Get indices of array at certain value python

Question

I have an array like this:

[[0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0]]

I like to loop through the array. Everywhere where the array has a value of 1 I like to get the index of the array and perform a operation.

Roughly something like this:

for value in array:
   if value ==1:
     print arrayIndexX, arrayIndexY

What's the operation? What are you trying to do? From your other questions it looks like you are using NumPy. You probably want to read : scipy-lectures.github.io/intro/numpy/… — YXD, Commented Feb 3, 2014 at 17:51
Yeah, I am using numpy. I converted a GIS raster to a numpy array and like to get the coordinates. — ustroetz, Commented Feb 3, 2014 at 17:52
@ustroetz sure but what do you want to do with the coordinates? Do you want to call a function on another array of the same shape based on these coordinates? Do you want to use the indices to generate a polygon of the outline? Sometimes these questions are a case of meta.stackexchange.com/a/66378/160939 and you can sometimes get a much more useful answer with some broader details :) — YXD, Commented Feb 3, 2014 at 17:56

Ashwini Chaudhary · Accepted Answer · 2014-02-03 17:50:29Z

6

You can use numpy.where with numpy.column_stack here. Example:

>>> import numpy as np
>>> a = np.array([[1, 0, 0], [0, 1, 0], [1, 1, 1]])
>>> np.column_stack(np.where(a==1))
array([[0, 0],
       [1, 1],
       [2, 0],
       [2, 1],
       [2, 2]])

answered Feb 3, 2014 at 17:50

Ashwini Chaudhary

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Comments

mhlester · Accepted Answer · 2014-02-03 17:47:44Z

5

Using enumerate, and a nested loop through rows and columns:

for y, row in enumerate(array):
    for x, val in enumerate(row):
        if val == 1:
            print x, y

answered Feb 3, 2014 at 17:47

mhlester

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Comments

Magellan88 · Accepted Answer · 2014-02-04 06:54:14Z

2

In this case you could also use the np.nonzero( YourArray ) function, this will give you exactly what you want.

edited Feb 4, 2014 at 6:54

answered Feb 3, 2014 at 20:28

Magellan88

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2 Comments

TimCera Over a year ago

Apologies - I missed your answer before giving mine.

Magellan88 Over a year ago

That's allright, I think it is important that this function popped up here somewhere, as is is most probably the fastest one.

inspectorG4dget · Accepted Answer · 2014-02-03 17:50:32Z

1

for r, row in enumerate(array):
    for c, val in enumerate(row):
        if val == 1:
            print r,c

Alternatively, you could build a list containing the required coordinate values:

[(r,c) for r,row in enumerate(array) for c,val in enumerate(row) if val==1]

answered Feb 3, 2014 at 17:50

inspectorG4dget

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Comments

TimCera · Accepted Answer · 2014-02-04 00:07:02Z

But, I think maybe far better is to use masked arrays...

marray = np.ma.array(b, mask=(b == 0))
print(marray)
[[-- -- -- -- -- 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 -- -- -- -- -- --]]

With masked array you could then just make the manipulations that you want and the only elements that are used are the unmasked ones.

CCP · Accepted Answer · 2014-02-03 17:57:58Z

0

Maybe this could help, where "a" its your array:

for i in range(len(a)):
    for j in range(len(a[i])):
        if(a[i][j]==1):
            print i,j

answered Feb 3, 2014 at 17:57

CCP

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Comments

TimCera · Accepted Answer · 2014-02-03 23:59:35Z

Probably numpy.nonzero is the easiest...

b = np.array(
[[0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0]])

ind = np.nonzero(b)
print(ind)
(array([ 0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2,  2,  2,  2,  2,  3,
    3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  4,  4,  4,  4,  4,
    4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5,  5,
    5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  6,  6,  6,  6,
    6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,
    6,  6,  6,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,
    7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8,  8,  8,
    8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,
    9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,
    9,  9,  9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
   10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11,
   11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11,
   11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12,
   12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13,
   13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13,
   13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14,
   14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15,
   15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16,
   16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16,
   16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
   17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18,
   18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18,
   18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19,
   19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20,
   21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22,
   22, 22, 23, 23, 23, 23]), array([ 5,  6,  5,  6,  7,  8,  9,  5,  6,  7,  8,  9, 10, 11, 12, 13,  4,
    5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,  4,  5,  6,  7,  8,
    9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,  3,  4,  5,  6,
    7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
   24, 25, 27,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
   17, 18, 19, 20, 21, 22, 23, 24, 25, 27,  3,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27,
    3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
   20, 21, 22, 23, 24, 25, 27,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11,
   12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,  2,  3,  4,
    5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
   22, 23, 24, 25,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14,
   15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,  1,  2,  3,  4,  5,  6,
    7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
   24,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
   17, 18, 19, 20, 21, 22, 23, 24,  1,  2,  3,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,  0,  1,
    2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
   19, 20, 21, 22, 23, 24,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,
   11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,  2,  3,  4,  5,
    6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
   23,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
   21, 22, 23,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
   12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 15, 16, 17, 18, 19, 20,
   21, 22, 18, 19, 20, 21]))

Andrew Swann · Accepted Answer · 2021-02-25 10:37:05Z

The answer depends on what operation you want to perform. For a standard enumeration over indices and values there is a built in numpy iterator for this, namely numpy.ndenumerate, that is dimension agnostic.

import numpy as np

rng = np.random.default_rng()
a = rng.integers(2, size=(5,5))
print(a)

for ind, val in np.ndenumerate(a):
    if val == 1:
        print(ind)

giving

[[0 0 1 1 0]
 [1 1 1 1 0]
 [0 1 0 0 0]
 [1 1 0 0 1]
 [0 0 1 0 0]]
(0, 2)
(0, 3)
(1, 0)
(1, 1)
(1, 2)
(1, 3)
(2, 1)
(3, 0)
(3, 1)
(3, 4)
(4, 2)

Note that masked arrays are often a better way to operate on a certain set of entries.

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