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python Programming Glossary: np.bincountIncrement Numpy array with repeated indices http://stackoverflow.com/questions/2004364/increment-numpy-array-with-repeated-indices 2 5 2 # indices to increment by one each replicate bbins np.bincount b b.sort # sort b because bincount is sorted incr bbins np.nonzero.. best way Is there are risk involved with assuming that the np.bincount and np.unique operations would result in the same sorted order.. numpy share improve this question After you do bbins np.bincount b why not do a len bbins bbins Edited for further simplification...
Is it possible to specify your own distance function using Scikits.Learn K-Means Clustering? http://stackoverflow.com/questions/5529625/is-it-possible-to-specify-your-own-distance-function-using-scikits-learn-k-means 0 if verbose print kmeans d iterations cluster sizes jiter np.bincount xtoc if verbose 2 r50 np.zeros k r90 np.zeros k for j in range..
find the most frequent number in a numpy vector http://stackoverflow.com/questions/6252280/find-the-most-frequent-number-in-a-numpy-vector use np.argmax a np.array 1 2 3 1 2 1 1 1 3 2 2 1 counts np.bincount a print np.argmax counts For a more complicated list that perhaps..
Python group by array a, and summarize array b - Performance http://stackoverflow.com/questions/7538382/python-group-by-array-a-and-summarize-array-b-performance SO FAR but a as integer only def unique_Unutbu a b x np.bincount a weights b aResult np.unique a bResult x aResult Maybe someone.. has values that can fit in dtype int32 then you could use np.bincount with weights import numpy as np a 7 3 5 7 5 7 b 0.2 0.1 0.3.. numpy as np a 7 3 5 7 5 7 b 0.2 0.1 0.3 0.1 0.1 0.2 x np.bincount a weights b print x # 0. 0. 0. 0.1 0. 0.4 0. 0.5 print x 7 3..
Numpy accumulating one array in another using index array http://stackoverflow.com/questions/9790436/numpy-accumulating-one-array-in-another-using-index-array numpy.bincount was introduced for this purpose tmp np.bincount idx w v len tmp tmp I think as of 1.6 you can also pass a minlength..
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