How to Perform a Log Rank Test in R, The most frequent technique to compare survival curves between two groups is to use a log-rank test.
The following hypotheses are used in this test
H0: There is no difference in survival between the two groups.
HA: There is a difference in survival between the two groups.
We can reject the null hypothesis and conclude that there is enough evidence to claim there is a difference in survival between the two groups if the p-value of the test is less than some significance level (e.g. =0.05).
In R, we may use the survdiff() function from the survival package to do a log-rank test, which has the following syntax.
How to Perform a Log Rank Test in R
survdiff(Surv(time, status) ~ predictors, data)
The Chi-Squared test statistic and related p-value are returned by this function.
This function is used to execute a log-rank test in R, as seen in the example below.
Example: Log Rank Test in R
We’ll use the ovarian dataset from the survival package for this example. The following information about 26 patients may be found in this dataset:
After being diagnosed with ovarian cancer, how long do you live (in months)?
Whether or whether the time spent surviving was censored
(rx = 1 or rx = 2) Type of treatment received
The following code demonstrates how to inspect the dataset’s first six rows.
Let’s view the first six rows of the dataset
head(ovarian) futime fustat age resid.ds rx ecog.ps 1 59 1 72.3315 2 1 1 2 115 1 74.4932 2 1 1 3 156 1 66.4658 2 1 2 4 421 0 53.3644 2 2 1 5 431 1 50.3397 2 1 1 6 448 0 56.4301 1 1 2
The following code demonstrates how to use a log-rank test to see if patients who got various treatments had different survival rates.
make a log-rank test
survdiff(Surv(futime, fustat) ~ rx, data=ovarian) Call: survdiff(formula = Surv(futime, fustat) ~ rx, data = ovarian) N Observed Expected (O-E)^2/E (O-E)^2/V rx=1 13 7 5.23 0.596 1.06 rx=2 13 5 6.77 0.461 1.06 Chisq= 1.1 on 1 degrees of freedom, p= 0.3
With one degree of freedom, the Chi-Squared test statistic is 1.1, and the associated p-value is 0.3. We cannot reject the null hypothesis because the p-value is not smaller than 0.05.
To put it another way, we don’t have enough evidence to establish that the two treatments have a statistically significant difference in survival.
The survival curves for each group can alternatively be plotted using the following syntax.
For each therapy group, draw a survival curve.
plot(survfit(Surv(futime, fustat) ~ rx, data = ovarian), xlab = "Time", ylab = "Survival probability")
In R, a plot of survival curves.
Although the survival curves change somewhat, the difference is not statistically significant, according to the log-rank test.