python - 对于数组中的所有唯一值，计算两个数组中相同值的数量

Question

我有两个数组 A 和 B。A 有多个值(这些值可以是字符串、整数或 float )，B 的值是 0 和 1。对于 A 中的每个唯一值，我需要重合点的计数B 中的 1 和 B 中的 0。这两个计数都需要存储为单独的变量。例如:

A = [1, 1, 3, 2, 2, 1, 1, 3, 3] # input multivalue array; it has three unique values – 1,2,3
B = [0, 0, 0, 1, 1, 1, 0, 1, 0] # input binary array
#Desired result: 
countA1_B1 = 1 #for unique value of '1' in A the count of places where there is '1' in B
countA1_B0 = 3 #for unique value of '1' in A the count of places where there is '0' in B
countAno1_B1 = 3 #for unique value of '1' in A the count of places where there is no '1' in A but there is '1' in B 
countAno1_B0 = 2 #for unique value of '1' in A the count of places where there is no '1' in A and there is '0' in B 

我需要这个来表示 A 中的所有唯一值。A 数组/列表将是一个栅格，因此将不知道唯一值。因此代码将首先提取 A 中的唯一值，然后进行剩余的计算我解决这个问题的方法(参见帖子 previous question :)

Import numpy as np
A = [1, 1, 3, 2, 2, 1, 1, 3, 3] # input array
B = [0, 0, 0, 1, 1, 1, 0, 1, 0] # input binary array
A_arr = np.array(A)
A_unq = np.unique(A_arr)
#code 1    
A_masked_arrays = np.array((A_arr[None, :] == A_unq[:, None]).astype(int)) 
#code 2
# A_masked_arrays = [(A==unique_val).astype(int) for unique_val in
np.unique(A)]
print(A_masked_arrays) 
out = {val: arr for val, arr in zip(list(A_unq), list(A_arr))} 
#zip() throws error
#TypeError: 'zip' object is not callable. 
dict = {}
for i in A_unq:
    for j in A_masked_arrays:
        dict = i, j
        print(dict)

得到的结果:

# from code 1
[[1 1 0 0 0 1 1 0 0]
 [0 0 0 1 1 0 0 0 0]
 [0 0 1 0 0 0 0 1 1]]
# from code 2
[array([1, 1, 0, 0, 0, 1, 1, 0, 0]), array([0, 0, 0, 1, 1, 0, 0, 0, 0]), 
array([0, 0, 1, 0, 0, 0, 0, 1, 1])]

使用字典创建我得到这个结果

(1, array([1, 1, 0, 0, 0, 1, 1, 0, 0]))
(1, array([0, 0, 0, 1, 1, 0, 0, 0, 0]))
(1, array([0, 0, 1, 0, 0, 0, 0, 1, 1]))
(2, array([1, 1, 0, 0, 0, 1, 1, 0, 0]))
(2, array([0, 0, 0, 1, 1, 0, 0, 0, 0]))
(2, array([0, 0, 1, 0, 0, 0, 0, 1, 1]))
(3, array([1, 1, 0, 0, 0, 1, 1, 0, 0]))
(3, array([0, 0, 0, 1, 1, 0, 0, 0, 0]))
(3, array([0, 0, 1, 0, 0, 0, 0, 1, 1]))

这就是我被困住的地方。从这里如何获得 A 中每个唯一值的最终计数，如 countA1_B1、countA1_B0、countAno1_B1、countAno1_B0 等。需要帮助。提前致谢。

Answer 1

选择性使用np.bincount应该可以解决问题

Au, Ai = np.unique(A, return_index = True)

out = np.empty((2, Au.size))
out[0] = np.bincount(Ai, weight = 1-np.array(B), size = Au.size)
out[1] = bp.bincount(Ai, weight = np.array(B),   size = Au.size)

outdict = {}

for i in range(Au.size):
    for j in [0, 1]:
        outdict[(Au(i), j)] = out[j, i]