如何使用 LabelBinarizer 对正确的训练和测试进行一次热编码

Question

假设我有这样的火车：

Name   | day
------------
First  |  0
Second |  1
Third  |  1
Forth  |  2

还有一个不包含所有这些名称或日期的测试集。像这样：

Name   | day
------------
First  |  2
Second |  1
Forth  |  0

我有以下代码来转换编码特征中的这些列：

features_to_encode = ['Name', 'day']
label_final = pd.DataFrame()

for feature in features_to_encode:
    label_campaign = LabelBinarizer()
    label_results = label_campaign.fit_transform(df[feature])
    label_results = pd.DataFrame(label_results, columns=label_campaign.classes_)
    label_final = pd.concat([label_final, label_results], axis=1)

df_encoded = label_final.join(df)

在火车上产生以下输出（效果很好）：

First  | Second  | Third  | Forth | 0 | 1 | 2 | 
-----------------------------------------------
  1    |    0    |   0    |   0   | 1 | 0 | 0 |
  0    |    1    |   0    |   0   | 0 | 1 | 0 |
  0    |    0    |   1    |   0   | 0 | 1 | 0 |
  0    |    0    |   0    |   1   | 0 | 0 | 1 |

但是，当我在测试数据（新数据）上运行此程序时，如果测试数据不包含与训练数据完全相同的名称和日期，我会得到不匹配的特征。所以如果我在这个测试样本上运行类似的代码，我会得到：

First  | Second  | Forth | 0 | 1 | 2 | 
--------------------------------------
  1    |    0    |   0   | 0 | 0 | 1 |
  0    |    1    |   0   | 0 | 1 | 0 |
  0    |    0    |   1   | 1 | 0 | 0 |

我可以做些什么来保留来自训练数据的相同转换并将其正确应用于测试数据，从而产生这个所需的输出：

First  | Second  | Third  | Forth | 0 | 1 | 2 | 
-----------------------------------------------
  1    |    0    |   0    |   0   | 0 | 0 | 1 |
  0    |    1    |   0    |   0   | 0 | 1 | 0 |
  0    |    0    |   0    |   1   | 1 | 0 | 0 |

我已经尝试添加一个 dict 来捕获 fit_transform 结果，但我不确定这是否有效或之后如何处理：

features_to_encode = ['Name', 'day']
label_final = pd.DataFrame()

labels = {}--------------------------------------------------------------------> TRIED THIS
for feature in features_to_encode:
    label_campaign = LabelBinarizer()
    label_results = label_campaign.fit_transform(df[feature])
    labels[feature] = label_results--------------------------------------------> WITH THIS
    label_results = pd.DataFrame(label_results, columns=label_campaign.classes_)
    label_final = pd.concat([label_final, label_results], axis=1)

df_encoded = label_final.join(df)

感谢任何帮助。谢谢=)

Answer 1

这样的事情应该可以工作。我通常使用数据框直到最后一次，因为它们更容易使用。 X 应该是您预测之前的测试数据框。 original_cols 应该是您的训练集列的列表。让我知道它是否适合您。

def normalize_X(X, original_cols):


    missing_cols= set(original_cols) - set(X.columns)
    extra_cols= set(X.columns) - set(original_cols)
    if len(missing_cols)>0:

        print(f'missing columns: {", ".join(missing_cols)}')
        for col in (missing_cols):

            X[col] = 0
    if len(extra_cols)>0:

        print(f'Columns to drop: {", ".join(extra_cols)} ',)
        X = X.drop(columns = extra_cols)
    X = X[original_cols]
    return X

Answer 2

pd.CategoricalDtype 和 pd.get_dummies

names_cat = pd.CategoricalDtype(['First', 'Second', 'Third', 'Forth'])
days_cat = pd.CategoricalDtype([0, 1, 2, 3, 4])

dumb_names = pd.get_dummies(df.Name.astype(names_cat))
dumb_names.columns = dumb_names.columns.to_numpy()

dumb_days = pd.get_dummies(df.day.astype(days_cat))
dumb_days.columns = dumb_days.columns.to_numpy()

   First  Second  Third  Forth  0  1  2  3  4
0      1       0      0      0  0  0  1  0  0
1      0       1      0      0  0  1  0  0  0
2      0       0      0      1  1  0  0  0  0

---

LabelBinarizer.classes_

from sklearn.preprocessing import LabelBinarizer

lb_0 = LabelBinarizer()
lb_1 = LabelBinarizer()

lb_0.classes_ = ['First', 'Second', 'Third', 'Forth']
lb_1.classes_ = [0, 1, 2, 3, 4]

a = lb_0.transform(df.Name)
b = lb_1.transform(df.day)

data = np.column_stack([a, b])
idx = df.index
col = np.concatenate([lb_0.classes_, lb_1.classes_])

result = pd.DataFrame(data, idx, col)
result

   First  Second  Third  Forth  0  1  2  3  4
0      1       0      0      0  0  0  1  0  0
1      0       1      0      0  0  1  0  0  0
2      0       0      0      1  1  0  0  0  0

---

reindex

cols = ['First', 'Second', 'Third', 'Forth', 0, 1, 2]
result = pd.concat(map(pd.get_dummies, map(df.get, df)), axis=1)
result.reindex(columns=cols, fill_value=0)

   First  Second  Third  Forth  0  1  2
0      1       0      0      0  0  0  1
1      0       1      0      0  0  1  0
2      0       0      0      1  1  0  0

Answer 3

另一种方法，可能更适合您在不同变量之间具有共同值的情况，并且如果您计划为多个列自动编码以进行编码：

df_train = pd.DataFrame({'Name': ['First', 'Second', 'Third', 'Fourth'], 'Day': [2,1,1,2]})
df_test = pd.DataFrame({'Name': ['First', 'Second', 'Fourth'], 'Day': [2,1,0]})
categories = []

cols_to_encode = ['Name', 'Day']
# Union of all values in both training and testing datasets:
for col in cols_to_encode:
    categories.append(list(set(df_train[col].unique().tolist() + df_test[col].unique().tolist())))

# Sorts the class names under each variable    
for cat in categories:
    cat.sort()

for col_name, cat in zip(cols_to_encode, categories):
    df_test[col_name] =  pd.Categorical(df_test[col_name], categories=cat)
df_test = pd.get_dummies(df_test, columns=cols_to_encode)

df_test

>>

    Name_First  Name_Second Name_Third  Name_Fourth Day_0   Day_1   Day_2   Day_3   Day_4
0   1           0           0           0           0       0       1       0    0 
1   0           1           0           0           0       1       0       0    0
2   0           0           0           1           1       0       0       0    0