assigning points to bins

Question

What is a good way to bin numerical values into a certain range? For example, suppose I have a list of values and I want to bin them into N bins by their range. Right now,

Answer 1

numpy.histogram() does exactly what you want.

The function signature is:

numpy.histogram(a, bins=10, range=None, normed=False, weights=None, new=None)

We're mostly interested in a and bins. a is the input data that needs to be binned. bins can be a number of bins (your num_bins), or it can be a sequence of scalars, which denote bin edges (half open).

import numpy
values = numpy.arange(10, dtype=int)
bins = numpy.arange(-1, 11)
freq, bins = numpy.histogram(values, bins)
# freq is now [0 1 1 1 1 1 1 1 1 1 1]
# bins is unchanged

To quote the documentation:

All but the last (righthand-most) bin is half-open. In other words, if bins is:

[1, 2, 3, 4]

then the first bin is [1, 2) (including 1, but excluding 2) and the second [2, 3). The last bin, however, is [3, 4], which includes 4.

Edit: You want to know the index in your bins of each element. For this, you can use numpy.digitize(). If your bins are going to be integral, you can use numpy.bincount() as well.

>>> values = numpy.random.randint(0, 20, 10)
>>> values
array([17, 14,  9,  7,  6,  9, 19,  4,  2, 19])
>>> bins = numpy.linspace(-1, 21, 23)
>>> bins
array([ -1.,   0.,   1.,   2.,   3.,   4.,   5.,   6.,   7.,   8.,   9.,
        10.,  11.,  12.,  13.,  14.,  15.,  16.,  17.,  18.,  19.,  20.,
        21.])
>>> pos = numpy.digitize(values, bins)
>>> pos
array([19, 16, 11,  9,  8, 11, 21,  6,  4, 21])

Since the interval is open on the upper limit, the indices are correct:

>>> (bins[pos-1] == values).all()
True
>>> import sys
>>> for n in range(len(values)):
...     sys.stdout.write("%g <= %g < %g\n"
...             %(bins[pos[n]-1], values[n], bins[pos[n]]))
17 <= 17 < 18
14 <= 14 < 15
9 <= 9 < 10
7 <= 7 < 8
6 <= 6 < 7
9 <= 9 < 10
19 <= 19 < 20
4 <= 4 < 5
2 <= 2 < 3
19 <= 19 < 20