Find the row indexes of several values in a numpy array

Question

I have an array X:

X = np.array([[4,  2],
              [9,  3],
              [8,  5],
              [3,  3],
              [5,  6]])

And

Answer 1

Here is a pretty fast solution that scales up well using numpy and hashlib. It can handle large dimensional matrices or images in seconds. I used it on 520000 X (28 X 28) array and 20000 X (28 X 28) in 2 seconds on my CPU

Code:

import numpy as np
import hashlib


X = np.array([[4,  2],
              [9,  3],
              [8,  5],
              [3,  3],
              [5,  6]])

searched_values = np.array([[4, 2],
                            [3, 3],
                            [5, 6]])

#hash using sha1 appears to be efficient
xhash=[hashlib.sha1(row).digest() for row in X]
yhash=[hashlib.sha1(row).digest() for row in searched_values]

z=np.in1d(xhash,yhash)  

##Use unique to get unique indices to ind1 results
_,unique=np.unique(np.array(xhash)[z],return_index=True)

##Compute unique indices by indexing an array of indices
idx=np.array(range(len(xhash)))
unique_idx=idx[z][unique]

print('unique_idx=',unique_idx)
print('X[unique_idx]=',X[unique_idx])

Output:

unique_idx= [4 3 0]
X[unique_idx]= [[5 6]
 [3 3]
 [4 2]]

Answer 2

The numpy_indexed package (disclaimer: I am its author) contains functionality for performing such operations efficiently (also uses searchsorted under the hood). In terms of functionality, it acts as a vectorized equivalent of list.index:

import numpy_indexed as npi
result = npi.indices(X, searched_values)

Note that using the 'missing' kwarg, you have full control over behavior of missing items, and it works for nd-arrays (fi; stacks of images) as well.

Update: using the same shapes as @Rik X=[520000,28,28] and searched_values=[20000,28,28], it runs in 0.8064 secs, using missing=-1 to detect and denote entries not present in X.

Answer 3

X = np.array([[4,  2],
              [9,  3],
              [8,  5],
              [3,  3],
              [5,  6]])

S = np.array([[4, 2],
              [3, 3],
              [5, 6]])

result = [[i for i,row in enumerate(X) if (s==row).all()] for s in S]

or

result = [i for s in S for i,row in enumerate(X) if (s==row).all()]

if you want a flat list (assuming there is exactly one match per searched value).

Answer 4

Another way is to use cdist function from scipy.spatial.distance like this:

    np.nonzero(cdist(X, searched_values) == 0)[0]

Basically, we get row numbers of X which have distance zero to a row in searched_values, meaning they are equal. Makes sense if you look on rows as coordinates.

Answer 5

Another alternative is to use asvoid (below) to view each row as a single value of void dtype. This reduces a 2D array to a 1D array, thus allowing you to use np.in1d as usual:

import numpy as np

def asvoid(arr):
    """
    Based on http://stackoverflow.com/a/16973510/190597 (Jaime, 2013-06)
    View the array as dtype np.void (bytes). The items along the last axis are
    viewed as one value. This allows comparisons to be performed which treat
    entire rows as one value.
    """
    arr = np.ascontiguousarray(arr)
    if np.issubdtype(arr.dtype, np.floating):
        """ Care needs to be taken here since
        np.array([-0.]).view(np.void) != np.array([0.]).view(np.void)
        Adding 0. converts -0. to 0.
        """
        arr += 0.
    return arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))

X = np.array([[4,  2],
              [9,  3],
              [8,  5],
              [3,  3],
              [5,  6]])

searched_values = np.array([[4, 2],
                            [3, 3],
                            [5, 6]])

idx = np.flatnonzero(np.in1d(asvoid(X), asvoid(searched_values)))
print(idx)
# [0 3 4]

Answer 6

Approach #1

One approach would be to use NumPy broadcasting, like so -

np.where((X==searched_values[:,None]).all(-1))[1]

Approach #2

A memory efficient approach would be to convert each row as linear index equivalents and then using np.in1d, like so -

dims = X.max(0)+1
out = np.where(np.in1d(np.ravel_multi_index(X.T,dims),\
                    np.ravel_multi_index(searched_values.T,dims)))[0]

Approach #3

Another memory efficient approach using np.searchsorted and with that same philosophy of converting to linear index equivalents would be like so -

dims = X.max(0)+1
X1D = np.ravel_multi_index(X.T,dims)
searched_valuesID = np.ravel_multi_index(searched_values.T,dims)
sidx = X1D.argsort()
out = sidx[np.searchsorted(X1D,searched_valuesID,sorter=sidx)]

Please note that this np.searchsorted method assumes there is a match for each row from searched_values in X.

