我有数据集我想在拆分时运行logistic regression
如何运行一个循环，以便我可以使用 SkLearn 的 X_train、X_test 拆分函数运行自动逻辑回归，并为每个数据集打印 3 个单独的准确度结果？
要在 X 的一个数据集上运行，y 如下：
X_train, X_test, y_train, y_test = train_test_split(X, y,
test_size=0.25, shuffle=False)
logreg = LogisticRegression()
logreg.fit(X_train,y_train)
y_pred = logreg.predict(X_test)
print('Accuracy:',metrics.accuracy_score(y_test, y_pred))
保存在字典中：
dictionary = {X1:y1, X2:y2, X3:y3}
accuracies = []
for k,v in dictionary.items():
X_train, X_test, y_train, y_test = train_test_split(k, v,
test_size=0.25, shuffle=False)
logreg = LogisticRegression()
logreg.fit(X_train,y_train)
y_pred = logreg.predict(X_test)
accuracies.append(metrics.accuracy_score(y_test, y_pred))
但这并不是一个好方法，尤其是当数据集太大时。
所以：
accuracies = []
for i in range(1,4):
X = 'X{}'.format(i)
y = 'y{}'.format(i)
X_train, X_test, y_train, y_test = train_test_split(vars()[X], vars()[y],
test_size=0.25, shuffle=False)
logreg = LogisticRegression()
logreg.fit(X_train,y_train)
y_pred = logreg.predict(X_test)
accuracies.append(metrics.accuracy_score(y_test, y_pred))
注意：为了访问小部分不同的变量名，比如这里的 X1,X2,...，我们可以使用上面提到的方法：
import numpy as np
X1= np.arange(1,10)
y1=[i**2 for i in X1]
X2= np.arange(-5,5)
y2=[i**2 for i in X2]
for i in range(1,3):
X = 'X{}'.format(i)
y = 'y{}'.format(i)
print('X_{}'.format(i) , vars()[X])
print('y_{}'.format(i) , vars()[y])
输出：
X_1 [1 2 3 4 5 6 7 8 9]
y_1 [1, 4, 9, 16, 25, 36, 49, 64, 81]
X_2 [-5 -4 -3 -2 -1 0 1 2 3 4]
y_2 [25, 16, 9, 4, 1, 0, 1, 4, 9, 16]