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。