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up vote 2 down vote favorite

I have an ordered array like this: numpy.array([1, 2, 5, 10, 25, 36, 66, 90, 121, 230, 333, 500])

Suppose I want all values up to 60 (if 60 isn't in, i want to stop at the first value greater than 60), so I want [1, 2, 5, 10, 25, 36, 66]. If I use numpy.where() with <= 60, it stops before 66.

My solution 

from numpy import *
x = array([1, 2, 5, 10, 25, 36, 66, 90, 121, 230, 333, 500])
print x[:where(x >= 60)[0][0]+1]
>>>[ 1  2  5 10 25 36 66]
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please correct your question. you probably meant to search for 66 instead of 60. –  Andrea Zonca Mar 10 '11 at 23:17
 
it would have been better style to do "import numpy as np" rather than importing * - also maybe clearer to non-numpy users what the problem is. –  DangerMouse Apr 20 '12 at 6:17
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2 Answers

up vote 3 down vote accepted

You don't need anything special in numpy for this.

import numpy, bisect
a = numpy.array([1, 2, 5, 10, 25, 36, 66, 90, 121, 230, 333, 500])
idx = bisect.bisect(a, 60)
print a[:idx]
share|improve this answer
 
this gives me [ 1 2 5 10 25 36], stops before 60. i guess it's because i worded it out wrong before sorry. –  canadian milk bag Mar 10 '11 at 20:21
 
Of course it stops before 60. The number 60 isn't in that list. –  Glenn Maynard Mar 10 '11 at 20:22
 
And if you do a[:idx + 1], you will get the number after your cutoff to stay, as you indicated in one of your comments –  Benjamin Mar 10 '11 at 20:23
 
If you want one more, then ... add one. –  Glenn Maynard Mar 10 '11 at 20:23
 
I would see some sense in using bisect only if you have a list instead of an array, so that you avoid using numpy completely. Instead, you you have a numpy array, you should use searchsorted. –  Andrea Zonca Mar 12 '11 at 4:53
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up vote 8 down vote

there is a specific numpy function to do this, np.searchsorted, which is much faster than bisect.

a=np.arange(1e7)
c=2e6
%timeit bisect.bisect(a,c)
10000 loops, best of 3: 31.6 us per loop
%timeit np.searchsorted(a,c)
100000 loops, best of 3: 6.77 us per loop

More remarkably ,it has also a specific keyword side for including or not the last point:

In [23]: a[:a.searchsorted(66,side='right')]
Out[23]: array([ 1,  2,  5, 10, 25, 36, 66])

In [24]: a[:a.searchsorted(66,side='left')]
Out[24]: array([ 1,  2,  5, 10, 25, 36])
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