kaplanMeierEstimates

Computes Kaplan-Meier estimates of survival probabilities in stratified samples.

Synopsis

kaplanMeierEstimates (x)

Required Arguments

float x[[]] (Input)

Two-dimensional data array of size nObservations×ncol.

Return Value

An array of length nObservations×2. The first column contains the estimated survival probabilities, and the second column contains Greenwood’s estimate of the standard deviation of these probabilities. If the i-th observation contains censor codes out of range or if a variable is missing, then the corresponding elements of the return value are set to missing (NaN, not a number). Similarly, if an element in the return value is not defined, then it is set to missing.

Optional Arguments

t_print, (Input)

Print Kaplan-Meier estimates of survival probabilities in stratified samples.

xResponseCol, int (Input)

Column index for the response times in the data array, x. The interpretation of these times as either right-censored or exact failure times depends on censorCodesCol.

Default: xResponseCol = 0.

censorCodesCol, int (Input)

Column index for the optional censoring codes in the data array, x. If x[i][censorCodesCol]= 0, the failure time x[i] [xResponseCol] is treated as an exact time of failure. Otherwise it is treated as a right-censored time.

Default: It is assumed that there is no censor code column in x. All observations are assumed to be exact failure times.

freqResponseCol, int (Input)

Column index for the number of responses associated with each row in the data array, x.

Default: It is assumed that there is no frequency response column in x. Each observation in the data array is assumed to be for a single failure.

stratumNumberCol, int (Input)

Column index for the stratum number for each observation in the data array, x. Column stratumNumberCol of x contains a unique value for each stratum in the data. Kaplan-Meier estimates are computed within each stratum.

Default: It is assumed that there is no stratum number column in x. The data is assumed to come from one stratum.

sorted, (Input)

If this option is used, column xResponseCol of x is assumed to be sorted in ascending order within each stratum. Otherwise, a detached sort is conducted prior to analysis. If sorting is performed, all censored individuals are assumed to follow tied failures.

Default: Column xResponseCol of x is not sorted.

nMissing (Output)

Number of rows of data in x containing missing values.

Description

Function kaplanMeierEstimates computes Kaplan-Meier (or product-limit) estimates of survival probabilities for a sample of failure times that can be right censored or exact times. A survival probability S(t) is defined as 1 - F(t), where F(t) is the cumulative distribution function of the failure times (t). Greenwood’s estimate of the standard errors of the survival probability estimates are also computed. (See Kalbfleisch and Prentice, 1980, pages 13 and 14.)

Let \((t_i,\delta_i)\), for \(i=1,\ldots,n\) denote the failure censoring times and the censoring codes for the n observations in a single sample. Here, \(t_i=\text{x}_{i-1,irt}\) is a failure time if \(\delta_i\) is 0, where \(\delta_I=\text{x}_{i-1,censorCodesCol}\). Also, \(t_i\) is a right censoring time if \(\delta_i\) is 1. Rows in x containing values other than 0 or 1 for \(\delta_i\) are ignored. Let the number of observations in the sample that have not failed by time \(s_{(i)}\) be denoted by \(n_{(i)}\), where \(s_{(i)}\) is an ordered (from smallest to largest) listing of the distinct failure times (censoring times are omitted). Then the Kaplan-Meier estimate of the survival probabilities is a step function, which in the interval from \(s_{(i)}\) to \(s_{(i+1)}\) (including the lower endpoint) is given by

\[\hat{S}(t) = \prod_{j=1}^{i} \left(\frac{n_{(j)} - d_{(j)}}{n_{(j)}}\right)\]

where \(d_{(j)}\) denotes the number of failures occurring at time \(s_{(j)}\), and \(n_{(j)}\) is the number of observation that have not failed prior to \(s_{(j)}\).

Note that one row of X may correspond to more than one failed (or censored) observation when the frequency option is in effect (freqResponseCol is specified). The Kaplan-Meier estimate of the survival probability prior to time \(s_{(1)}\) is 1.0, while the Kaplan-Meier estimate of the survival probability after the last failure time is not defined.

Greenwood’s estimate of the variance of

\[\hat{S}(t)\]

in the interval from \(s_{(i)}\) to \(s_{(i+1)}\) is given as

\[\mathrm{est.var}\left(\hat{S}(t)\right) = \hat{S}^2(t) \sum_{j=1}^{i} \frac{d_{(j)}}{n_{(j)} \left(n_{(j)} - d_{(j)}\right)}\]

Function kaplanMeierEstimates computes the single sample estimates of the survival probabilities for all samples of data included in x during a single call. This is accomplished through the stratumNumberCol column of x, which if present, must contain a distinct code for each sample of observations. If stratumNumberCol is not specified, there is no grouping column, and all observations are assumed to come from the same sample.

