python中变量中的函数参数名称

我正在使用

sklearn

来训练不同的模型。我想通过

sklearn

的决策树分类器，同一参数的不同值，并绘制一个图。我想对许多这样的参数执行此操作。因此，我想创建一个通用函数，它可以处理所有参数及其值

我的问题是，有没有一种方法可以将参数名（而不是值）分配给变量并将其传递给我的函数

---

例如，-决策树采用

max\u depth

，

min\u samples\u leaf

等参数。我想尝试两个参数的不同值，一次一个，并分别为

max\u depth

和

min\u samples\u leaf

绘制结果。

使用字典并将其与

**

一起传递

kwargs = {
    "max_depth": value,
    "min_samples_leaf": value,
}
fun(**kwargs)

---

使用字典并将其与

**

一起传递

kwargs = {
    "max_depth": value,
    "min_samples_leaf": value,
}
fun(**kwargs)

这个解决方案不是很“Pythonic”，但很容易遵循。您可以在循环、嵌套循环或类似的方式中调用函数

dt = DecisionTreeClassifier(criterion='entropy', min_samples_leaf=150, min_samples_split=100)

是使用决策树的标准调用，只需循环使用并替换

minu samples\u leaf

和

minu samples\u split

from sklearn.tree import DecisionTreeClassifier
from sklearn.metrics import accuracy_score, roc_curve, auc
from sklearn.model_selection import train_test_split

min_samples_leafs = [50, 100, 150]
min_samples_splits =[50, 100, 150]

for sample_leafs in min_samples_leafs:
    for sample_splits in min_samples_splits:

        dt = DecisionTreeClassifier(criterion='entropy', min_samples_leaf=sample_leafs, min_samples_split=sample_splits)

        X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=101)

        dt = dt.fit(X_train, y_train)
        y_pred_train = dt.predict(X_train)
        y_pred_test = dt.predict(X_test)


        print("Training Accuracy: %.5f" %accuracy_score(y_train, y_pred_train))
        print("Test Accuracy: %.5f" %accuracy_score(y_test, y_pred_test))
        print('sample_leafs: ', sample_leafs)
        print('sample_leafs: ', sample_splits)
        print('\n')

输出：

Training Accuracy: 0.96689
Test Accuracy: 0.96348
sample_leafs:  50
sample_leafs:  50


Training Accuracy: 0.96689
Test Accuracy: 0.96348
sample_leafs:  50
sample_leafs:  100


Training Accuracy: 0.96509
Test Accuracy: 0.96293
sample_leafs:  50
sample_leafs:  150


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  50


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  100


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  150


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  50


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  100


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  150

您可以通过像这样传递列表来实现此功能

def do_decision_tree_stuff(min_samples_leafs, min_samples_splits):

您可以这样调用函数

 do_decision_tree_stuff([50, 100, 150], [50, 100, 150])

这个解决方案不是很“Pythonic”，但很容易遵循。您可以在循环、嵌套循环或类似的方式中调用函数

dt = DecisionTreeClassifier(criterion='entropy', min_samples_leaf=150, min_samples_split=100)

是使用决策树的标准调用，只需循环使用并替换

minu samples\u leaf

和

minu samples\u split

from sklearn.tree import DecisionTreeClassifier
from sklearn.metrics import accuracy_score, roc_curve, auc
from sklearn.model_selection import train_test_split

min_samples_leafs = [50, 100, 150]
min_samples_splits =[50, 100, 150]

for sample_leafs in min_samples_leafs:
    for sample_splits in min_samples_splits:

        dt = DecisionTreeClassifier(criterion='entropy', min_samples_leaf=sample_leafs, min_samples_split=sample_splits)

        X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=101)

        dt = dt.fit(X_train, y_train)
        y_pred_train = dt.predict(X_train)
        y_pred_test = dt.predict(X_test)


        print("Training Accuracy: %.5f" %accuracy_score(y_train, y_pred_train))
        print("Test Accuracy: %.5f" %accuracy_score(y_test, y_pred_test))
        print('sample_leafs: ', sample_leafs)
        print('sample_leafs: ', sample_splits)
        print('\n')

输出：

Training Accuracy: 0.96689
Test Accuracy: 0.96348
sample_leafs:  50
sample_leafs:  50


Training Accuracy: 0.96689
Test Accuracy: 0.96348
sample_leafs:  50
sample_leafs:  100


Training Accuracy: 0.96509
Test Accuracy: 0.96293
sample_leafs:  50
sample_leafs:  150


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  50


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  100


Training Accuracy: 0.96313
Test Accuracy: 0.96256
sample_leafs:  100
sample_leafs:  150


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  50


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  100


Training Accuracy: 0.96188
Test Accuracy: 0.96037
sample_leafs:  150
sample_leafs:  150

您可以通过像这样传递列表来实现此功能

def do_decision_tree_stuff(min_samples_leafs, min_samples_splits):

您可以这样调用函数

 do_decision_tree_stuff([50, 100, 150], [50, 100, 150])

---

这只解决了两个变量的问题，但不适用于多个变量。然后传递更多列表、列表列表或字典，如下所示。这只解决了两个变量的问题，但不适用于多个变量。然后传递更多列表或列表，问题是“param1”是一个字符串，而DecisionTreeClassifier（或任何函数）不接受字符串作为参数。这相当于使用DecisionTreeClassifier（“param1”=value），但我想要DecisionTreeClassifier（param1=value），但Python将字符串转换为参数名。这就是它的工作原理：在打开kwargs（使用kwargs.items（））后，我将max_depth存储在fun中的变量名param1中，然后使用DecisionTreeClassifier（param1=value）。但这会给出一个错误：“param1不是有效参数”。不要这样做，只需执行DecisionTreeClassifier（**kwargs）。这就是你想要的python方式。问题是“param1”是一个字符串，而decisiontreeclassier（或任何函数）不接受字符串作为参数。这相当于使用DecisionTreeClassifier（“param1”=value），但我想要DecisionTreeClassifier（param1=value），但Python将字符串转换为参数名。这就是它的工作原理：在打开kwargs（使用kwargs.items（））后，我将max_depth存储在fun中的变量名param1中，然后使用DecisionTreeClassifier（param1=value）。但这会给出一个错误：“param1不是有效参数”。不要这样做，只需执行DecisionTreeClassifier（**kwargs）。这是做你想做的事的蟒蛇式方法。