On Sat, Feb 4, 2012 at 2:35 PM, Benjamin Root <ben.r...@ou.edu> wrote:
> > > On Saturday, February 4, 2012, Naresh Pai <n...@uark.edu> wrote: > > I am somewhat new to Python (been coding with Matlab mostly). I am > trying to > > simplify (and expedite) a piece of code that is currently a bottleneck > in a larger > > code. > > I have a large array (7000 rows x 4500 columns) titled say, abc, and I > am trying > > to find a fast method to count the number of instances of each unique > value within > > it. All unique values are stored in a variable, say, unique_elem. My > current code > > is as follows: > > import numpy as np > > #allocate space for storing element count > > elem_count = zeros((len(unique_elem),1)) > > #loop through and count number of unique_elem > > for i in range(len(unique_elem)): > > elem_count[i]= np.sum(reduce(np.logical_or,(abc== x for x > in [unique_elem[i]]))) > > This loop is bottleneck because I have about 850 unique elements and it > takes > > about 9-10 minutes. Can you suggest a faster way to do this? > > Thank you, > > Naresh > > > > no.unique() can return indices and reverse indices. It would be trivial > to histogram the reverse indices using np.histogram(). > > Instead of histogram(), you can use bincount() on the inverse indices: u, inv = np.unique(abc, return_inverse=True) n = np.bincount(inv) u will be an array of the unique elements, and n will be an array of the corresponding number of occurrences. Warren
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