我想在我的 DecisionTreeClassifier 实现中创建 Predict 和 Predict_proba 方法，但它给出了错误
Traceback (most recent call last):
File "c:\Users\Nijat\project.py", line 136, in <module>
print(model.predict(X))
^^^^^^^^^^^^^^^^
File "c:\Users\Nijat\project.py", line 128, in predict
return [1 if p[0] > 0.5 else 0 for p in self.predict_proba(X)]
^^^^^^^^^^^^^^^^^^^^^
File "c:\Users\Nijat\project.py", line 121, in predict_proba
class1_proba = self.bypass_tree(self.tree, sample)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Users\Nijat\project.py", line 105, in bypass_tree
while node['type'] == 'node':
~~~~^^^^^^^^
KeyError: 'type'
这是我的代码：
import numpy as np
import pandas as pd
class MyTreeClf:
def __init__(self, max_depth=5, min_samples_split=2, max_leafs=20):
self.max_depth = max_depth
self.min_samples_split = min_samples_split
self.max_leafs = max_leafs
self.tree = None
self.leafs_cnt = 0
def node_entropy(self, probs):
return -np.sum([p * np.log2(p) for p in probs if p > 0])
def node_ig(self, x_col, y, split_value):
left_mask = x_col <= split_value
right_mask = x_col > split_value
if len(x_col[left_mask]) == 0 or len(x_col[right_mask]) == 0:
return 0
left_counts = np.bincount(y[left_mask])
right_counts = np.bincount(y[right_mask])
left_probs = left_counts / len(y[left_mask]) if len(y[left_mask]) > 0 else np.zeros_like(left_counts)
right_probs = right_counts / len(y[right_mask]) if len(y[right_mask]) > 0 else np.zeros_like(right_counts)
entropy_after = (len(y[left_mask]) / len(y) * self.node_entropy(left_probs) +
len(y[right_mask]) / len(y) * self.node_entropy(right_probs))
entropy_before = self.node_entropy(np.bincount(y) / len(y))
return entropy_before - entropy_after
def get_best_split(self, X: pd.DataFrame, y: pd.Series):
best_col, best_split_value, best_ig = None, None, -np.inf
for col in X.columns:
sorted_unique_values = np.sort(X[col].unique())
for i in range(1, len(sorted_unique_values)):
split_value = (sorted_unique_values[i - 1] + sorted_unique_values[i]) / 2
ig = self.node_ig(X[col], y, split_value)
if ig > best_ig:
best_ig = ig
best_col = col
best_split_value = split_value
return best_col, best_split_value
def fit(self, X: pd.DataFrame, y: pd.Series, depth=1, node=None):
if self.max_leafs < 2:
self.leafs_cnt = 2
return
if node is None:
node = {}
self.tree = node
best_col, best_split_value = self.get_best_split(X, y)
node['type'] = None
node['feature'] = best_col
node['threshold'] = best_split_value
if len(y.unique()) == 1:
self.leafs_cnt += 1
node['type'] = 'leaf'
node['class_counts'] = {y.unique()[0]: len(y)}
return
if len(y) == 1:
self.leafs_cnt += 1
node['type'] = 'leaf'
node['class_counts'] = {y.values[0]: 1}
return
if depth >= self.max_depth or len(y) < self.min_samples_split or (self.leafs_cnt + 2 > self.max_leafs):
self.leafs_cnt += 1
node['type'] = 'leaf'
node['class_counts'] = y.value_counts().to_dict()
return
if best_col is None:
node['type'] = 'leaf'
node['class_counts'] = y.value_counts().to_dict()
self.leafs_cnt += 1
return
node['type'] = 'node'
node['feature'] = best_col
node['threshold'] = best_split_value
left_mask = X[best_col] <= best_split_value
right_mask = X[best_col] > best_split_value
node['left'] = {}
node['right'] = {}
self.fit(X[left_mask], y[left_mask], depth + 1, node['left'])
self.fit(X[right_mask], y[right_mask], depth + 1, node['right'])
def bypass_tree(self, node, sample):
while node['type'] == 'node':
feature_value = sample[node['feature']]
if feature_value <= node['threshold']:
node = node['left']
else:
node = node['right']
total_count = sum(node['class_counts'].values())
class_1_count = node['class_counts'].get(1, 0)
class1_proba = class_1_count / total_count if total_count > 0 else 0
return class1_proba
def predict_proba(self, X: pd.DataFrame):
proba = []
for _, sample in X.iterrows():
class1_proba = self.bypass_tree(self.tree, sample)
proba.append(class1_proba)
return np.array(proba)
def predict(self, X: pd.DataFrame):
return [1 if p[0] > 0.5 else 0 for p in self.predict_proba(X)]
df = pd.read_csv('c:\\Users\\Nijat\\Downloads\\banknote+authentication.zip', header=None)
df.columns = ['variance', 'skewness', 'curtosis', 'entropy', 'target']
X, y = df.iloc[:, :4], df['target']
model = MyTreeClf(max_depth=3, min_samples_split=2, max_leafs=1)
model.fit(X, y)
print(model.predict(X))
predict和predict_proba方法采用pandas数据帧形式的特征矩阵。 对于数据框中的每一行： 遍历树的节点直到其中一片叶子。 写出第一类的概率。 返回预测列表 - 与数据框中的行数一样多： Predict_proba - 返回概率（对于第一类）。 预测 - 按阈值 > 0.5
将概率转换为二元类验证：
输入数据：决策树的两组参数
输出：返回的概率和标签的预测（它们的总和）
我完全不知道用于检查代码的数据集，但我认为它是“钞票认证”
输入示例： {“最大深度”：3，“最小样本分割”：2，“最大叶数”：1} 示例输出： 11.2443438914
这里有两点：
model = MyTreeClf(max_depth=3, min_samples_split=2, max_leafs=1)
max_leafs <2
条件，将触发拟合返回 将其设置为两个将允许构建树
我没有您的数据集，所以我们将使用随机数据集进行测试
df = pd.DataFrame(columns=['variance', 'skewness', 'curtosis', 'entropy', 'target'],data=np.random.random(size=(500, 5)))
df['target'] =df['target'].apply(lambda x: 0 if x<0.5 else 1)
X, y = df.iloc[:, :4], df['target']
然后，查看模型的树，看看它是否已构建
model = MyTreeClf(max_depth=3, min_samples_split=2, max_leafs=2)
model.fit(X, y)
print(model.tree)
它给了我们一棵树，因此节点结构将起作用 您现在可以运行
model.predict_proba(X)def predict(self, X: pd.DataFrame):
return [1 if p > 0.5 else 0 for p in self.predict_proba(X)]