如何使用python将对数概率转换为0到1之间的简单概率

Question

我正在使用高斯混合模型进行说话人识别。我使用此代码来预测每个语音剪辑的说话者。

for path in file_paths:   
path = path.strip()   
print (path)
sr,audio = read(source + path)
vector   = extract_features(audio,sr)
#print(vector)
log_likelihood = np.zeros(len(models))
#print(len(log_likelihood))
for i in range(len(models)):
    gmm1   = models[i]  #checking with each model one by one
    #print(gmm1)
    scores = np.array(gmm1.score(vector)) 
    #print(scores)
    #print(len(scores))
    log_likelihood[i] = scores.sum()
    print(log_likelihood)
    winner = np.argmax(log_likelihood)
    #print(winner)
print ("\tdetected as - ", speakers[winner])

它给了我这样的输出：

[ 311.79769716    0.            0.            0.            0.        ]
[  311.79769716 -5692.56559902     0.             0.             0.        ]
[  311.79769716 -5692.56559902 -6170.21460788     0.             0.        ]
[  311.79769716 -5692.56559902 -6170.21460788 -6736.73192695     0.        ]
[  311.79769716 -5692.56559902 -6170.21460788 -6736.73192695 -6753.00196447]
    detected as -  bart

在这里，得分函数为我提供了每个说话者的对数概率。现在，我想确定阈值，为此，我需要将这些对数概率值转换为简单概率值（0到1之间）。我怎样才能做到这一点？我正在使用python软件。

Answer 1

您必须采用np.exp()对数概率的指数（）才能取回实际概率。这是一个例子：

# some array
In [9]: a
Out[9]: array([1, 2, 3, 4, 5, 6, 7, 8, 9])

# softmax
In [10]: probs = np.exp(a) / (np.exp(a)).sum()

In [11]: probs.sum()
Out[11]: 1.0

# log probabilities
In [12]: log_probs = np.log(probs)

In [13]: log_probs
Out[13]: 
array([-8.45855173, -7.45855173, -6.45855173, -5.45855173, -4.45855173,
       -3.45855173, -2.45855173, -1.45855173, -0.45855173])

# mostly, won't sum to 1.0
In [14]: log_probs.sum()
Out[14]: -40.126965551706405

# get the probabilities back
In [15]: probabilities = np.exp(log_probs)

In [16]: probabilities.sum()
Out[16]: 1.0

In [17]: probabilities
Out[17]: 
array([  2.12078996e-04,   5.76490482e-04,   1.56706360e-03,
         4.25972051e-03,   1.15791209e-02,   3.14753138e-02,
         8.55587737e-02,   2.32572860e-01,   6.32198578e-01])