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up vote 1 down vote favorite

Given a numpy array

A = np.array([[[29, 64, 83],
               [17, 92, 38],
               [67, 34, 20]],
              [[73, 28, 45],
               [19, 84, 61],
               [22, 63, 49]],
              [[48, 30, 13],
               [11, 52, 86],
               [62, 25, 12]]])

I want the index of a certain value, say 63

There is no possibility that the value will be duplicated or missing

I did

idx = np.where(A == 63)

print(idx)

I got 

(array([1], dtype=int32), array([2], dtype=int32), array([1], dtype=int32))

What I want is

[1, 2, 1]

as a list or other iterable without all that array, dtype=int32 etc.

How do I do this?

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3 Answers 3

up vote 2 down vote

If you want to get a numpy array back, just use the concatenate function:

In [30]: np.concatenate(idx)
Out[30]: array([1, 2, 1])

If you really have your heart set on a Python list, then just:

In [31]: np.concatenate(idx).tolist()
Out[31]: [1, 2, 1]
share|improve this answer
up vote 1 down vote

What about this approach using list-comprehensions? 

   idx = [x[0] for x in np.where(A==63)]
share|improve this answer
    
That seems to work. Thanks. I will have to study it to understand what it is doing. –  Barry Andersen 17 hours ago
    
Check out list comprehension if yo want to know how it works. –  memecs 17 hours ago
    
What I don't understand is what the 0 in x[0] is doing –  Barry Andersen 16 hours ago
    
It's pulling of the value from the array[...] –  memecs 16 hours ago
up vote 0 down vote

Numpy arrays support returning elements where a condition is true. You can use np.where(..) or use:

>>> A==63
array([[[False, False, False],
        [False, False, False],
        [False, False, False]],

       [[False, False, False],
        [False, False, False],
        [False,  True, False]],

       [[False, False, False],
        [False, False, False],
        [False, False, False]]], dtype=bool)

You can then flatten that index array to only the True values using an array's .nonzero() method:

>>> (A==63).nonzero()
(array([1]), array([2]), array([1]))

Note that is a Python tuple of numpy arrays with the first being the X index, the second being the Y and then Z in A[X,Y,Z] form.

For one element only, you could then flatten that using .r_:

>>> np.r_[(A==63).nonzero()]
array([1, 2, 1])

And you can produce a Python list if you wish:

>>> np.r_[(A==63).nonzero()].tolist()
[1, 2, 1]

A more interesting use case is when you have more than one index in the matrix that is True. Consider all values >63:

>>> A>63
array([[[False,  True,  True],
        [False,  True, False],
        [ True, False, False]],

       [[ True, False, False],
        [False,  True, False],
        [False, False, False]],

       [[False, False, False],
        [False, False,  True],
        [False, False, False]]], dtype=bool)

You can also use .nonzero() method or the nonzero function:

>>> np.nonzero(A>63)
(array([0, 0, 0, 0, 1, 1, 2]), array([0, 0, 1, 2, 0, 1, 1]), array([1, 2, 1, 0, 0, 1, 2]))
     ^^^ X's                      ^^^ Y's                              ^^^ Z's

Notice now that this is a tuple of 3 arrays (in this case) of all X's, all Y's, all Z's in that order.

You can use np.transpose to produce an array of those element indexes of the form [[X, Y, Z],...] like so:

>>> np.transpose((A>63).nonzero())
array([[0, 0, 1],
       [0, 0, 2],
       [0, 1, 1],
       [0, 2, 0],
       [1, 0, 0],
       [1, 1, 1],
       [2, 1, 2]])

Or (suitable for printing for human eyes for example) you can use zip:

>>> zip(*(A>63).nonzero())
[(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 2, 0), (1, 0, 0), (1, 1, 1), (2, 1, 2)]

Or, to print:

>>> print '\n'.join([str(e) for e in zip(*(A>63).nonzero())])
(0, 0, 1)
(0, 0, 2)
(0, 1, 1)
(0, 2, 0)
(1, 0, 0)
(1, 1, 1)
(2, 1, 2)

Which, of course, would work for a single element just as well:

>>> zip(*(A==63).nonzero())[0]
(1, 2, 1)

Or numpy way:

>>> np.transpose((A==63).nonzero())[0]
array([1, 2, 1])

All the methods here works with np.where(A==63) in place of (A==63).nonzero() as an example.

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