CODEWITHSUNDEEP
pythonpython-3.xnumpyisinstance
zeniapy
zeniapy

Reputation: 175

np.isin - testing whether a Numpy array contains a given row considering the order

I am using the following line to find if the rows of b are in a

 a[np.all(np.isin(a[:, 0:3], b[:, 0:3]), axis=1), 3]

The arrays have more entries along axis=1, I only compare the first 3 entries and return the fourth entry (idx=3) of a.

The possible error I realized is, that the order of the entries is not considered. Therefore, the following example for a and b:

a = np.array([[...],
              [1, 2, 3, 1000],
              [2, 1, 3, 2000],
              [...]])

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

would return [1000, 2000] instead of the of only [1000].

How can I consider the order of the rows as well?

Upvotes: 0

Views: 1917

Answers (1)

Pierre D
Pierre D

Reputation: 26321

For small b (less than 100 rows), try this instead:

a[(a[:, :3] == b[:, None]).all(axis=-1).any(axis=0)]

Example:

a = np.array([[1, 0, 5, 0],
              [1, 2, 3, 1000],
              [2, 1, 3, 2000],
              [0, 0, 1, 1]])

b = np.array([[1, 2, 3], [0, 0, 1]])

>>> a[(a[:, :3] == b[:, None]).all(axis=-1).any(axis=0), 3]
array([1000,    1])

Explanation:

The key is to "distribute" equality tests for all rows of a (the first 3 columns) to all rows of b:

# on the example above

>>> a[:, :3] == b[:, None]
array([[[ True, False, False],
        [ True,  True,  True],  # <-- a[1,:3] matches b[0]
        [False, False,  True],
        [False, False, False]],

       [[False,  True, False],
        [False, False, False],
        [False, False, False],
        [ True,  True,  True]]])  # <-- a[3, :3] matches b[1]

Be warned that this can be large: the shape is (len(b), len(a), 3).

Then the first .all(axis=-1) means that we want all entire rows to match:

>>> (a[:, :3] == b[:, None]).all(axis=-1)
array([[False,  True, False, False],
       [False, False, False,  True]])

The final bit .any(axis=0) means: "match any row in b":

>>> (a[:, :3] == b[:, None]).all(axis=-1).any(axis=0)
array([False,  True, False,  True])

I.e.: "a[2, :3] matches some row(s) of b as well as a[3, :3]".

Finally, use this as a mask in a and take the column 3.

Note on performance

The technique above distributes equality for the product of the rows of a over the rows of b. This can be slow and use a large amount of memory if both a and b have many rows.

As an alternative, you may use set membership in pure Python (without subsetting of columns --that can be done by the caller):

def py_rows_in(a, b):
    z = set(map(tuple, b))
    return [row in z for row in map(tuple, a)]

When b has more than 50~100 rows, then this may be faster, compared to the np version above, written here as a function:

def np_rows_in(a, b):
    return (a == b[:, None]).all(axis=-1).any(axis=0)

import perfplot

fig, axes = plt.subplots(ncols=2, figsize=(16, 5))
plt.subplots_adjust(wspace=.5)
for ax, alen in zip(axes, [100, 10_000]):
    a = np.random.randint(0, 20, (alen, 4))
    plt.sca(ax)
    ax.set_title(f'a: {a.shape[0]:_} rows')
    perfplot.show(
        setup=lambda n: np.random.randint(0, 20, (n, 3)),
        kernels=[
            lambda b: np_rows_in(a[:, :3], b),
            lambda b: py_rows_in(a[:, :3], b),
        ],
        labels=['np_rows_in', 'py_rows_in'],
        n_range=[2 ** k for k in range(10)],
        xlabel='len(b)',
    )
plt.show()

comparative performance

Upvotes: 1

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