assigning points to bins

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, I 开发者_如何学编程do something like this:

from scipy import *
num_bins = 3 # number of bins to use
values = # some array of integers...
min_val = min(values) - 1
max_val = max(values) + 1
my_bins = linspace(min_val, max_val, num_bins)
# assign point to my bins
for v in values:
  best_bin = min_index(abs(my_bins - v))

where min_index returns the index of the minimum value. The idea is that you can find the bin the point falls into by seeing what bin it has the smallest difference with.

But I think this has weird edge cases. What I am looking for is a good representation of bins, ideally ones that are half closed half open (so that there is no way of assigning one point to two bins), i.e.

bin1 = [x1, x2)
bin2 = [x2, x3)
bin3 = [x3, x4)
etc...

what is a good way to do this in Python, using numpy/scipy? I am only concerned here with binning integer values.

thanks very much for your help.

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

This is fairly straightforward in numpy using broadcasting--my example below is four lines of code (not counting first two lines to create bins and data points, which would of course ordinarily be supplied.)

import numpy as NP
# just creating 5 bins at random, each bin expressed as (x, y, z) although, this code
# is not limited by bin number or bin dimension
bins = NP.random.random_integers(10, 99, 15).reshape(5, 3) 
# creating 30 random data points
data = NP.random.random_integers(10, 99, 90).reshape(30, 3)
# for each data point i want the nearest bin, but before i can generate a distance
# matrix, i need to 'conform' the array dimensions
# 'broadcasting' is an excellent and concise way to do this
bins = bins[:, NP.newaxis, :]
data2 = data[NP.newaxis, :, :]
# now i can calculate the distance matrix
dist_matrix = NP.sqrt(NP.sum((data - bins)**2, axis=-1)) 
bin_assignments = NP.argmin(dist_matrix, axis=0)

'bin_assignments' is a 1d array of indices comprised of integer values from 0 to 4, corresponding to the five bins--the bin assignments for each of the 30 original points in the 'data' matrix above.