如何在LogisticRegression中使用交叉验证来解决分类问题？

Question

• 我想建立一个数据框架，在句子中使用最多重复的单词，并通过Logistic-Regression进行分类。
• 我试着用密码把步骤写清楚。
• 列B是我的目标。

我所拥有的：(示例)

raw_data={"A":["This is yellow","That is green","These are orange","This is a pen","This is an Orange"],
          "B":["Yes","No","Yes","No","No"]   }
df=pd.DataFrame(raw_data)
df

    A                   B
0   This is yellow      Yes
1   That is green       No
2   These are orange    Yes
3   This is a pen       No
4   This is an Orange   No

我所做的：

### 1-Import Libraries:
import numpy as np 
import pandas as pd

### 2- Create data set:
raw_data={"A":["This is yellow","That is green","These are orange","This is a pen","This is an Orange"],
          "B":["Yes","No","Yes","No","No"]   }
df=pd.DataFrame(raw_data)
df


          A              B
0   This is yellow      Yes
1   That is green       No
2   These are orange    Yes
3   This is a pen       No
4   This is an Orange   No

### 3- Count the word and charachters 
df['word_count'] = df['A'].agg(lambda x: len(x.split(" ")))
df['char_count'] = df['A'].agg(lambda x:len(x))
df
             A         B    word_count  char_count
0   This is yellow     Yes  3           14
1   That is green      No   3           13
2   These are orange   Yes  3           16
3   This is a pen      No   4           13
4   This is an Orange  No   4           17

### 4- Count the most repeated words in column "A"
df_word_count=pd.DataFrame(df.A.str.split('').explode().value_counts()).reset_index().rename({'index':"A,"A":"Count"},axis=1)
display(df_word_count)
list_word_count=list(df_word_count["A"])
len(list_word_count)

    A       Count
0   is      4
1   This    3
2   yellow  1
3   These   1
4   orange  1
5   green   1
6   That    1
7   are     1
8   a       1
9   pen     1
10  Orange  1
11  an      1

### 5- Make a ZERO-Matrix
allfeatures=np.zeros((df.shape[0],len(list_word_count)))
allfeatures.shape

### 6- Create a data frame
for i in range(len(list_word_count)):
  allfeatures[:,i]=df['A'].agg(lambda x:x.split().count(list_word_count[i]))
Complete_data=pd.concat([df,pd.DataFrame(allfeatures)],axis=1)
display(Complete_data)

           A            B   word_count  char_count  0   1   2   3   4   5   6   7   8   9   10  11
0   This is yellow      Yes 3           14          1.0 1.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
1   That is green       No  3           13          1.0 0.0 0.0 0.0 0.0 1.0 1.0 0.0 0.0 0.0 0.0 0.0
2   These are orange    Yes 3           16          0.0 0.0 0.0 1.0 1.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0
3   This is a pen       No  4           13          1.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0 1.0 0.0 0.0
4   This is an Orange   No  4           17          1.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0 1.0


### 7- change columns name from list
#This creates a list of the words 
l = list(df_word_count["A"])

l.insert(0,"char_count")
l.insert(0,"word_count")
l.insert(0,"B")
l.insert(0,"A")    
# Finally, I rename all the columns with the names that I have in the list l
Complete_data.columns = l

### 8- Define X and Y
x=Complete_data.drop(["A","B"],axis=1) # Features
y=Complete_data["B"] # Target


### 9- Encoding of Target
from sklearn.preprocessing import LabelEncoder
le = LabelEncoder()
y = le.fit_transform(y)


### 10- Train|Test split
from sklearn.model_selection import train_test_split
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size = 0.2, random_state = 0)

### 11- Import Sklearn needed packages
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import r2_score

from sklearn.model_selection import cross_val_score
from sklearn.model_selection import cross_val_predict

### 12- Prediction and Regression with Cross-Validation
LogReg=LogisticRegression()
LogReg.fit(x_train,y_train)

cv_LogReg=cross_val_score(LogReg,x_train,y_train,cv=2)
cv_LogReg_pred=cross_val_predict(LogReg,x_train,y_train,cv=2)

print("Score: ",r2_score(y_train,cv_LogReg_pred))

错误：

虽然我使用了LabelEncoder，但是算法找不到任何分类(0,1)

ValueError                                Traceback (most recent call last)
<ipython-input-127-2d7e54ebfd6c> in <module>
      4 #LogReg_pred=LogReg.predict(x_test)
      5 cv_LogReg=cross_val_score(LogReg,x_train,y_train,cv=2)
----> 6 cv_LogReg_pred=cross_val_predict(LogReg,x_train,y_train,cv=2)
      7 
      8 print("Score: ",r2_score(y_train,cv_LogReg_pred))

.
.
.

This solver needs samples of at least 2 classes in the data, but the data contains only one class: 0

我不知道我做错了什么，‍♂️

Answer 1

因为您正在对一个样本进行交叉验证，所以在选择一个样本时，它是以这样的方式划分的，即样本只包含一个类，因此您将得到该错误。如果您有更多的数据，您不应该得到这个错误。我已经对这5条记录进行了简单的Logistic回归，我能够创建一个模型，所以你能增加你的数据和检查吗？我增加了这样的数据：

2-创建数据集: raw_data={"A":“这是黄色的”，“那是绿色的”，“这些是橙色的”，“这>是一支钢笔”，“这是一支橙色的”，“这是一支橙色的”，“这是一只绿色的”，“这是一只鸟”，“这是一只鸟”，“这是一支钢笔”，“这是一支橙色的”，“这是一只绿的”，“这是一只鸟”，“这是一支钢笔”，“这是一支橘子”，“这是一只鸟”，“这是一只猩猩”。，"B":“是”、“否”、“是”、“否”、“否”、“是”、“否”、“是”、“否”、“否” }

还有一件事，我把最后一行的r2_score改成了准确性评分。

Answer 2

这似乎是因为您的y_train数据集(或其中一个折叠)只包含一个类，在本例中，所有示例都属于类0。另见这个stackexchange答案。您可以通过增加样本数量(拥有同一类的所有样本的可能性较低)或使用分层数据拆分/交叉验证策略来确保值为0和1的样本数在所有子集之间大致相同。