python-2.7 - 如何使用 python scikit-learn 执行欠采样(正确的方式)？

Question

我正在尝试使用 python scikit learn 对多数类执行欠采样。目前，我的代码查找少数类的 N，然后尝试对多数类的完全相同的 N 进行欠采样。结果测试数据和训练数据都具有 1:1 的分布。但我真正想要的是仅在训练数据上执行此 1:1 分布，但在测试数据中的原始分布上进行测试。

我不太确定如何执行后者，因为中间有一些字典向量化，这让我感到困惑。

# Perform undersampling majority group
minorityN = len(df[df.ethnicity_scan == 1]) # get the total count of low-frequency group
minority_indices = df[df.ethnicity_scan == 1].index
minority_sample = df.loc[minority_indices]

majority_indices = df[df.ethnicity_scan == 0].index
random_indices = np.random.choice(majority_indices, minorityN, replace=False) # use the low-frequency group count to randomly sample from high-frequency group
majority_sample = data.loc[random_indices]

merged_sample = pd.concat([minority_sample, majority_sample], ignore_index=True) # merging all the low-frequency group sample and the new (randomly selected) high-frequency sample together
df = merged_sample
print 'Total N after undersampling:', len(df)

# Declaring variables
X = df.raw_f1.values
X2 = df.f2.values
X3 = df.f3.values
X4 = df.f4.values
y = df.outcome.values

# Codes skipped ....
def feature_noNeighborLoc(locString):
    pass
my_dict16 = [{'location': feature_noNeighborLoc(feature_full_name(i))} for i in X4]
# Codes skipped ....

# Dict vectorization
all_dict = []
for i in range(0, len(my_dict)):
    temp_dict = dict(
        my_dict[i].items() + my_dict2[i].items() + my_dict3[i].items() + my_dict4[i].items()
        + my_dict5[i].items() + my_dict6[i].items() + my_dict7[i].items() + my_dict8[i].items()
        + my_dict9[i].items() + my_dict10[i].items()
        + my_dict11[i].items() + my_dict12[i].items() + my_dict13[i].items() + my_dict14[i].items()
        + my_dict19[i].items()
        + my_dict16[i].items() # location feature
        )
all_dict.append(temp_dict)

newX = dv.fit_transform(all_dict)

X_train, X_test, y_train, y_test = cross_validation.train_test_split(newX, y, test_size=testTrainSplit)

# Fitting X and y into model, using training data
classifierUsed2.fit(X_train, y_train)

# Making predictions using trained data
y_train_predictions = classifierUsed2.predict(X_train)
y_test_predictions = classifierUsed2.predict(X_test)

Answer 1

您想要对某个类别的训练样本进行二次采样，因为您需要一个对所有标签进行相同处理的分类器。

如果您想这样做而不是二次采样，您可以将分类器的“class_weight”参数的值更改为“平衡”(或某些分类器的“自动”)，这可以完成您想要做的工作。

您可以阅读 LogisticRegression 分类器的文档作为示例。请注意“class_weight”参数的描述here .

通过将该参数更改为“平衡”，您将不再需要进行子采样。