When failures and right-censored observations are tied and the data are to be sorted by kaplanMeierEstimates (sorted optional argument is not used), kaplanMeierEstimates assumes that the time of censoring for the tied-censored observations is immediately after the tied failure (within the same sample). When the sorted optional argument is used, the data are assumed to be sorted from smallest to largest according to column xResponseCol of x within each stratum. Furthermore, a small increment of time is assumed (theoretically) to elapse between the failed and censored observations that are tied (in the same sample). Thus, when the sorted optional argument is used, the user must sort all of the data in x from smallest to largest according to column xResponseCol (and column stratumNumberCol, if present). By appropriate sorting of the observations, the user can handle censored and failed observations that are tied in any manner desired.

The t_print option prints life tables. One table for each stratum is printed. In addition to the survival probabilities at each failure point, the following is also printed: the number of individuals remaining at risk, Greenwood’s estimate of the standard errors for the survival probabilities, and the Kaplan-Meier log-likelihood. The Kaplan-Meier log-likelihood is computed as:

\[\ell = \sum_j d_{(j)} \ln d_{(j)} + \left(n_{(j)} - d_{(j)}\right) \ln \left(n_{(j)} - d_{(j)}\right) - n_{(j)} \ln n_{(j)}\]

where the sum is with respect to the distinct failure times \(s_{(j)}\), \(d_{(j)}\).

Example

The following example is taken from Kalbfleisch and Prentice (1980, page 1). The first column in x contains the death/censoring times for rats suffering from vaginal cancer. The second column contains information as to which of two forms of treatment were provided, while the third column contains the censoring code. Finally, the fourth column contains the frequency of each observation. The product-limit estimates of the survival probabilities are computed for both groups with one call to kaplanMeierEstimates.

Function kaplanMeierEstimates could have been called with the sorted optional argument if the censored observations had been sorted with respect to the failure time variable. t_print option is used to print the life tables.

from numpy import *
from pyimsl.stat.kaplanMeierEstimates import kaplanMeierEstimates

icen = 2
ifrq = 3
igrp = 1

x = array([
    [143, 5, 0, 1],
    [164, 5, 0, 1],
    [188, 5, 0, 2],
    [190, 5, 0, 1],
    [192, 5, 0, 1],
    [206, 5, 0, 1],
    [209, 5, 0, 1],
    [213, 5, 0, 1],
    [216, 5, 0, 1],
    [220, 5, 0, 1],
    [227, 5, 0, 1],
    [230, 5, 0, 1],
    [234, 5, 0, 1],
    [246, 5, 0, 1],
    [265, 5, 0, 1],
    [304, 5, 0, 1],
    [216, 5, 1, 1],
    [244, 5, 1, 1],
    [142, 7, 0, 1],
    [156, 7, 0, 1],
    [163, 7, 0, 1],
    [198, 7, 0, 1],
    [205, 7, 0, 1],
    [232, 7, 0, 2],
    [233, 7, 0, 4],
    [239, 7, 0, 1],
    [240, 7, 0, 1],
    [261, 7, 0, 1],
    [280, 7, 0, 2],
    [296, 7, 0, 2],
    [323, 7, 0, 1],
    [204, 7, 1, 1],
    [344, 7, 1, 1]])

kaplanMeierEstimates(x,
                     t_print=True,
                     freqResponseCol=ifrq,
                     censorCodesCol=icen,
                     stratumNumberCol=igrp)

Output

                   Kaplan Meier Survival Probabilities
                      For Group Value = 5
 
       Number      Number                 Survival     Estimated
      at risk     Failing        Time  Probability    Std. Error
           19           1         143      0.94737      0.051228
 
           18           1         164      0.89474      0.070406
 
           17           2         188      0.78947      0.093529
 
           15           1         190      0.73684       0.10102
 
           14           1         192      0.68421       0.10664
 
           13           1         206      0.63158       0.11066
 
           12           1         209      0.57895       0.11327
 
           11           1         213      0.52632       0.11455
 
           10           1         216      0.47368       0.11455
 
            8           1         220      0.41447       0.11452
 
            7           1         227      0.35526       0.11243
 
            6           1         230      0.29605       0.10816
 
            5           1         234      0.23684       0.10145
 
            3           1         246      0.15789      0.093431
 
            2           1         265     0.078947      0.072792
 
            1           1         304            0  ............

 Total number in group    =      19
 Total number failing     =      17
 Product Limit Likelihood = -49.1692

                   Kaplan Meier Survival Probabilities
                      For Group Value = 7
 
       Number      Number                 Survival     Estimated
      at risk     Failing        Time  Probability    Std. Error
           21           1         142      0.95238      0.046471
 
           20           1         156      0.90476      0.064056
 
           19           1         163      0.85714       0.07636
 
           18           1         198      0.80952      0.085689
 
           16           1         205      0.75893      0.094092
 
           15           2         232      0.65774       0.10529
 
           13           4         233      0.45536       0.11137
 
            9           1         239      0.40476       0.10989
 
            8           1         240      0.35417       0.10717
 
            7           1         261      0.30357       0.10311
 
            6           2         280      0.20238      0.090214
 
            4           2         296      0.10119      0.067783
 
            2           1         323     0.050595      0.049281

 Total number in group    =      21
 Total number failing     =      19
 Product Limit Likelihood = -50.4277