Hi, some time ago I needed the same thing and hacked the function
histogram (from numpy). Here goes my function, I hope it will result useful
Cheers,
Juan
## Calculates the histogram allowing for overlapping bins, which are
given by
#
# @param a
# @param bins a sequence of pairs (left,right), limits for each bin
#
# @return hist (numpy array)
# bin_centers (numpy array)
def myhistogram(a, bins):
Compute the histogram of a set of data.
Parameters
--
a : array_like
Input data.
bins : sequence of pairs
It defines the bin edges (left,right), allowing for non-uniform
bin widths.
Returns
---
hist : array
The values of the histogram. See `normed` and `weights` for a
description of the possible semantics.
bin_centers : array of dtype float
Return the bin centers ``(length(hist))``.
Notes
-
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.
Examples
myhistogram([1,2,1], bins=[(0,1),(1,1.5),(1.5,2.5),(2,3)])
(array([0.5, 1.25, 2, 2.5]), array([0, 0, 1, 2, 3]))
bins = np.asarray(bins) # bins are 2-dimensional arrays
of shape (n,2)
if len(bins.shape) != 2 or bins.shape[1] != 2:
raise AttributeError, 'bins must be a list/array of 2-tuples.'
a = np.asarray(a)
a = a.ravel()
n = np.zeros(len(bins), int)
block = 65536
for i in np.arange(0, len(a), block):
sa = np.sort(a[i:i+block])
n += np.r_[sa.searchsorted(bins[:-1,1], 'left'),
sa.searchsorted(bins[-1,1], 'right')]\
- np.r_[sa.searchsorted(bins[:-1,0], 'left'),
sa.searchsorted(bins[-1,0], 'right')]
return n, (bins[:,0]+bins[:,1])/2.
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