Sklearn - 从Logistic回归中返回前三类

Posted 2021-05-03

我正在尝试创建一个模型，将客户的电子邮件分类（“案例原因”）。我已经清理了停止词等，并测试了几个不同的模型，Logistic回归是最准确的。问题是它只有70％的时间是准确的。这主要是因为数据的扩展问题（有少数案例原因可以获得大部分电子邮件。

我不想仅仅预测单一结果，而是尝试给代理商提供前三名（或者可能是5名）的选择。

这是我已经拥有的：

# vectorize the text
tfidf = TfidfVectorizer(sublinear_tf=True, min_df=5, norm='l2', encoding='latin-1', 
ngram_range=(1, 2), stop_words=internal_stop_words)

features = tfidf.fit_transform(df.Description).toarray()
labels = df.category_id
features.shape

在对所有内容进行矢量化后，我通过以下块运行它来测试哪4个模型最适合。这表明Logistic回归率为70％，是四者中最好的：

models = [
    RandomForestClassifier(n_estimators=200, max_depth=3, random_state=0),
    LinearSVC(),
    MultinomialNB(),
    LogisticRegression(random_state=0),
]
CV = 5
cv_df = pd.DataFrame(index=range(CV * len(models)))
entries = []
for model in models:
  model_name = model.__class__.__name__
  accuracies = cross_val_score(model, features, labels, scoring='accuracy', cv=CV)
  for fold_idx, accuracy in enumerate(accuracies):
    entries.append((model_name, fold_idx, accuracy))
cv_df = pd.DataFrame(entries, columns=['model_name', 'fold_idx', 'accuracy'])

我创建了分类器，它是传递值的函数：

X_train, X_test, y_train, y_test = train_test_split(df['Description'], df['Reason'], 
       random_state = 0)
count_vect = CountVectorizer()
X_train_counts = count_vect.fit_transform(X_train)
tfidf_transformer = TfidfTransformer()
X_train_tfidf = tfidf_transformer.fit_transform(X_train_counts)

clf = LogisticRegression(solver='saga',multi_class='multinomial').fit(X_train_tfidf, y_train)

print(clf.predict(count_vect.transform(["""i dont know my password"""])))

['Reason #1']

在这种情况下，这不是正确的原因。我可以运行以下命令来获得一个表格，显示每个分类的概率：

#Test log res
probs = clf.predict_proba(count_vect.transform(["""I dont know my password"""]))
classes = clf.classes_

probs.shape = (len(category_to_id),1)
output = pd.DataFrame(data=[classes,probs]).T
output.columns= ['reason','prob']
output.sort_values(by='prob', ascending=False)

返回：

index       reason        prob
7           Reason #7     [0.6036937161535804]
6           Reason #6     [0.1576980112870697]
3           Reason #3     [0.13221805369421305]
13          Reason #13    [0.028848040868062686]
8           Reason #8     [0.02264491874676607]
9           Reason #9     [0.01725043255540921]
0           Reason #0     [0.01600640516713904]
10          Reason #10    [0.005444588928021622]
4           Reason #4     [0.0052240828713529894]
5           Reason #5     [0.0048409867159243045]
2           Reason #2     [0.0024794864823573935]
1           Reason #1     [0.0014065266971805264]
11          Reason #11    [0.001393613395266496]
12          Reason #12    [0.0008511364376563769]

所以我按最可能的原因排序，在这种情况下，＃3是正确的答案。

如何将前N个结果返回到输入，以及测试N个结果之一中存在的实际原因的模型精度？

答案

您可以按降序对概率进行排序并检索top-n。要计算准确度，您可以定义自定义函数，如果y_true属于top-n，则会将预测视为正确。沿着这些方向的东西应该有效：

probs = clf.predict_proba(X_test)
# Sort desc and only extract the top-n
top_n = np.argsort(probs)[:,:-n-1:-1]

# Calculate accuracy
true_preds = 0
for i in range(len(y_test)):
    if y_test[i] in top_n[i]:
        true_preds += 1

accuracy = true_preds/len(y_test)