np.mean(y_train == y_train_pred)

这段代码的作用是计算训练集上预测准确率。其中，y_train是训练集上的真实标签，y_train_pred是模型在训练集上的预测标签。np.mean()函数计算数组元素的平均值，而(y_train == y_train_pred)会返回一个由True和False组成的布尔数组，表示模型预测是否正确。因此，np.mean(y_train == y_train_pred)等价于正确预测的样本数除以总样本数，即训练集上的预测准确率。

相关问题

TypeError Traceback (most recent call last) Cell In[62], line 3 1 train_x,test_x,train_y,test_y=train_test_split(X,Y,test_size=0.2,random_state=0) 2 model=LinearRegression().fit(train_x,train_y) ----> 3 y_pred=reg.predict(train_x) 4 mse_train=mean_square_error(train_y,y_pred) 5 print(f"模型偏置：{np.round(reg.intercept_,2)}") TypeError: LinearModel.predict() missing 1 required positional argument: 'X'

在你的代码中，出现了一个错误。在第3行中，你使用了 `reg.predict(train_x)` ，但是你之前定义的模型是 `model` 而不是 `reg`。因此，你需要将这行代码修改为 `y_pred = model.predict(train_x)`。

修正后的代码如下所示：

train_x, test_x, train_y, test_y = train_test_split(X, Y, test_size=0.2, random_state=0)
model = LinearRegression().fit(train_x, train_y)
y_pred = model.predict(train_x)
mse_train = mean_square_error(train_y, y_pred)
print(f"模型偏置：{np.round(model.intercept_, 2)}")

请检查你的代码并使用正确的模型对象进行预测。

brier <- brier_efron(y_train_true = y_dat, y_train_pred = y_dat_pred, y_newdata = y_val, y_newdata_pred = y_val_pred, times = c(1:10)) brier$bs改成python

在Python中，你可以使用上面提供的`brier_efron`函数来计算Brier分数。以下是将R代码转换为Python代码的版本：

import pandas as pd
import numpy as np
from lifelines.utils import concordance_index

def brier_efron(y_train_true, y_train_pred, y_newdata, y_newdata_pred, times):

    baseline = base_efron(y_train_true, y_train_pred)

    y_newdata = pd.DataFrame(y_newdata, columns=["time", "event"])
    y_newdata = y_newdata.sort_values(by="time")
    y_newdata_pred = y_newdata_pred[y_newdata.index]

    Y_x = np.zeros((len(y_newdata), len(times)))
    for i, x in enumerate(times):
        Y_x[:, i] = np.where(y_newdata["time"] > x, 1, 0)

    kmf = KaplanMeierFitter().fit(y_newdata["time"], event_observed=y_newdata["event"])
    G_t = kmf.survival_function_at_times(times).values.flatten()
    G_x = kmf.predict(y_newdata["time"]).values.flatten()

    W_x = np.zeros((len(y_newdata), len(times)))
    for t in range(len(times)):
        W_x[:, t] = (1 - Y_x[:, t]) * y_newdata["event"] / G_x + Y_x[:, t] / G_t[t]

    Lambda_t = [baseline.baseline_hazard_[baseline.baseline_survival_.index(baseline.baseline_survival_.index <= x)].sum() for x in times]
    S_x = np.exp(-np.exp(y_newdata_pred) @ np.array(Lambda_t))

    BS_t = [np.mean(W_x[:, x] * (Y_x[:, x] - S_x[:, x])**2) for x in range(len(times))]

    return pd.DataFrame({"time": times, "bs": BS_t})

# 使用示例
y_dat = [...]  # y_train_true的值
y_dat_pred = [...]  # y_train_pred的值
y_val = [...]  # y_newdata的值
y_val_pred = [...]  # y_newdata_pred的值

times = np.arange(1, 11)
brier_result = brier_efron(y_train_true=y_dat, y_train_pred=y_dat_pred, y_newdata=y_val, y_newdata_pred=y_val_pred, times=times)

print(brier_result)

请注意替换`[...]`部分的数据为实际的数据值。`brier_result`将是一个包含时间和Brier分数的数据框。你可以根据需要对结果进行进一步操作。