98

Assume the following arrays are given:

a = array([1,3,5])
b = array([2,4,6])

How would one interweave them efficiently so that one gets a third array like this

c = array([1,2,3,4,5,6])

It can be assumed that length(a)==length(b).

D R
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    How about, same question, but you are trying to interleave matrices. That is a and b are 3 dimensional, and not necessarily the same size in the first dimension. Note:Only the first dimension should be interleaved. – Geronimo Nov 17 '17 at 21:29

11 Answers11

173

I like Josh's answer. I just wanted to add a more mundane, usual, and slightly more verbose solution. I don't know which is more efficient. I expect they will have similar performance.

import numpy as np
a = np.array([1,3,5])
b = np.array([2,4,6])

c = np.empty((a.size + b.size,), dtype=a.dtype)
c[0::2] = a
c[1::2] = b
Paul
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    Unless speed is really really important, I would go with this as it's much more comprehensible which is important if anyone is ever going to look at it again. – John Salvatier Mar 18 '11 at 03:02
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    +1 I played around with timings and your code surprisingly seems to be 2-5x faster depending on inputs. I still find the efficiency of these types of operations to be nonintuitive, so it's always worth it to use `timeit` to test things out if a particular operation is a bottleneck in your code. There are usually more than one way to do things in numpy, so definitely profile code snippets. – JoshAdel Mar 18 '11 at 03:04
  • @JoshAdel: I guess if `.reshape` creates an additional copy of the array, then that would explain a 2x performance hit. I don't think it always makes a copy, however. I'm guessing the 5x difference is only for small arrays? – Paul Mar 18 '11 at 03:39
  • looking at `.flags` and testing `.base` for my solution, it looks like the reshape to 'F' format creates a hidden copy of the vstacked data, so it's not a simple view as I thought it would be. And strangely the 5x is only for intermediate sized arrays for some reason. – JoshAdel Mar 18 '11 at 14:52
  • Another advantage of this answer is it's not limited to arrays of the same length. It could weave `n` items with `n-1` items. – EliadL Sep 25 '19 at 11:48
84

I thought it might be worthwhile to check how the solutions performed in terms of performance. And this is the result:

enter image description here

This clearly shows that the most upvoted and accepted answer (Pauls answer) is also the fastest option.

The code was taken from the other answers and from another Q&A:

# Setup
import numpy as np

def Paul(a, b):
    c = np.empty((a.size + b.size,), dtype=a.dtype)
    c[0::2] = a
    c[1::2] = b
    return c

def JoshAdel(a, b):
    return np.vstack((a,b)).reshape((-1,),order='F')

def xioxox(a, b):
    return np.ravel(np.column_stack((a,b)))

def Benjamin(a, b):
    return np.vstack((a,b)).ravel([-1])

def andersonvom(a, b):
    return np.hstack( zip(a,b) )

def bhanukiran(a, b):
    return np.dstack((a,b)).flatten()

def Tai(a, b):
    return np.insert(b, obj=range(a.shape[0]), values=a)

def Will(a, b):
    return np.ravel((a,b), order='F')

# Timing setup
timings = {Paul: [], JoshAdel: [], xioxox: [], Benjamin: [], andersonvom: [], bhanukiran: [], Tai: [], Will: []}
sizes = [2**i for i in range(1, 20, 2)]

# Timing
for size in sizes:
    func_input1 = np.random.random(size=size)
    func_input2 = np.random.random(size=size)
    for func in timings:
        res = %timeit -o func(func_input1, func_input2)
        timings[func].append(res)

%matplotlib notebook

import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(1)
ax = plt.subplot(111)

for func in timings:
    ax.plot(sizes, 
            [time.best for time in timings[func]], 
            label=func.__name__)  # you could also use "func.__name__" here instead
ax.set_xscale('log')
ax.set_yscale('log')
ax.set_xlabel('size')
ax.set_ylabel('time [seconds]')
ax.grid(which='both')
ax.legend()
plt.tight_layout()

Just in case you have numba available you could also use that to create a function:

import numba as nb

@nb.njit
def numba_interweave(arr1, arr2):
    res = np.empty(arr1.size + arr2.size, dtype=arr1.dtype)
    for idx, (item1, item2) in enumerate(zip(arr1, arr2)):
        res[idx*2] = item1
        res[idx*2+1] = item2
    return res

It could be slightly faster than the other alternatives:

enter image description here

MSeifert
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    Also of note, the accepted answer is _way_ faster than the a native Python solution with [`roundrobin()`](https://docs.python.org/3/library/itertools.html#itertools-recipes) from the itertools recipes. – Brad Solomon Jan 28 '18 at 17:41
46

