计算每行特定值内的numpy数组中的实例

Question

I have a numpy array such as this

[[ 0, 57],
 [ 7, 72],
 [ 2, 51],
 [ 8, 67],
 [ 4, 42]]

I want to find out for each row, how many elements in the 2nd column are within a certain distance (say, 10) of the 2nd column value for that row. So in this example, here the solution would be

[[ 0, 57, 3],
 [ 7, 72, 2],
 [ 2, 51, 3],
 [ 8, 67, 3],
 [ 4, 42, 2]]

So [first row, third column] is 3, because there are 3 elements in the 2nd column (57,51,67) which are within distance 10 from 57. Similarly for each row

Any help would be appreciated!

Answer 1

Here's one approach leveraging broadcasting with outer-subtraction -

(np.abs(a[:,1,None] - a[:,1]) <= 10).sum(1)

With outer subtract builtin and count_nonzero for counting -

np.count_nonzero(np.abs(np.subtract.outer(a[:,1],a[:,1]))<=10,axis=1)

Sample run -

# Input array
In [23]: a
Out[23]: 
array([[ 0, 57],
       [ 7, 72],
       [ 2, 51],
       [ 8, 67],
       [ 4, 42]])

# Get count
In [24]: count = (np.abs(a[:,1,None] - a[:,1]) <= 10).sum(1)

In [25]: count
Out[25]: array([3, 2, 3, 3, 2])

# Stack with input
In [26]: np.c_[a,count]
Out[26]: 
array([[ 0, 57,  3],
       [ 7, 72,  2],
       [ 2, 51,  3],
       [ 8, 67,  3],
       [ 4, 42,  2]])

Alternatively with SciPy's cdist -

In [53]: from scipy.spatial.distance import cdist

In [54]: (cdist(a[:,None,1],a[:,1,None], 'minkowski', p=2)<=10).sum(1)
Out[54]: array([3, 2, 3, 3, 2])

For million rows in the input, we might want to resort to a loopy one -

n = len(a)
count = np.empty(n, dtype=int)
for i in range(n):
    count[i] = np.count_nonzero(np.abs(a[:,1]-a[i,1])<=10)

Answer 2

Here's a non-broadcasting approach, which takes advantage of the fact that to know how many numbers are within 3 of 10, you can subtract the number of numbers <= 13 from those strictly less than 7.

import numpy as np

def broadcast(x, width):
    # for comparison
    return (np.abs(x[:,None] - x) <= width).sum(1)


def largest_leq(arr, x, allow_equal=True):
    maybe = np.searchsorted(arr, x)
    maybe = maybe.clip(0, len(arr) - 1)
    above = arr[maybe] > x if allow_equal else arr[maybe] >= x
    maybe[above] -= 1
    return maybe

def faster(x, width):
    uniq, inv, counts = np.unique(x, return_counts=True, return_inverse=True)
    counts = counts.cumsum()

    low_bounds = uniq - width
    low_ix = largest_leq(uniq, low_bounds, allow_equal=False)
    low_counts = counts[low_ix]
    low_counts[low_ix < 0] = 0

    high_bounds = uniq + width
    high_counts = counts[largest_leq(uniq, high_bounds)]

    delta = high_counts - low_counts
    out = delta[inv]
    return out

This passes my tests:

for width in range(1, 10):
    for window in range(5):
        for trial in range(10):
            x = np.random.randint(0, 10, width)
            b = broadcast(x, window).tolist()
            f = faster(x, window).tolist()
            assert b == f

and behaves pretty well even at larger sizes:

In [171]: x = np.random.random(10**6)

In [172]: %time faster(x, 0)
Wall time: 386 ms
Out[172]: array([1, 1, 1, ..., 1, 1, 1], dtype=int64)

In [173]: %time faster(x, 1)
Wall time: 372 ms
Out[173]: array([1000000, 1000000, 1000000, ..., 1000000, 1000000, 1000000], dtype=int64)

In [174]: x = np.random.randint(0, 10, 10**6)

In [175]: %timeit faster(x, 3)
10 loops, best of 3: 83 ms per loop