逻辑回归的准确性
Accuracy in logistic regression
这是我发现的稍微修改过的代码 here...
我使用的逻辑与原作者相同,但仍然没有得到很好的准确性。平均倒数排名接近(我的:52.79,示例:48.04)
cv = CountVectorizer(binary=True, max_df=0.95)
feature_set = cv.fit_transform(df["short_description"])
X_train, X_test, y_train, y_test = train_test_split(
feature_set, df["category"].values, random_state=2000)
scikit_log_reg = LogisticRegression(
verbose=1, solver="liblinear", random_state=0, C=5, penalty="l2", max_iter=1000)
model = scikit_log_reg.fit(X_train, y_train)
target = to_categorical(y_test)
y_pred = model.predict_proba(X_test)
label_ranking_average_precision_score(target, y_pred)
>> 0.5279108613021547
model.score(X_test, y_test)
>> 0.38620071684587814
但是笔记本样本的准确性 (59.80) 与我的代码 (38.62) 不匹配
示例笔记本中使用的以下函数是否正确返回准确度?
def compute_accuracy(eval_items:list):
correct=0
total=0
for item in eval_items:
true_pred=item[0]
machine_pred=set(item[1])
for cat in true_pred:
if cat in machine_pred:
correct+=1
break
accuracy=correct/float(len(eval_items))
return accuracy
笔记本代码正在检查实际类别是否在模型返回的前 3 名中:
def get_top_k_predictions(model, X_test, k):
probs = model.predict_proba(X_test)
best_n = np.argsort(probs, axis=1)[:, -k:]
preds=[[model.classes_[predicted_cat] for predicted_cat in prediction] for prediction in best_n]
preds=[item[::-1] for item in preds]
return preds
如果您将代码的评估部分替换为以下代码,您将看到您的模型 returns 的前 3 准确度也为 0.5980:
...
model = scikit_log_reg.fit(X_train, y_train)
top_preds = get_top_k_predictions(model, X_test, 3)
pred_pairs = list(zip([[v] for v in y_test], top_preds))
print(compute_accuracy(pred_pairs))
# below is a simpler & more Pythonic version of compute_accuracy
print(np.mean([actual in pred for actual, pred in zip(y_test, top_preds)]))
这是我发现的稍微修改过的代码 here...
我使用的逻辑与原作者相同,但仍然没有得到很好的准确性。平均倒数排名接近(我的:52.79,示例:48.04)
cv = CountVectorizer(binary=True, max_df=0.95)
feature_set = cv.fit_transform(df["short_description"])
X_train, X_test, y_train, y_test = train_test_split(
feature_set, df["category"].values, random_state=2000)
scikit_log_reg = LogisticRegression(
verbose=1, solver="liblinear", random_state=0, C=5, penalty="l2", max_iter=1000)
model = scikit_log_reg.fit(X_train, y_train)
target = to_categorical(y_test)
y_pred = model.predict_proba(X_test)
label_ranking_average_precision_score(target, y_pred)
>> 0.5279108613021547
model.score(X_test, y_test)
>> 0.38620071684587814
但是笔记本样本的准确性 (59.80) 与我的代码 (38.62) 不匹配
示例笔记本中使用的以下函数是否正确返回准确度?
def compute_accuracy(eval_items:list):
correct=0
total=0
for item in eval_items:
true_pred=item[0]
machine_pred=set(item[1])
for cat in true_pred:
if cat in machine_pred:
correct+=1
break
accuracy=correct/float(len(eval_items))
return accuracy
笔记本代码正在检查实际类别是否在模型返回的前 3 名中:
def get_top_k_predictions(model, X_test, k):
probs = model.predict_proba(X_test)
best_n = np.argsort(probs, axis=1)[:, -k:]
preds=[[model.classes_[predicted_cat] for predicted_cat in prediction] for prediction in best_n]
preds=[item[::-1] for item in preds]
return preds
如果您将代码的评估部分替换为以下代码,您将看到您的模型 returns 的前 3 准确度也为 0.5980:
...
model = scikit_log_reg.fit(X_train, y_train)
top_preds = get_top_k_predictions(model, X_test, 3)
pred_pairs = list(zip([[v] for v in y_test], top_preds))
print(compute_accuracy(pred_pairs))
# below is a simpler & more Pythonic version of compute_accuracy
print(np.mean([actual in pred for actual, pred in zip(y_test, top_preds)]))