python - 从另一个数组中的一个numpy数组中获取值的频率

Question

我有两个 numpy 数组，例如：

import numpy as np
a1 = np.linspace(0,2*np.pi,101)
a2 = np.random.choice(a1, 60)

我需要计算每个值a1出现在a2. 我可以用循环来做到这一点，但我希望有一个更好的解决方案。

带循环的解决方案：

a3 = np.zeros_like(a1)
for i in range(len(a1)):
    a3[i] = np.sum(a2==a1[i])

Answer 1

这是使用 perfplot ( https://github.com/nschloe/perfplot ) 的不同解决方案的性能比较：

import perfplot
import numpy as np
from collections import Counter

def count_a_in_b_loop(b, a = np.linspace(0,2*np.pi,101)):
    c = np.zeros_like(a)
    for i in range(len(a)):
        c[i] = np.sum(b==a[i])
    return c

def count_a_in_b_counter(b, a=np.linspace(0,2*np.pi,101)):
    c = Counter(b)
    c = np.array([(c[k] if k in c else 0)  for k in a])
    return c

def count_occ(a2,a1=np.linspace(0,2*np.pi,101),use_closeness=True):
    # Trace back indices for each elem of a2 in a1
    idx = np.searchsorted(a1,a2)
    # Set out of bounds indices to something within
    idx[idx==len(a1)] = 0
    # Check for the matches
    if use_closeness==1:
        mask = np.isclose(a1[idx],a2)
    else:
        mask = a1[idx] == a2
    # Get counts
    return np.bincount((idx+1)*mask,minlength=len(a1)+1)[1:]

def count_broadcasting(a2, a1=np.linspace(0,2*np.pi,101)):
    (a1[:,None]==a2).sum(1) # For exact matches
    return np.isclose(a1[:,None],a2).sum(1) # For close matches

def count_occ_rounding(a2, a1_lastnum=2*np.pi, a1_num_sample=101):
    s = a1_lastnum/(a1_num_sample-1)
    p = np.round(a2/s).astype(int)
    return np.bincount(p,minlength=a1_num_sample)

def count_add_to_unique(a2, a1=np.linspace(0,2*np.pi,101)):
    unq, idx, cnts = np.unique(np.concatenate([a1, a2]), return_inverse = True, return_counts = True)
    #assert np.all(unq[idx[:len(a1)]] == a1)
    return cnts[idx[:len(a1)]] - 1

perfplot.show(
        setup=lambda n: np.random.choice(np.linspace(0,2*np.pi,101), n),
        kernels=[
            count_a_in_b_loop, count_a_in_b_counter, count_occ, count_broadcasting, count_occ_rounding, add_to_unique
            ],
        labels=['loop', 'counter','searchsorted','broadcasting','occ_rounding','add_to_unique'],
        n_range=[2**k for k in range(15)],
        xlabel='len(a)'
        )

Answer 2

另一种np.unique方法：

>>> import numpy as np
>>> a1 = np.linspace(0,2*np.pi,101)
>>> a2 = np.random.choice(a1, 60)
>>> 
>>> unq, idx, cnts = np.unique(np.concatenate([a1, a2]), return_inverse=True, return_counts=True)
>>> assert np.all(unq[idx[:len(a1)]] == a1)
>>> result = cnts[idx[:len(a1)]] - 1
>>> result
array([0, 0, 2, 0, 2, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0,
       0, 1, 0, 0, 0, 1, 1, 1, 2, 0, 0, 1, 2, 1, 0, 2, 0, 0, 0, 1, 0, 2,
       0, 1, 2, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 4, 0, 0, 0, 1, 1, 1, 0, 0,
       2, 0, 0, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 2, 0, 0, 0, 1, 1, 1, 1,
       0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 2, 1])

Answer 3

如果我正确理解您的问题，这是使用np.uniqueand的一种方法np.isin：

import numpy as np

a1 = np.linspace(0,2*np.pi,101)
a2 = np.random.choice(a1, 60)

vals_counts = np.unique(a2, return_counts=True)

arr = np.array(list(zip(*vals_counts)))

print(arr.shape)
# (46, 2)

res = arr[np.where(np.isin(arr[:, 0], a1))]

print(res.shape)
# (46, 2)

print(res)

[[ 0.06283185  1.        ]
 [ 0.12566371  1.        ]
 ...
 [ 5.65486678  3.        ]
 [ 5.96902604  2.        ]
 [ 6.09468975  1.        ]
 [ 6.1575216   1.        ]
 [ 6.28318531  1.        ]]

Answer 4

方法#1

一种基于np.searchsorted浮点数并结合其接近性的矢量化解决方案是 -

def count_occ(a1,a2,use_closeness=True):
    # Trace back indices for each elem of a2 in a1
    idx = np.searchsorted(a1,a2)

    # Set out of bounds indices to something within
    idx[idx==len(a1)] = 0

    # Check for the matches
    if use_closeness==1:
        mask = np.isclose(a1[idx],a2)
    else:
        mask = a1[idx] == a2

    # Get counts
    return np.bincount((idx+1)*mask,minlength=len(a1)+1)[1:]

样品运行 -

In [154]: a1 = np.array([1.0000001,4,5,6])

In [155]: a2 = np.array([2,5,8,5,8,5,0.999999999999])

In [156]: count_occ(a1,a2)
Out[156]: array([1, 0, 3, 0])

In [157]: count_occ(a1,a2,use_closeness=False)
Out[157]: array([0, 0, 3, 0])

方法#2

或者，我们也可以使用broadcasting一种简短但占用大量内存的方法，如下所示 -

(a1[:,None]==a2).sum(1) # For exact matches
np.isclose(a1[:,None],a2).sum(1) # For close matches

a1方法 #3：使用lin-spaced 数据的特定情况

对于特定情况下a1是 lin-spaced array 并再次考虑closeness，我们可以优化数据的进一步使用rounding，a2如下所示 -

def count_occ_rounding(a2, a1_startnum=0,a1_lastnum=2*np.pi, a1_num_sample=101):
    s = (a1_lastnum-a1_startnum)/(a1_num_sample-1)
    p = np.round((a2 - a1_startnum)/s).astype(int)    
    return np.bincount(p,minlength=a1_num_sample)

示例运行以验证具有通用开始、结束范围数组的输出a1-

In [284]: a1 = np.linspace(-2*np.pi,2*np.pi,201)
     ...: a2 = np.random.choice(a1, 60)
     ...: out1 = count_occ_rounding(a2, -2*np.pi, 2*np.pi, 201)
     ...: out2 = np.isclose(a1[:,None],a2).sum(1)
     ...: print np.allclose(out1, out2)
True