Append values of two strings into pairs

Question

I start with two numpy arrays, the "x values" and the "y values":

import numpy as np
x = np.arange(100)
y = np.arange(100)

The output is

[ 0  1  2  3  4 ..... 96 97 98 99]
[ 0  1  2  3  4 ..... 96 97 98 99]

I would like to append these values together into an array of len() = 100 such that the output is

[ (0,0) (1,1) (2,2) (3,3) .... (98,98) (99,99) ]

How does one use indexing to both (A) put the pairs in the correct order and (B) put the paratheses ( and comma , in the correct order?

I'm fairly certain you can use zip(x,y) but I'm not 100% sure that will work properly with numpy — kylieCatt
– kylieCatt, Commented Jun 29, 2015 at 16:27
Why does your title mention strings, the question has nothing to do with strings. — Akavall
– Akavall, Commented Jun 29, 2015 at 16:34
@PadraicCunningham Couldn't I just convert output list into a np.ndarray? — ShanZhengYang
– ShanZhengYang, Commented Jun 29, 2015 at 17:15
@ShanZhengYang, you can do it all in numpy, no point creating a list to turn it back into a numpy array — Padraic Cunningham
– Padraic Cunningham, Commented Jun 29, 2015 at 17:16

Anand S Kumar · Accepted Answer · 2015-06-29 16:35:35Z

2

For your particular requirement, you can use the built-in zip function, which combines multiple lists at their corresponding indexes (that is ith index of all lists that are parameter to it in combined in the returned iterator).

Example -

import numpy as np
x = np.arange(100)
y = np.arange(100)
print(list(zip(x,y)))
>>> [(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12), (13, 13), (14, 14), (15, 15), (16, 16), (17, 17), (18, 18), (19, 19), (20, 20), (21, 21), (22, 22), (23, 23), (24, 24), (25, 25), (26, 26), (27, 27), (28, 28), (29, 29), (30, 30), (31, 31), (32, 32), (33, 33), (34, 34), (35, 35), (36, 36), (37, 37), (38, 38), (39, 39), (40, 40), (41, 41), (42, 42), (43, 43), (44, 44), (45, 45), (46, 46), (47, 47), (48, 48), (49, 49), (50, 50), (51, 51), (52, 52), (53, 53), (54, 54), (55, 55), (56, 56), (57, 57), (58, 58), (59, 59), (60, 60), (61, 61), (62, 62), (63, 63), (64, 64), (65, 65), (66, 66), (67, 67), (68, 68), (69, 69), (70, 70), (71, 71), (72, 72), (73, 73), (74, 74), (75, 75), (76, 76), (77, 77), (78, 78), (79, 79), (80, 80), (81, 81), (82, 82), (83, 83), (84, 84), (85, 85), (86, 86), (87, 87), (88, 88), (89, 89), (90, 90), (91, 91), (92, 92), (93, 93), (94, 94), (95, 95), (96, 96), (97, 97), (98, 98), (99, 99)]

For Python 2.x , please note you do not need list(zip(...)) , since zip itself would return a list , but for Python 3.x , zip returns an iterator, and to print it we would need to convert it into a list.

edited Jun 29, 2015 at 16:35

answered Jun 29, 2015 at 16:27

Anand S Kumar

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3 Comments

Cleb Over a year ago

Is 'list' needed here? Should work without it as well or am I missing something?

kylieCatt Over a year ago

@Cleb depends. zip returns a generator object so no, you don't need list() right away but eventually you will need to consume the generator to get the values out.

Anand S Kumar Over a year ago

For python 3 , zip returns an iterator, if we just print it, we would get something like - <zip object at 0x00355D78> . For Python 2, it is not required I think

Kasravnd · Accepted Answer · 2015-06-29 16:28:28Z

You can use np.dstack to get the columns :

>>> np.dstack((x,y))
array([[[ 0,  0],
        [ 1,  1],
        [ 2,  2],
        [ 3,  3],
        [ 4,  4],
        [ 5,  5],
        [ 6,  6],
        [ 7,  7],
        [ 8,  8],
        [ 9,  9],
           ...
        [99, 99]]])

And if you want to get tuple instead of list you can use map to convert it to tuple:

>>> map(tuple,np.dstack((x,y))[0])
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12), (13, 13), (14, 14), (15, 15), (16, 16), (17, 17), (18, 18), (19, 19), (20, 20), (21, 21), (22, 22), (23, 23), (24, 24), (25, 25), (26, 26), (27, 27), (28, 28), (29, 29), (30, 30), (31, 31), (32, 32), (33, 33), (34, 34), (35, 35), (36, 36), (37, 37), (38, 38), (39, 39), (40, 40), (41, 41), (42, 42), (43, 43), (44, 44), (45, 45), (46, 46), (47, 47), (48, 48), (49, 49), (50, 50), (51, 51), (52, 52), (53, 53), (54, 54), (55, 55), (56, 56), (57, 57), (58, 58), (59, 59), (60, 60), (61, 61), (62, 62), (63, 63), (64, 64), (65, 65), (66, 66), (67, 67), (68, 68), (69, 69), (70, 70), (71, 71), (72, 72), (73, 73), (74, 74), (75, 75), (76, 76), (77, 77), (78, 78), (79, 79), (80, 80), (81, 81), (82, 82), (83, 83), (84, 84), (85, 85), (86, 86), (87, 87), (88, 88), (89, 89), (90, 90), (91, 91), (92, 92), (93, 93), (94, 94), (95, 95), (96, 96), (97, 97), (98, 98), (99, 99)]
>>>

Akavall · Accepted Answer · 2015-06-29 16:32:29Z

0

You could use vstack

In [36]: xy = np.vstack((x,y)).T

In [37]: xy.shape
Out[37]: (100, 2)

In [38]: xy[0]
Out[38]: array([0, 0])

In [39]: xy[1]
Out[39]: array([1, 1])

answered Jun 29, 2015 at 16:32

Akavall

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