<f4')])
In [408]: arr['f1'] = arr2D
In [409]: arr['f0'] = _400
In [410]: arr
Out[410]:
array([[(7. , 0.05889647), (7.5, -0.2381424 ), (8. , -0.49582142),
(8.5, -0.70726925)],
[(7. , 0.5336367 ), (7.5, 0.1654723 ), (8. , -0.16439857),
(8.5, -0.44880486)]], dtype=[('f0', '<f4'), ('f1', '<f4')])