计算numpy数组中每个实例的特定值

Question

我有一个像这样的纽比阵列

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

我想找出每一行，第二列中有多少元素在第二列值的某个距离内（比如10）。在这个例子中，解决方案是

^{pr2}$

所以[first row，third column]是3，因为第2列（57,51,67）中有3个元素，它们与57的距离在10之内。每一行的情况类似

任何帮助都将不胜感激！在

Answer 1

这里有一种利用^{}和outer-subtraction-

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

用outer subtract builtin和{}计数-

^{pr2}$

样本运行-

# 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]])

或者使用^{}-

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])

对于输入中的百万行，我们可能需要使用一个循环的-

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

这里有一个非广播的方法，它利用了这样一个事实：要知道有多少个数字在10的3/3范围内，可以从那些严格小于7的数字中减去<；=13。在

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

这通过了我的测试：

^{pr2}$

即使在大尺寸的情况下也表现良好：

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