计算 numpy 数组列的唯一值概率

Question

我想根据预测列表创建一个 scikit-learns predict_proba 版本。

我目前有一个如下所示的列表：

[[0,1,0,0,0,1,1,0,0,0],[0,1,0,1,0,1,1,1,0,0],[0,0,0,0,0,1,1,0,0,0]]

我想找出每个列表的第一个值是 0 或 1 的概率，然后每个连续值都相同。

即输出是这样的：

[[0.33,0.66],[0,1],[0.66,0.3]........etc

我写了下面的代码，它运行良好，但看起来很笨拙，我确定有更好的方法来实现我的目标吗？

有什么建议吗？

#create np array from list 
ar = np.array([[0,1,0,0,0,1,1,0,0,0],[0,1,0,1,0,1,1,1,0,0],[0,0,0,0,0,1,1,0,0,0]])

#calculate unique values and sort in order
uni = np.unique(ar)
uni.sort()

#create new pred list
new_pred = []

#transpose and iterate
for row in ar.transpose():
    # create dic with keys as unique values
    val_dic = {k: 0 for k in uni}    

    #create list for row probabilities
    row_pred = []

    #iterate row and incremnet dic if found
    for val in row:
        if val in val_dic.keys():
            val_dic[val] = val_dic.get(val, 0) + 1

    #calc row total 
    total = sum(val_dic.values())

    #append row list with probabilities
    for val in val_dic.values():
        row_pred.append(val/total)       

    #append final output list
    new_pred.append(row_pred)

print(new_pred)

输出：

[[1.0, 0.0], [0.3333333333333333, 0.6666666666666666], [1.0, 0.0], [0.6666666666666666, 0.3333333333333333], [1.0, 0.0], [0.0, 1.0], [0.0, 1.0], [0.6666666666666666, 0.3333333333333333], [1.0, 0.0], [1.0, 0.0]]

Answer 1

如果您的 ar 仅由 0、1 组成，就像您的问题一样，您可以这样做来简化您的代码：

import numpy as np

ar = np.array([[0,1,0,0,0,1,1,0,0,0],[0,1,0,1,0,1,1,1,0,0],[0,0,0,0,0,1,1,0,0,0]])

prob_1 = ar.T.sum(axis=1) / len(ar)          # <-- max sum of row is len(ar) == 3
prob_0 = 1.0 - prob_1
print(np.column_stack((prob_0, prob_1)))

打印：

[[1.         0.        ]
 [0.33333333 0.66666667]
 [1.         0.        ]
 [0.66666667 0.33333333]
 [1.         0.        ]
 [0.         1.        ]
 [0.         1.        ]
 [0.66666667 0.33333333]
 [1.         0.        ]
 [1.         0.        ]]

计算 numpy 数组列的唯一值概率

Calculate unique value probability over numpy array columns

python

numpy

probability