是否有可能 运行 一个 Cox-Proportional-Hazards-Model 对 `lifelines` 或其他包裹中的基线危险具有指数分布?

Is it possible to run a Cox-Proportional-Hazards-Model with an exponential distribution for the baseline hazard in `lifelines` or another package?

我考虑使用 lifelines 包来适应 Cox-Proportional-Hazards-Model. I read that lifelines uses a nonparametric approach to fit the baseline hazard,这会导致某些时间点的 baseline_hazards 不同(请参见下面的代码示例)。对于我的申请,我需要一个 exponential distribution leading to a baseline hazard h0(t) = lambda 随时间变化。

所以我的问题是:是否(与此同时)运行 一个考克斯比例风险模型,其基线风险呈指数分布 lifelines 或另一个 Python包?

示例代码:

from lifelines import CoxPHFitter
import pandas as pd

df = pd.DataFrame({'duration': [4, 6, 5, 5, 4, 6], 
                   'event': [0, 0, 0, 1, 1, 1], 
                   'cat': [0, 1, 0, 1, 0, 1]})

cph = CoxPHFitter()
cph.fit(df, duration_col='duration', event_col='event', show_progress=True)
cph.baseline_hazard_

给予

        baseline hazard
T   
4.0     0.160573
5.0     0.278119
6.0     0.658032

生命线作者在这里。

因此,此模型本身不在生命线中,但您可以轻松地自己实现它(也许我会为将来的版本做一些事情)。这个想法依赖于比例风险模型和 AFT(加速故障时间)模型的交集。在具有指数风险(即恒定基线风险)的 cox-ph 模型中,风险如下所示:

h(t|x) = lambda_0(t) * exp(beta * x) = lambda_0 * exp(beta * x)

在指数分布的 AFT 规范中,风险如下所示:

h(t|x) = exp(-beta * x - beta_0) = exp(-beta * x) * exp(-beta_0) = exp(-beta * x) * lambda_0

注意负号的区别!

因此,我们可以进行指数 AFT 拟合,而不是进行 CoxPH(如果我们想要与 CoxPH 相同的解释,则翻转符号)。我们可以使用自定义回归模型语法来执行此操作:

from lifelines.fitters import ParametricRegressionFitter
from autograd import numpy as np


class ExponentialAFTFitter(ParametricRegressionFitter):

    # this is necessary, and should always be a non-empty list of strings.
    _fitted_parameter_names = ['lambda_']

    def _cumulative_hazard(self, params, T, Xs):
        # params is a dictionary that maps unknown parameters to a numpy vector.
        # Xs is a dictionary that maps unknown parameters to a numpy 2d array
        lambda_ = np.exp(np.dot(Xs['lambda_'], params['lambda_']))
        return T / lambda_

测试这个,

from lifelines.datasets import load_rossi
from lifelines import CoxPHFitter

rossi = load_rossi()
rossi['intercept'] = 1
regressors = {'lambda_': rossi.columns}
eaf = ExponentialAFTFitter().fit(rossi, "week", "arrest", regressors=regressors)

eaf.print_summary()
"""
<lifelines.ExponentialAFTFitter: fitted with 432 observations, 318 censored>
         event col = 'arrest'
number of subjects = 432
  number of events = 114
    log-likelihood = -686.37
  time fit was run = 2019-06-27 15:13:18 UTC

---
                    coef exp(coef)  se(coef)     z      p  -log2(p)  lower 0.95  upper 0.95
lambda_ fin         0.37      1.44      0.19  1.92   0.06      4.18       -0.01        0.74
        age         0.06      1.06      0.02  2.55   0.01      6.52        0.01        0.10
        race       -0.30      0.74      0.31 -0.99   0.32      1.63       -0.91        0.30
        wexp        0.15      1.16      0.21  0.69   0.49      1.03       -0.27        0.56
        mar         0.43      1.53      0.38  1.12   0.26      1.93       -0.32        1.17
        paro        0.08      1.09      0.20  0.42   0.67      0.57       -0.30        0.47
        prio       -0.09      0.92      0.03 -3.03 <0.005      8.65       -0.14       -0.03
        _intercept  4.05     57.44      0.59  6.91 <0.005     37.61        2.90        5.20
_fixed  _intercept  0.00      1.00      0.00   nan    nan       nan        0.00        0.00
---
"""

CoxPHFitter().fit(load_rossi(), 'week', 'arrest').print_summary()
"""
<lifelines.CoxPHFitter: fitted with 432 observations, 318 censored>
      duration col = 'week'
         event col = 'arrest'
number of subjects = 432
  number of events = 114
partial log-likelihood = -658.75
  time fit was run = 2019-06-27 15:17:41 UTC

---
      coef exp(coef)  se(coef)     z      p  -log2(p)  lower 0.95  upper 0.95
fin  -0.38      0.68      0.19 -1.98   0.05      4.40       -0.75       -0.00
age  -0.06      0.94      0.02 -2.61   0.01      6.79       -0.10       -0.01
race  0.31      1.37      0.31  1.02   0.31      1.70       -0.29        0.92
wexp -0.15      0.86      0.21 -0.71   0.48      1.06       -0.57        0.27
mar  -0.43      0.65      0.38 -1.14   0.26      1.97       -1.18        0.31
paro -0.08      0.92      0.20 -0.43   0.66      0.59       -0.47        0.30
prio  0.09      1.10      0.03  3.19 <0.005      9.48        0.04        0.15
---
Concordance = 0.64
Log-likelihood ratio test = 33.27 on 7 df, -log2(p)=15.37
"""

注意符号的变化!所以如果你想要模型中的恒定基线风险,那就是 exp(-4.05).