问题描述
我有两个数组A和B.A有多个值(这些值可以是字符串，整数或浮点数)，B的值是0和1.对于A中的每个唯一值，我需要重合的点数B中的1和B中的0.两个计数都需要存储为单独的变量.例如:
I have two arrays A and B. A has multiple values (these values can be string or integer or float) and B has values 0 and 1. I need, for each unique value in A, the count of points that coincide with the 1s in B and the 0s in B. Both the counts need to be stored as separate variables.For example:
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，然后执行剩余的计算我解决这个问题的方法(请参见:)
I need this for all the unique values in A. The A array/list would be a raster and hence the unique values will not be known. So the code would first extract the unique values in A and then do the remaining calculationsMy approach to solving this (see post 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)
获得的结果
# 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
Using dictionary creation I get this result
(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.预先感谢.
This is where I am stuck up. From here how to get to the final count of each unique value in A as countA1_B1, countA1_B0, countAno1_B1, countAno1_B0 and so on. Need help with this. Thanks in advance.
推荐答案
使用熊猫进行这种分组操作要容易得多:
It's much easier to use pandas to do this kind of groupby operation:
In [11]: import pandas as pd
In [12]: df = pd.DataFrame({"A": A, "B": B})
In [13]: df
Out[13]:
A B
0 1 0
1 1 0
2 3 0
3 2 1
4 2 1
5 1 1
6 1 0
7 3 1
8 3 0
现在您可以使用groupby:
Now you can use groupby:
In [14]: gb = df.groupby("A")["B"]
In [15]: gb.count() # number of As
Out[15]:
A
1 4
2 2
3 3
Name: B, dtype: int64
In [16]: gb.sum() # number of As where B == 1
Out[16]:
A
1 1
2 2
3 1
Name: B, dtype: int64
In [17]: gb.count() - gb.sum() # number of As where B == 0
Out[17]:
A
1 3
2 0
3 2
Name: B, dtype: int64
您还可以通过Apply来更明确，更广泛地执行此操作(例如，如果它不仅是0和1)，
You can also do this more explicitly and more generally (e.g. if it's not just 0 and 1) with an apply:
In [18]: gb.apply(lambda x: (x == 1).sum())
Out[18]:
A
1 1
2 2
3 1
Name: B, dtype: int64
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