Survival Object - is 12month survival = 365days

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Survival Object - is 12month survival = 365days

R help mailing list-2
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?






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Re: Survival Object - is 12month survival = 365days

R help mailing list-2
On Sep 4, 2020, at 11:45 AM, POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST) via R-help <[hidden email]> 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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Re: Survival Object - is 12month survival = 365days

R help mailing list-2
Hi Mark

Huge thanks for the detailed answer.

At the moment, likely to be a mix of tabulation (30, 60, 90), a plot and some narrative with the 6mo, 12mo and so on. I think!

So it sounds like days is the answer, and then estimate months and years from days.

Sent from Nine<http://www.9folders.com/>
________________________________
From: Marc Schwartz <[hidden email]>
Sent: Friday, 4 September 2020 18:32
To: POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST)
Cc: R-help; Terry Therneau
Subject: Re: [R] Survival Object - is 12month survival = 365days

On Sep 4, 2020, at 11:45 AM, POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST) via R-help <[hidden email]> 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



********************************************************************************************************************

This message may contain confidential information. If yo...{{dropped:19}}

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.