Python/Numpy - Wrap Slice Around End of Array

Question

I have two 1D arrays, one that has some values of interest (a) and another that provides indices into that array (b). I know that the values in b always increase, except at one point (could be anywhere) where the number decreases since it rolls from the end to the beginning of array a. The method below seems to work, but I just think that a cleaner way must exist. Can anyone suggest something better? Thanks.

Code:

import numpy as np
a = np.arange(12)
b = np.array([5, 9, 2, 4])
#I want to generate these:
#[5,6,7,8,9]
#[9,10,11,0,1,2]
#[2,3,4]
#[4,5]

a = np.roll(a, -b[0], axis=0)
# Subtract off b[0] but ensure that all values are positive
b = (b-b[0]+len(a))%len(a)
for i, ind in enumerate(b):
   if i < len(b)-1:
      print a[b[i]:b[i+1]+1]
   else:
      print np.hstack((a[b[i]:len(a)], a[0]))

Benjamin · Accepted Answer · 2011-04-16 02:02:10Z

3

A bit shorter, but maybe I can still do better...

import numpy as np

a = np.arange(12)
b = np.array([5, 9, 2, 4])
b = np.append(b, b[0])

for i in range(0, len(b)-1):
    print np.roll(a, len(a)-b[i])[:b[i+1]-b[i]+1]

answered Apr 16, 2011 at 2:02

Benjamin

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4 Comments

JoshAdel Over a year ago

You could make it a little cleaner by using: np.roll(a,-b[i])[:b[i+1]-b[i]+1]

Scott B Over a year ago

Thanks, this is cleaner. Although it has potentially many more calls to the roll function, which I would assume could slow things down for very large arrays. Can you think of anyway to do this without using roll? Replacing your np.roll(... line with simply a[b[i]:b[i+1]+1] would work for all cases besides that one where the roll-over occurs.

Benjamin Over a year ago

You could append the whole array a to itself, and recalculate the indices for a[start:end], but I think that would be more complicated, not less. I'm sure there is a way to do this with strides, just haven't had the time to try...

Scott B Over a year ago

I was thinking that it could be done with strides as well. I don't know them well at all though. If I find time, I'll try to investigate them more.

maxz · Accepted Answer · 2014-05-14 16:52:19Z

1

Not sure if this helps, but a fast way without messing with the memory of a would be this:

import numpy as np

a = np.arange(12)
b = np.array([5, 9, 2, 4])
b = np.append(b, b[0])

b2 = b.copy()

b2[(np.diff(b)<0).nonzero()[0]+1:] += a.size

print [np.take(a, np.r_[b2[i]:b2[i+1]+1], mode='wrap') for i in range(b.size-1)]

print [np.roll(a, len(a)-b[i])[:b[i+1]-b[i]+1] for i in range(b.size-1)]

%timeit [np.take(a, np.r_[b2[i]:b2[i+1]+1], mode='wrap') for i in range(b.size-1)]
# 10000 loops, best of 3: 28.6 µs per loop

%timeit [np.roll(a, len(a)-b[i])[:b[i+1]-b[i]+1] for i in range(b.size-1)]
# 10000 loops, best of 3: 77.7 µs per loop

answered May 14, 2014 at 16:52

maxz

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