test for membership in a 2d numpy array

I have two 2D arrays of the same size

a = array([[1,2],[3,4],[5,6]])
b = array([[1,2],[3,4],[7,8]])

I want to know the rows of b that are in a.

So the output should be :

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

without making :

array([any(i == a) for i in b])

cause a and b are huge.

There is a function that does this but only for 1D arrays : in1d

Answers:

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Method 1

What we’d really like to do is use np.in1d… except that np.in1d only works with 1-dimensional arrays. Our arrays are multi-dimensional. However, we can view the arrays as a 1-dimensional array of strings:

arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))

For example,

In [15]: arr = np.array([[1, 2], [2, 3], [1, 3]])

In [16]: arr = arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))

In [30]: arr.dtype
Out[30]: dtype('V16')

In [31]: arr.shape
Out[31]: (3, 1)

In [37]: arr
Out[37]: 
array([[b'x01x00x00x00x00x00x00x00x02x00x00x00x00x00x00x00'],
       [b'x02x00x00x00x00x00x00x00x03x00x00x00x00x00x00x00'],
       [b'x01x00x00x00x00x00x00x00x03x00x00x00x00x00x00x00']],
      dtype='|V16')

This makes each row of arr a string. Now it is just a matter of hooking this up
to np.in1d:

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 on the entire row.
    """
    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])))


def inNd(a, b, assume_unique=False):
    a = asvoid(a)
    b = asvoid(b)
    return np.in1d(a, b, assume_unique)


tests = [
    (np.array([[1, 2], [2, 3], [1, 3]]),
     np.array([[2, 2], [3, 3], [4, 4]]),
     np.array([False, False, False])),
    (np.array([[1, 2], [2, 2], [1, 3]]),
     np.array([[2, 2], [3, 3], [4, 4]]),
     np.array([True, False, False])),
    (np.array([[1, 2], [3, 4], [5, 6]]),
     np.array([[1, 2], [3, 4], [7, 8]]),
     np.array([True, True, False])),
    (np.array([[1, 2], [5, 6], [3, 4]]),
     np.array([[1, 2], [5, 6], [7, 8]]),
     np.array([True, True, False])),
    (np.array([[-0.5, 2.5, -2, 100, 2], [5, 6, 7, 8, 9], [3, 4, 5, 6, 7]]),
     np.array([[1.0, 2, 3, 4, 5], [5, 6, 7, 8, 9], [-0.5, 2.5, -2, 100, 2]]),
     np.array([False, True, True]))
]

for a, b, answer in tests:
    result = inNd(b, a)
    try:
        assert np.all(answer == result)
    except AssertionError:
        print('''
a:
{a}
b:
{b}

answer: {answer}
result: {result}'''.format(**locals()))
        raise
else:
    print('Success!')

yields

Success!

Method 2

In [1]: import numpy as np

In [2]: a = np.array([[1,2],[3,4]])

In [3]: b = np.array([[3,4],[1,2]])

In [5]: a = a[a[:,1].argsort(kind='mergesort')]

In [6]: a = a[a[:,0].argsort(kind='mergesort')]

In [7]: b = b[b[:,1].argsort(kind='mergesort')]

In [8]: b = b[b[:,0].argsort(kind='mergesort')]

In [9]: bInA1 = b[:,0] == a[:,0]

In [10]: bInA2 = b[:,1] == a[:,1]

In [11]: bInA = bInA1*bInA2

In [12]: bInA
Out[12]: array([ True,  True], dtype=bool)

should do this… Not sure, whether this is still efficient. You need do mergesort, as other methods are unstable.

Edit:

If you have more than 2 columns and if the rows are sorted already, you can do

In [24]: bInA = np.array([True,]*a.shape[0])

In [25]: bInA
Out[25]: array([ True,  True], dtype=bool)

In [26]: for k in range(a.shape[1]):
    bInAk = b[:,k] == a[:,k]
    bInA = bInAk*bInA
   ....:     

In [27]: bInA
Out[27]: array([ True,  True], dtype=bool)

There is still space for speeding up, as in the iteration, you don’t have to check the entire column, but only the entries where the current bInA is True.

Method 3

If you have smth like a=np.array([[1,2],[3,4],[5,6]]) and b=np.array([[5,6],[1,2],[7,6]]), you can convert them into complex 1-D array:

c=a[:,0]+a[:,1]*1j
d=b[:,0]+b[:,1]*1j

This whole stuff in my Interpreter looks like this:

>>> c=a[:,0]+a[:,1]*1j
>>> c
array([ 1.+2.j,  3.+4.j,  5.+6.j])
>>> d=b[:,0]+b[:,1]*1j
>>> d
array([ 5.+6.j,  1.+2.j,  7.+6.j])

And now that you have just 1D array, you can easily do np.in1d(c,d), and the Python will give you:

>>> np.in1d(c,d)
array([ True, False,  True], dtype=bool)

And with this you don’t need any loops, at least with this data type

Method 4

the numpy module can actually broadcast through your array and tell what parts are the same as the other and return true if they are and false if they are not:

import numpy as np
a = np.array(([1,2],[3,4],[5,6])) #converting to a numpy array
b = np.array(([1,2],[3,4],[7,8])) #converting to a numpy array
new_array = a == b #creating a new boolean array from comparing a and b

now new_array looks like this:

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

but that is not what you want. So you can transpose (flip x and y) the array and then compare the two rows with an & gate. This will now create a 1-D array that will only return true if both columns in the row are true:

new_array = new_array.T #transposing
result = new_array[0] & new_array[1] #comparing rows

when you print result you now get what you’re looking for:

[ True  True False]

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0