Create list of tuples of1D numpy array with every row of a 2D numpy array

Question

I have a 1D numpy array of values, and a 2D numpy array of values

x = array([4,5,6,7],[8,9,10,11])
y = array([0,1,2,3])

I want to create tuples of each row of y with the row in x

End result being

array([(4,0),(5,1),(6,2),(7,3),(8,0),(9,1),(10,2),(11,3)])

Is there anyway to do this with numpy functions instead of for loops?

Do you expect to have array of tuples here? – mathfux Oct 13 '20 at 18:50 — mathfux, Oct 13 '20 at 18:50

Quang Hoang · Answer 1 · 2020-10-13T18:48:17.467

2

Let's try:

np.concatenate([list(zip(a,y)) for a in x ])

Output:

array([[ 4,  0],
       [ 5,  1],
       [ 6,  2],
       [ 7,  3],
       [ 8,  0],
       [ 9,  1],
       [10,  2],
       [11,  3]])

Or pure python:

list(zip(x.ravel(), np.tile(y,len(x))) )

Output:

[(4, 0), (5, 1), (6, 2), (7, 3), (8, 0), (9, 1), (10, 2), (11, 3)]

edited Oct 13 '20 at 18:48

answered Oct 13 '20 at 18:35

Quang Hoang

Isn't there a looping inside comprehension you did? Also, `np.repeat` works in a different manner than OP expects. – mathfux Oct 13 '20 at 18:47
@mathfux thanks for pointing out. Updated the answer. – Quang Hoang Oct 13 '20 at 18:48

mathfux · Accepted Answer · 2020-10-13T19:20:28.710

2

Assuming len(y) divides a size of x, you can merge a flatten view of x with an artificial padding with np.tile and then transpose result:

np.array([x.flatten(), np.tile(y, x.size//len(y))]).T

Note that x.flatten() is a synonym of x.ravel().

edited Oct 13 '20 at 19:20

answered Oct 13 '20 at 18:45

mathfux

1

Just a side note, `flatten()` is [not a synonym](https://stackoverflow.com/questions/28930465/what-is-the-difference-between-flatten-and-ravel-functions-in-numpy) of `ravel()`. – Quang Hoang Oct 13 '20 at 18:49
Thanks for a source – mathfux Oct 13 '20 at 18:50

2 Answers2