Here is a one-liner:

c = numpy.vstack((a,b)).reshape((-1,),order='F')
JoshAdel
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24

Here is a simpler answer than some of the previous ones

import numpy as np
a = np.array([1,3,5])
b = np.array([2,4,6])
inter = np.ravel(np.column_stack((a,b)))

After this inter contains:

array([1, 2, 3, 4, 5, 6])

This answer also appears to be marginally faster:

In [4]: %timeit np.ravel(np.column_stack((a,b)))
100000 loops, best of 3: 6.31 µs per loop

In [8]: %timeit np.ravel(np.dstack((a,b)))
100000 loops, best of 3: 7.14 µs per loop

In [11]: %timeit np.vstack((a,b)).ravel([-1])
100000 loops, best of 3: 7.08 µs per loop
Community
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xioxox
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10

This will interleave/interlace the two arrays and I believe it is quite readable:

a = np.array([1,3,5])      #=> array([1, 3, 5])
b = np.array([2,4,6])      #=> array([2, 4, 6])
c = np.hstack( zip(a,b) )  #=> array([1, 2, 3, 4, 5, 6])
andersonvom
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7

Improving @xioxox's answer:

import numpy as np
a = np.array([1,3,5])
b = np.array([2,4,6])
inter = np.ravel((a,b), order='F')
MSeifert
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Will
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6

Maybe this is more readable than @JoshAdel's solution:

c = numpy.vstack((a,b)).ravel([-1])
Benjamin
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    `ravel`'s `order` argument in [the documentation](http://docs.scipy.org/doc/numpy/reference/generated/numpy.ravel.html) is one of `C`, `F`, `A`, or `K`. I think you really want `.ravel('F')`, for FORTRAN order (column first) – Nick T Feb 11 '14 at 18:10
3

I needed to do this but with multidimensional arrays along any axis. Here's a quick general purpose function to that effect. It has the same call signature as np.concatenate, except that all input arrays must have exactly the same shape.

import numpy as np

def interleave(arrays, axis=0, out=None):
    shape = list(np.asanyarray(arrays[0]).shape)
    if axis < 0:
        axis += len(shape)
    assert 0 <= axis < len(shape), "'axis' is out of bounds"
    if out is not None:
        out = out.reshape(shape[:axis+1] + [len(arrays)] + shape[axis+1:])
    shape[axis] = -1
    return np.stack(arrays, axis=axis+1, out=out).reshape(shape)
clwainwright
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    +1 for such a generalized recipe (works with n-dim, interleaves along any axis, works for any number of input arrays, takes an optional `out` arg, and works for sub-classed arrays). Personally, I would prefer `axis` to default to `-1` rather than to `0`, but maybe that's just me. And you might want to link to this answer of yours, from [this question](https://stackoverflow.com/questions/42162300/how-to-interleave-numpy-ndarrays/42162780https://stackoverflow.com/questions/42162300/how-to-interleave-numpy-ndarrays/42162780), which actually asked for the input arrays to be n-dimensional. – fountainhead Dec 02 '20 at 00:11
2

One can also try np.insert. (Solution migrated from Interleave numpy arrays)

import numpy as np
a = np.array([1,3,5])
b = np.array([2,4,6])
np.insert(b, obj=range(a.shape[0]), values=a)

Please see the documentation and tutorial for more information.

Tai
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2

vstack sure is an option, but more straightforward solution for your case could be the hstack

>>> a = array([1,3,5])
>>> b = array([2,4,6])
>>> hstack((a,b)) #remember it is a tuple of arrays that this function swallows in.
>>> array([1, 3, 5, 2, 4, 6])
>>> sort(hstack((a,b)))
>>> array([1, 2, 3, 4, 5, 6])

and more importantly this works for arbitrary shapes of a and b

Also you may want to try out dstack

>>> a = array([1,3,5])
>>> b = array([2,4,6])
>>> dstack((a,b)).flatten()
>>> array([1, 2, 3, 4, 5, 6])

u've got options now!

bhanukiran
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    -1 to first answer because question has nothing to do with sorting. +1 to second answer, which is the best I've seen so far. This is why multiple solutions should be posted as multiple answers. Please split it into multiple answers. – endolith Jan 16 '13 at 17:57
2

Another one-liner: np.vstack((a,b)).T.ravel()
One more: np.stack((a,b),1).ravel()

Arty
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