rearranging an array of arrays (python)

Question

So I have a array of arrays, like

[array([-0.05504106,  4.21890792]), array([-0.05504106,  4.21890792]), array([-0.0533802 ,  4.10717668]), array([-0.0546635 ,  4.19501313])]

And what I'm attempting to do is make it into an array of 2 arrays, in which the components are put together like this:

[array([-0.05504106, -0.05504106, -.0533802, -.0546635]), array([4.21890792, 4.21890792, 4.10717668, 4.19501313])

I've tried a number of approaches. The first is as follows:

for i in range(len(array_1)-1):
zipped = zip(array_1[i], array_1[i+1], array_1[i+2])
print zipped

However, this approach has the drawback of having to add an array_1[i+n] for each additional array within array_1. Not at all practical if array_1 has many arrays within it.

The next thing I tried was attempting to use itertools.repeat in conjunction with the above code, like so:

for i in range(len(array_1)-1):
zipped = zip(itertools.repeat(array_1[i], len(array_1))
print zipped

however, this did not function the way I wanted it to.

Could you give me some idea of how I would accomplish this task? Should I be using zip and/or intertools.repeat?

so you want them to be grouped by their inner-index? basically: [(array[0][0], array[1][0], ..., array[n][0]), (array[0][1], ..., array[n][1]), ..., (array[0][m], ..., array[n][m])] ? — Adam Smith
– Adam Smith, Commented Mar 28, 2014 at 21:21

Hyperboreus · Accepted Answer · 2014-03-28 21:25:10Z

1

I am not sure which array class you are using, but with simple lists you can use zip:

a = [[-0.05504106, 4.21890792], [-0.05504106, 4.21890792], [-0.0533802, 4.10717668],[-0.0546635, 4.19501313]]

b = list(zip(*a)) #omit list in python2
print(b)

This prints:

[(-0.05504106, -0.05504106, -0.0533802, -0.0546635), (4.21890792, 4.21890792, 4.10717668, 4.19501313)]

answered Mar 28, 2014 at 21:25

Hyperboreus

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2 Comments

Philip R. Over a year ago

huh. well, thats alot easier than i thought it would be. I guess im not familiar enough with the use of the zip() function. Thanks!

Hyperboreus Over a year ago

@PhilipR. As a rule of thumb: zip(*x) unzips x.

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