Warren Weckesser <warren.weckesser <at> enthought.com> writes: > > > On Sat, Feb 4, 2012 at 2:35 PM, Benjamin Root <ben.root <at> ou.edu> wrote: > > > On Saturday, February 4, 2012, Naresh Pai <npai <at> 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 > > > > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion <at> scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
The histogram() solution works perfect since unique_elem is ordered. I appreciate everyone's help. _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
