How can I get an array of alternating values in python?

Question

Simple question here:

I\'m trying to get an array that alternates values (1, -1, 1, -1.....) for a given length. np.repeat just gives me (1, 1, 1, 1,-1, -1,-1, -1

Answer 1

Use numpy.tile!

import numpy
a = numpy.tile([1,-1], 15)

Answer 2

Maybe you're looking for itertools.cycle?

list_ = (1,-1,2,-2)  # ,3,-3, ...

for n, item in enumerate(itertools.cycle(list_)):
    if n==30:
        break

    print item

Answer 3

If you want a memory efficient solution, try this:

def alternator(n):
    for i in xrange(n):
        if i % 2 == 0:
            yield 1
        else:
            yield -1

Then you can iterate over the answers like so:

for i in alternator(n):
    # do something with i

Answer 4

I like @Benjamin's solution. An alternative though is:

import numpy as np
a = np.empty((15,))
a[::2] = 1
a[1::2] = -1

This also allows for odd-length lists.

EDIT: Also just to note speeds, for a array of 10000 elements

import numpy as np
from timeit import Timer

if __name__ == '__main__':

    setupstr="""
import numpy as np
N = 10000
"""

    method1="""
a = np.empty((N,),int)
a[::2] = 1
a[1::2] = -1
"""

    method2="""
a = np.tile([1,-1],N)
"""

    method3="""
a = np.array([1,-1]*N)   
"""

    method4="""
a = np.array(list(itertools.islice(itertools.cycle((1,-1)), N)))    
"""
    nl = 1000
    t1 = Timer(method1, setupstr).timeit(nl)
    t2 = Timer(method2, setupstr).timeit(nl)
    t3 = Timer(method3, setupstr).timeit(nl)
    t4 = Timer(method4, setupstr).timeit(nl)

    print 'method1', t1
    print 'method2', t2
    print 'method3', t3
    print 'method4', t4

Results in timings of:

method1 0.0130500793457
method2 0.114426136017
method3 4.30518102646
method4 2.84446692467

If N = 100, things start to even out but starting with the empty numpy arrays is still significantly faster (nl changed to 10000)

method1 0.05735206604
method2 0.323992013931
method3 0.556654930115
method4 0.46702003479

Numpy arrays are special awesome objects and should not be treated like python lists.

Answer 5

use multiplication:

[1,-1] * n

Answer 6

use resize():

In [38]: np.resize([1,-1], 10) # 10 is the length of result array
Out[38]: array([ 1, -1,  1, -1,  1, -1,  1, -1,  1, -1])

it can produce odd-length array:

In [39]: np.resize([1,-1], 11)
Out[39]: array([ 1, -1,  1, -1,  1, -1,  1, -1,  1, -1,  1])