---

How does np.ravel_multi_index work?

This function gives us the linear index equivalent numbers. It accepts a 2D array of n-dimensional indices, set as columns and the shape of that n-dimensional grid itself onto which those indices are to be mapped and equivalent linear indices are to be computed.

Let's use the inputs we have for the problem at hand. Take the case of input X and note the first row of it. Since, we are trying to convert each row of X into its linear index equivalent and since np.ravel_multi_index assumes each column as one indexing tuple, we need to transpose X before feeding into the function. Since, the number of elements per row in X in this case is 2, the n-dimensional grid to be mapped onto would be 2D. With 3 elements per row in X, it would had been 3D grid for mapping and so on.

To see how this function would compute linear indices, consider the first row of X -

In [77]: X
Out[77]: 
array([[4, 2],
       [9, 3],
       [8, 5],
       [3, 3],
       [5, 6]])

We have the shape of the n-dimensional grid as dims -

In [78]: dims
Out[78]: array([10,  7])

Let's create the 2-dimensional grid to see how that mapping works and linear indices get computed with np.ravel_multi_index -

In [79]: out = np.zeros(dims,dtype=int)

In [80]: out
Out[80]: 
array([[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, 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, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0]])

Let's set the first indexing tuple from X, i.e. the first row from X into the grid -

In [81]: out[4,2] = 1

In [82]: out
Out[82]: 
array([[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, 0, 0],
       [0, 0, 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, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0]])

Now, to see the linear index equivalent of the element just set, let's flatten and use np.where to detect that 1.

In [83]: np.where(out.ravel())[0]
Out[83]: array([30])

This could also be computed if row-major ordering is taken into account.

Let's use np.ravel_multi_index and verify those linear indices -

In [84]: np.ravel_multi_index(X.T,dims)
Out[84]: array([30, 66, 61, 24, 41])

Thus, we would have linear indices corresponding to each indexing tuple from X, i.e. each row from X.

Choosing dimensions for np.ravel_multi_index to form unique linear indices

Now, the idea behind considering each row of X as indexing tuple of a n-dimensional grid and converting each such tuple to a scalar is to have unique scalars corresponding to unique tuples, i.e. unique rows in X.

Let's take another look at X -

In [77]: X
Out[77]: 
array([[4, 2],
       [9, 3],
       [8, 5],
       [3, 3],
       [5, 6]])

Now, as discussed in the previous section, we are considering each row as indexing tuple. Within each such indexing tuple, the first element would represent the first axis of the n-dim grid, second element would be the second axis of the grid and so on until the last element of each row in X. In essence, each column would represent one dimension or axis of the grid. If we are to map all elements from X onto the same n-dim grid, we need to consider the maximum stretch of each axis of such a proposed n-dim grid. Assuming we are dealing with positive numbers in X, such a stretch would be the maximum of each column in X + 1. That + 1 is because Python follows 0-based indexing. So, for example X[1,0] == 9 would map to the 10th row of the proposed grid. Similarly, X[4,1] == 6 would go to the 7th column of that grid.

So, for our sample case, we had -

In [7]: dims = X.max(axis=0) + 1 # Or simply X.max(0) + 1

In [8]: dims
Out[8]: array([10,  7])

Thus, we would need a grid of at least a shape of (10,7) for our sample case. More lengths along the dimensions won't hurt and would give us unique linear indices too.

Concluding remarks : One important thing to be noted here is that if we have negative numbers in X, we need to add proper offsets along each column in X to make those indexing tuples as positive numbers before using np.ravel_multi_index.