Count all values in a matrix less than a value

Question

I have to count all the values in a matrix (2-d array) that are less than 200.

The code I wrote down for this is:

za=0   
p31 = numpy.asarray(o31)   
for i in range(o31.size[0]):   
    for j in range(o32.size[1]):   
        if p31[i,j]<200:   
            za=za+1   
print za

o31 is an image and I am converting it into a matrix and then finding the values.

Is there a simpler way to do this?

Doesn't this just print the number of values that are less than 200, and not the actual values? — Burhan Khalid, Oct 21 '12 at 07:46
Yes.. I just need the total of all the values instead of the actual values themselves. — gran_profaci, Oct 21 '12 at 08:20
You aren't getting the total here either. Set `za` to an empty list `za = []`, then `za.append(p31[i,j])`, finally out of your for loop, `print sum(za)`; but I'm sure there is a better way since you are using numpy. — Burhan Khalid, Oct 21 '12 at 08:25

score 127 · Answer 1 · answered Oct 21 '12 at 08:32

127

This is very straightforward with boolean arrays:

p31 = numpy.asarray(o31)
za = (p31 < 200).sum() # p31<200 is a boolean array, so `sum` counts the number of True elements

answered Oct 21 '12 at 08:32

nneonneo

171,345
36
312
383

4

I think this should be the top answer. The `np.where` also works obviously, but gives additional information which is not required here and makes the answer a bit less straightforward, and performance a bit slower – Iddo weiner May 04 '21 at 10:47

score 110 · Accepted Answer · answered Oct 21 '12 at 08:11

110

The numpy.where function is your friend. Because it's implemented to take full advantage of the array datatype, for large images you should notice a speed improvement over the pure python solution you provide.

Using numpy.where directly will yield a boolean mask indicating whether certain values match your conditions:

>>> data
array([[1, 8],
       [3, 4]])
>>> numpy.where( data > 3 )
(array([0, 1]), array([1, 1]))

And the mask can be used to index the array directly to get the actual values:

>>> data[ numpy.where( data > 3 ) ]
array([8, 4])

Exactly where you take it from there will depend on what form you'd like the results in.

answered Oct 21 '12 at 08:11

abought

2,652
1
18
13

12

Thanks... But I need the total number of values, not the values themselves.. should I do sum(numpy.where(data<200))? – gran_profaci Oct 21 '12 at 08:21
9

If you want the number OF values, not the sum, you would do len(numpy.where(data<200)) – jimh Oct 12 '17 at 19:28
1

if condition changed to > 0, then it will return (array([0, 0, 1, 1]), array([0, 1, 0, 1])), what does it mean? or is it a bug? – hihell Oct 19 '17 at 16:46
1

This method is unnecessarily complex. @neonneo's method is much more straightforward and does a better job of answering the question. – waterproof Mar 20 '19 at 00:11
@jimh for me np.where returns a tuple, gotta do len(np.where(...)[0]) – user1114 Oct 30 '20 at 05:39

K Z · Answer 3 · 2012-10-21T08:41:48.780

There are many ways to achieve this, like flatten-and-filter or simply enumerate, but I think using Boolean/mask array is the easiest one (and iirc a much faster one):

>>> y = np.array([[123,24123,32432], [234,24,23]])
array([[  123, 24123, 32432],
       [  234,    24,    23]])
>>> b = y > 200
>>> b
array([[False,  True,  True],
       [ True, False, False]], dtype=bool)
>>> y[b]
array([24123, 32432,   234])
>>> len(y[b])
3
>>>> y[b].sum()
56789

Update:

As nneonneo has answered, if all you want is the number of elements that passes threshold, you can simply do:

>>>> (y>200).sum()
3

which is a simpler solution.

Speed comparison with filter:

### use boolean/mask array ###

b = y > 200

%timeit y[b]
100000 loops, best of 3: 3.31 us per loop

%timeit y[y>200]
100000 loops, best of 3: 7.57 us per loop

### use filter ###

x = y.ravel()
%timeit filter(lambda x:x>200, x)
100000 loops, best of 3: 9.33 us per loop

%timeit np.array(filter(lambda x:x>200, x))
10000 loops, best of 3: 21.7 us per loop

%timeit filter(lambda x:x>200, y.ravel())
100000 loops, best of 3: 11.2 us per loop

%timeit np.array(filter(lambda x:x>200, y.ravel()))
10000 loops, best of 3: 22.9 us per loop

*** use numpy.where ***

nb = np.where(y>200)
%timeit y[nb]
100000 loops, best of 3: 2.42 us per loop

%timeit y[np.where(y>200)]
100000 loops, best of 3: 10.3 us per loop

`timeit y[b]` omits half the code from the computation done in filter. `%timeit y[y>200]` is equivalent-ish. — Mad Physicist, Apr 06 '16 at 18:14

score 10 · Answer 4 · answered Oct 21 '12 at 14:35

10

Here's a variant that uses fancy indexing and has the actual values as an intermediate:

p31 = numpy.asarray(o31)
values = p31[p31<200]
za = len(values)

answered Oct 21 '12 at 14:35

Karol

1,246
2
13
20

This is just regular boolean indexing. – Mad Physicist Apr 06 '16 at 18:16

Ricoter · Answer 5 · 2019-06-15T13:04:28.253

6

To count the number of values larger than x in any numpy array you can use:

n = len(matrix[matrix > x])

The boolean indexing returns an array that contains only the elements where the condition (matrix > x) is met. Then len() counts these values.

edited Jun 15 '19 at 13:04

answered Dec 03 '18 at 19:39

Ricoter

665
5
17

1

You probably need to explain why this works. Overloaded `__gt__` I guess, but that's not obvious. – Adrian W Dec 03 '18 at 20:03

score 1 · Answer 6 · answered Feb 17 '21 at 17:31

1

You can use numpy.count_nonzero, converting the whole into a one-liner:

za = numpy.count_nonzero(numpy.asarray(o31)<200)  #as written in the code

answered Feb 17 '21 at 17:31

André Anjos

4,641
2
27
34

You could avoid an explicit call to `asarray` with `np.count_nonzero(np.less_than(o3, 200))`. – Mad Physicist Jan 06 '22 at 22:56

Count all values in a matrix less than a value

6 Answers6

Linked

Related