我想知道是否有人可以快速浏览以下代码片段,并指出一个方向,以发现我在计算模型中每个类的样本概率以及相关代码错误时的误解。我尝试手动计算sklearn函数lm.predict_proba(X)提供的结果,可悲的是结果是不同的,所以我做错了。
我认为该错误将在以下代码演练的“ d”部分中。也许在数学上,但我不明白为什么。
a)创建和训练逻辑回归模型(效果很好)
lm = LogisticRegression(random_state=413, multi_class='multinomial', solver='newton-cg')
lm.fit(X, train_labels)
b)节省系数和偏差(效果很好)
W = lm.coef_
b = lm.intercept_
c)使用lm.predict_proba(X)(工作正常)
def reshape_single_element(x,num):
singleElement = x[num]
nx,ny = singleElement.shape
return singleElement.reshape((1,nx*ny))
select_image_number = 6
X_select_image_data=reshape_single_element(train_dataset,select_image_number)
Y_probabilities = lm.predict_proba(X_select_image_data)
Y_pandas_probabilities = pd.Series(Y_probabilities[0], index=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'])
print"estimate probabilities for each class: \n" ,Y_pandas_probabilities , "\n"
print"all probabilities by lm.predict_proba(..) sum up to ", np.sum(Y_probabilities) , "\n"
输出为:
estimate probabilities for each class:
a 0.595426
b 0.019244
c 0.001343
d 0.004033
e 0.017185
f 0.004193
g 0.160380
h 0.158245
i 0.003093
j 0.036860
dtype: float64
all probabilities by lm.predict_proba(..) sum up to 1.0
d)手动执行lm.predict_proba完成的计算(无错误/警告,但结果不相同)
manual_calculated_probabilities = []
for select_class_k in range(0,10): #a=0. b=1, c=3 ...
z_for_class_k = (np.sum(W[select_class_k] *X_select_image_data) + b[select_class_k] )
p_for_class_k = 1/ (1 + math.exp(-z_for_class_k))
manual_calculated_probabilities.append(p_for_class_k)
print "formula: ", manual_calculated_probabilities , "\n"
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
e = np.exp(x)
dist = e / np.sum(np.exp(x),axis=0)
return dist
abc = softmax(manual_calculated_probabilities)
print "softmax:" , abc
输出为:
formula: [0.9667598370531315, 0.48453459121301334, 0.06154496922245115, 0.16456194859398865, 0.45634781280053394, 0.16999340794727547, 0.8867996361191054, 0.8854473986336552, 0.13124464656251109, 0.642913996162282]
softmax: [ 0.15329642 0.09464644 0.0620015 0.0687293 0.0920159 0.069103610.14151607 0.14132483 0.06647715 0.11088877]
由于在github logistic.py发表评论,所以使用了Softmax
For a multi_class problem, if multi_class is set to be "multinomial" the softmax function is used to find the predicted probability of each class.
注意:
print "shape of X: " , X_select_image_data.shape
print "shape of W: " , W.shape
print "shape of b: " , b.shape
shape of X: (1, 784)
shape of W: (10, 784)
shape of b: (10,)
我发现了一个非常相似的问题here,但遗憾的是我无法将其调整为我的代码,因此预测也相同。我尝试了许多不同的组合来计算变量'z_for_class_k'和'p_for_class_k',但遗憾的是,从'predict_proba(X)'再现预测值没有成功。
p_for_class_k = 1 /(1 + math.exp(-z_for_class_k))
1 / (1 + exp(-logit))
是仅用于二进制问题的简化。
在简化之前,实际方程看起来像这样:
p_for_classA =
exp(logit_classA) /
[1 + exp(logit_classA) + exp(logit_classB) ... + exp(logit_classC)]
换句话说,在计算特定类别的概率时,您必须将其他类别的所有权重和偏差也纳入公式中。
我没有数据可以验证这一点,但是希望这可以为您指明正确的方向。