[Numpy-discussion] fast method to to count a particular value in a large matrix
Naresh
npai@uark....
Sat Feb 4 15:20:59 CST 2012
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
>
>
>
>
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The histogram() solution works perfect since unique_elem is ordered. I
appreciate everyone's help.
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