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Piecewise exponential

model="piecewise_exponential" — a baseline hazard that is constant inside intervals you choose and jumps between them. The cheapest way to get an arbitrary hazard shape.

The model

Given breakpoints \(0 < \tau_1 < \dots < \tau_k\) and rates \(\lambda_0, \dots, \lambda_k\), the baseline hazard is

\[ h_0(t) = \lambda_j \quad \text{for } t \in [\tau_j, \tau_{j+1}), \]

with \(\tau_0 = 0\) and \(\tau_{k+1} = \infty\). Covariates act proportionally:

\[ h(t \mid X) = h_0(t)\exp(X^\top\beta). \]

There is always one more rate than there are breakpoints — two breakpoints cut the timeline into three pieces.

Parameters

gen_piecewise_exponential(n, breakpoints, hazard_rates, betas=None,
                          n_covariates=2, covariate_dist="normal",
                          covariate_params=None, model_cens="uniform",
                          cens_par=5.0, seed=None)
Parameter Type Default Meaning
n int — number of subjects
breakpoints list[float] — increasing times where the hazard changes
hazard_rates list[float] — one rate per interval — len(breakpoints) + 1 of them
betas list[float] | None None coefficients; random values are drawn when omitted
n_covariates int 2 number of covariates when betas is not given
covariate_dist "normal" | "uniform" | "binary" "normal" covariate distribution — see Covariates
covariate_params dict | None None distribution parameters
model_cens str "uniform" censoring mechanism
cens_par float 5.0 censoring parameter
seed int | Generator | None None reproducibility

Pin betas when you are validating

With betas=None the coefficients are drawn at random, so you no longer know the truth you are testing against. Pass them explicitly — betas=[0.0, 0.0] isolates the baseline hazard entirely.

Example

from gen_surv import generate

df = generate(model="piecewise_exponential", n=6,
              breakpoints=[1.0, 3.0], hazard_rates=[0.5, 1.0, 0.2],
              seed=42)
print(df)

   id      time  status        X0        X1
0   0  1.682976       1  0.750451  0.940565
1   1  1.513450       1 -1.951035 -1.302180
2   2  0.980989       1  0.127840 -0.316243
3   3  2.270415       1 -0.016801 -0.853044
4   4  0.495887       1  0.879398  0.777792
5   5  0.973194       0  0.066031  1.127241

Check: does each interval have the hazard you asked for?

The occurrence/exposure estimate — events in an interval divided by the time subjects spent at risk in it — should return each rate:

import numpy as np
from gen_surv import generate

breakpoints, rates = [1.0, 3.0], [0.5, 2.0, 0.2]
df = generate(model="piecewise_exponential", n=40000,
              breakpoints=breakpoints, hazard_rates=rates,
              betas=[0.0, 0.0], model_cens="uniform", cens_par=100.0, seed=7)

edges = [0.0] + breakpoints + [float(df["time"].max())]
for lo, hi, true_h in zip(edges[:-1], edges[1:], rates):
    at_risk = df[df["time"] > lo]
    exposure = float(np.minimum(at_risk["time"], hi).sub(lo).clip(lower=0).sum())
    events = int(((df["time"] > lo) & (df["time"] <= hi) & (df["status"] == 1)).sum())
    print(f"[{lo}, {hi:.2f})  declared={true_h}  empirical={events / exposure:.3f}")

[0.0, 1.00)   declared=0.5  empirical=0.501
[1.0, 3.00)   declared=2.0  empirical=2.011
[3.0, 37.31)  declared=0.2  empirical=0.202

Each interval carries the rate it was given. The open-ended last interval is populated only by the survivors of the earlier ones, so it is the noisiest of the three — widen n before reading much into it.

Approximating a smooth hazard

Piecewise-constant hazards approximate any shape if you use enough pieces — this is the standard trick behind piecewise-exponential survival models:

import numpy as np

# Approximate a Weibull hazard, h(t) = 1.5 * t^0.5, on [0, 5]
grid = np.linspace(0, 5, 11)
breakpoints = list(grid[1:-1])
midpoints = (grid[:-1] + grid[1:]) / 2
hazard_rates = list(1.5 * midpoints ** 0.5)

Ten pieces reproduce a Weibull hazard closely enough for most purposes; the approximation error is bounded by how much the true hazard moves within a piece, so refine the grid where it is steepest rather than everywhere.