The logrank test
The logrank test
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DOI：
10.1136/bmj.328.7447.1073
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发表时间：
2004-05-01
影响因子：
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通讯作者：
Altman, DG
中科院分区：
文献类型：
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作者：
Bland, JM;Altman, DG
我们经常希望比较两组（或更多组）个体的生存经验。例如，该表显示了51例复发性恶性神经胶质瘤1成人患者的生存时间，按肿瘤类型列表，并指出患者在分析时是已经死亡还是仍然存活-即，他们的生存时间被删失。[2]如图所示，生存曲线不同，但这是否足以得出结论，在人群中，间变性星形细胞瘤患者的生存率低于胶质母细胞瘤患者？我们可以计算每组的存活曲线，并比较在任何特定时间存活的比例。这种方法的缺点是，它没有提供两组总生存经验的比较，而是提供了一些任意时间点的比较。在图中，存活率的差异在某些时候比其他时候更大，最终变为零。我们在这里描述了对数秩检验，这是比较组间生存率的最常用方法，它考虑了整个随访期。它有一个相当大的优点，即它不需要我们知道生存曲线的形状或生存时间的分布。对数秩检验用于检验零假设，即在任何时间点，群体之间的事件（此处为死亡）概率均无差异。分析是基于事件发生的时间（这里是死亡）。对于每一个这样的时间，我们计算每个组中观察到的死亡人数，以及如果组间实际上没有差异时的预期人数。第1例死亡发生在第6周，第1组中有1例患者死亡。本周开始时，共有51例受试者存活，因此本周死亡风险为1/51。第1组有20例患者，因此，如果零假设为真，则第1组的预期死亡人数为20× 1/51= 0.39。同样，在第2组中，预期死亡人数为31× 1/51= 0.61。第二起事件发生在第10周，当时有两人死亡。两组中分别有19例和31例患者处于风险中（存活），1例在第6周死亡，因此第10周死亡的概率为2/50。预期死亡人数分别为19× 2/50= 0.76和31× 2/50= 1.24。每次事件发生时都执行相同的计算。如果对生存时间进行删失，则认为该个体在删失的那一周有死亡风险，但在随后的几周内没有。这种处理删失观察值的方式与Kaplan-Meier生存曲线相同。3根据各死亡时间的计算，第1组的预期死亡总数为22.48人，第2组为19.52人，观察到的死亡人数为14人和28人。我们现在可以使用零假设的χ2检验。检验统计量是每组的（O-E）2/E之和，其中O和E是观察到的和预期的事件的总和。这里（14 - 22.48）2/22.48+（28 - 19.52）2/19.52= 6.88。自由度是组数减1，即2 − 1= 1。从χ2分布表中，我们得到P< 0.01，因此组间差异具有统计学显著性。有一种不同的方法来计算检验统计量，4但我们更喜欢这种方法，因为它很容易扩展到多个组。也可以测试有序组之间的生存趋势。[4]虽然我们已经介绍了计算方法，但我们强烈建议使用统计软件。
We often wish to compare the survival experience of two (or more) groups of individuals. For example, the table shows survival times of 51 adult patients with recurrent malignant gliomas1 tabulated by type of tumour and indicating whether the patient had died or was still alive at analysis—that is, their survival time was censored. 2 As the figure shows, the survival curves differ, but is this sufficient to conclude that in the population patients with anaplastic astrocytoma have worse survival than patients with glioblastoma? We could compute survival curves3 for each group and compare the proportions surviving at any specific time. The weakness of this approach is that it does not provide a comparison of the total survival experience of the two groups, but rather gives a comparison at some arbitrary time point (s). In the figure the difference in survival is greater at some times than others and eventually becomes zero. We describe here the logrank test, the most popular method of comparing the survival of groups, which takes the whole follow up period into account. It has the considerable advantage that it does not require us to know anything about the shape of the survival curve or the distribution of survival times. The logrank test is used to test the null hypothesis that there is no difference between the populations in the probability of an event (here a death) at any time point. The analysis is based on the times of events (here deaths). For each such time we calculate the observed number of deaths in each group and the number expected if there were in reality no difference between the groups. The first death was in week 6, when one patient in group 1 died. At the start of this week, there were 51 subjects alive in total, so the risk of death in this week was 1/51. There were 20 patients in group 1, so, if the null hypothesis were true, the expected number of deaths in group 1 is 20× 1/51= 0.39. Likewise, in group 2 the expected number of deaths is 31× 1/51= 0.61. The second event occurred in week 10, when there were two deaths. There were now 19 and 31 patients at risk (alive) in the two groups, one having died in week 6, so the probability of death in week 10 was 2/50. The expected numbers of deaths were 19× 2/50= 0.76 and 31× 2/50= 1.24 respectively. The same calculations are performed each time an event occurs. If a survival time is censored, that individual is considered to be at risk of dying in the week of the censoring but not in subsequent weeks. This way of handling censored observations is the same as for the Kaplan-Meier survival curve. 3 From the calculations for each time of death, the total numbers of expected deaths were 22.48 in group 1 and 19.52 in group 2, and the observed numbers of deaths were 14 and 28. We can now use a χ2 test of the null hypothesis. The test statistic is the sum of (O–E) 2/E for each group, where O and E are the totals of the observed and expected events. Here (14− 22.48) 2/22.48+(28− 19.52) 2/19.52= 6.88. The degrees of freedom are the number of groups minus one, ie2− 1= 1. From a table of the χ2 distribution we get P< 0.01, so that the difference between the groups is statistically significant. There is a different method of calculating the test statistic, 4 but we prefer this approach as it extends easily to several groups. It is also possible to test for a trend in survival across ordered groups. 4 Although we have shown how the calculation is made, we strongly recommend the use of statistical software.