On Sep 4, 2020, at 11:45 AM, POLWART, Calum (SOUTH TEES HOSPITALS NHS 
FOUNDATION TRUST) via R-help <r-help@r-project.org> wrote:
> 
> Using survfit I can get the '1 year' Survival from this dataset which holds 
> survival in days:
> 
> require (survival)
> survfit( Surv(time, status) ~sex, data=colon)
> summary (fit, 365)
> 
> My current real world data I'm calculating time using lubridate to calculate 
> time and since it made the axis easy I just told it to do and so my "time" 
> appears to be  a float in months.
> 
> time <- time_length(interval(startDate, endDate), "months")
> 
> Is there a "right" approach to this (as in a convention). If I use 12months 
> as a year and describe it in the write up as 12, 24 and 36 month survival 
> rather than 1, 2 and 3 year presumably that is OK..
> 
> I've been asked to report 30, 60 & 90day. Then 6month, 1, 2 and 3 year 
> survival.
> 
> Should I calculate time 3 times, (interval day, month and year) and run the 
> survival on each to get the requested outputs or would people just provide 
> something close.
> 
> Should I run a campaign to decimilise time?

Hi,

The answer may depend upon whether you are presenting the results in a tabular 
fashion, in the body of a manuscript, or in a figure. Also, what may be the 
community conventions in your domain. 

If you want to get the irregular time points out in a single output, you can 
use the times argument to do this, remembering that the default time intervals 
are in days for this dataset:

> summary(fit, times = c(30, 60, 90, 180, 365.25, 2 * 365.25, 3 * 365.25))
Call: survfit(formula = Surv(time, status) ~ sex, data = colon)

                sex=0 
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
   30    887       2    0.998 0.00159        0.995        1.000
   60    880       6    0.991 0.00317        0.985        0.997
   90    869      11    0.979 0.00485        0.969        0.988
  180    827      42    0.931 0.00849        0.915        0.948
  365    731      94    0.825 0.01274        0.801        0.851
  730    595     135    0.673 0.01576        0.643        0.705
 1096    536      57    0.608 0.01641        0.577        0.641

                sex=1 
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
   30    962       5    0.995 0.00230        0.990        0.999
   60    955       6    0.989 0.00341        0.982        0.995
   90    947       8    0.980 0.00446        0.972        0.989
  180    906      41    0.938 0.00776        0.923        0.953
  365    819      85    0.850 0.01150        0.828        0.873
  730    679     133    0.711 0.01462        0.683        0.740
 1096    592      84    0.623 0.01566        0.593        0.654


Now, the time output there is arguably a bit cumbersome to read...but, at least 
you get the relevant values in a single output. You can transform those values 
as you may require.

Another option is to use the scale argument, but I just noted that, unless I am 
missing something, I think that there may be a lingering buglet in the code for 
summary.survfit(), and I am adding Terry Therneau here as a cc:, if that is 
correct. The behavior of the interaction between the times and scale arguments 
changed in 2009 after an exchange I had with Thomas Lumley: 

  https://stat.ethz.ch/pipermail/r-devel/2009-April/052901.html

and it is not clear to me if the current behavior is or is not intended after 
all this time. Albeit, it may be the defacto behavior at this point in either 
case, given some volume of code written over the years that may depend upon 
this behavior.

Thus, this may be better for you, using the current behavior:

> summary(fit, scale = 30.44, times = c(1, 2, 3, 6, 12, 24, 36) * 30.44)
Call: survfit(formula = Surv(time, status) ~ sex, data = colon)

                sex=0 
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
    1    887       2    0.998 0.00159        0.995        1.000
    2    880       6    0.991 0.00317        0.985        0.997
    3    868      12    0.977 0.00498        0.968        0.987
    6    826      42    0.930 0.00855        0.914        0.947
   12    731      93    0.825 0.01274        0.801        0.851
   24    595     135    0.673 0.01576        0.643        0.705
   36    536      57    0.608 0.01641        0.577        0.641

                sex=1 
 time n.risk n.event survival std.err lower 95% CI upper 95% CI
    1    962       5    0.995 0.00230        0.990        0.999
    2    955       6    0.989 0.00341        0.982        0.995
    3    946       9    0.979 0.00458        0.970        0.988
    6    906      40    0.938 0.00776        0.923        0.953
   12    819      85    0.850 0.01150        0.828        0.873
   24    679     133    0.711 0.01462        0.683        0.740
   36    592      84    0.623 0.01566        0.593        0.654


where the times values are now in months over the range of values, instead of 
days.

I don't use the lubridate package, so there may be other options for you there, 
but the above will work, if your underlying time intervals in the source data 
frame for the model are still in days as a unit of measurement. 

Using the base graphics functions, albeit perhaps you are using ggplot or 
similar, you can plot the above model with axis markings at the irregular time 
intervals, using something like the following:

plot(fit, xaxt = "n", las = 1, xlim = c(0, 36 * 30.44))
axis(1, at = c(1, 2, 3, 6, 12, 24, 36) * 30.44, labels = c(1, 2, 3, 6, 12, 24, 
36), cex.axis = 0.65)

This essentially truncates the x axis to 36 months, since the intervals in the 
example colon dataset go to about 9 years or so, and does not label the x axis. 
Bearing in mind that the underlying x axis unit is in days, the axis() function 
then places labels at the irregular intervals. You could then annotate the plot 
further as you may desire.

Regards,

Marc Schwartz

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  • [R] ... POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST) via R-help
    • ... Marc Schwartz via R-help
      • ... POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST) via R-help

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