[R] How to generate SE for the proportion value using a randomization process in R?
Rui Barradas
ru|pb@rr@d@@ @end|ng |rom @@po@pt
Thu Jan 28 21:21:36 CET 2021
Hello,
I don't know why you would need to see the indices but rewrite the
function bootprop as
bootprop_ind <- function(data, index){
d <- data[index, ]
#sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]], na.rm = TRUE)
index
}
and call in the same way. It will now return a matrix of indices with R
= 1000 rows and 19 columns.
Hope this helps,
Rui Barradas
Às 19:29 de 28/01/21, Marna Wagley escreveu:
> Hi Rui,
> I am sorry for asking you several questions.
>
> In the given example, randomizations (reshuffle) were done 1000 times,
> and its 1000 proportion values (results) are stored and it can be seen
> using b$t; but I was wondering how the table was randomized (which rows
> have been missed/or repeated in each randomizing procedure?).
>
> Is there any way we can see the randomized table and its associated
> results? Here in this example, I randomized (or bootstrapped) the table
> into three times (R=3) so I would like to store these three tables and
> look at them later to know which rows were repeated/missed. Is there any
> possibility?
> The example data and the code is given below.
>
> Thank you for your help.
>
> ####
> library(boot)
> dat<-structure(list(Sample = structure(c(1L, 12L, 13L, 14L, 15L, 16L,
> 17L, 18L, 19L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L), .Label = c("id1",
> "id10", "id11", "id12", "id13", "id14", "id15", "id16", "id17",
> "id18", "id19", "Id2", "id3", "id4", "id5", "id6", "id7", "id8",
> "id9"), class = "factor"), Time1 = c(0L, 1L, 1L, 1L, 0L, 0L,
> 1L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 1L, 0L, 1L, 0L), Time2 = c(1L,
> 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 1L, 0L,
> 1L, 1L)), .Names = c("Sample", "Time1", "Time2"), class = "data.frame",
> row.names = c(NA,
> -19L))
> daT<-data.frame(dat %>%
> mutate(Time1.but.not.in.Time2 = case_when(
> Time1 %in% "1" & Time2 %in% "0" ~ "1"),
> Time2.but.not.in.Time1 = case_when(
> Time1 %in% "0" & Time2 %in% "1" ~ "1"),
> BothTimes = case_when(
> Time1 %in% "1" & Time2 %in% "1" ~ "1")))
> cols.num <- c("Time1.but.not.in.Time2","Time2.but.not.in.Time1",
> "BothTimes")
> daT[cols.num] <- sapply(daT[cols.num],as.numeric)
> summary(daT)
>
> bootprop <- function(data, index){
> d <- data[index, ]
> sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]], na.rm = TRUE)
> }
>
> R <- 3
> set.seed(2020)
> b <- boot(daT, bootprop, R)
> b
> b$t0 # original
> b$t
> sd(b$t) # bootstrapped estimate of the SE of the sample prop.
> hist(b$t, freq = FALSE)
>
> str(b)
> b$data
> b$seed
> b$sim
> b$strata
> ################
>
>
> On Sat, Jan 23, 2021 at 12:36 AM Marna Wagley <marna.wagley using gmail.com
> <mailto:marna.wagley using gmail.com>> wrote:
>
> Yes Rui, I can see we don't need to divide by square root of sample
> size. The example is great to understand it.
> Thank you.
> Marna
>
>
> On Sat, Jan 23, 2021 at 12:28 AM Rui Barradas <ruipbarradas using sapo.pt
> <mailto:ruipbarradas using sapo.pt>> wrote:
>
> Hello,
>
> Inline.
>
> Às 07:47 de 23/01/21, Marna Wagley escreveu:
> > Dear Rui,
> > I was wondering whether we have to square root of SD to find
> SE, right?
>
> No, we don't. var already divides by n, don't divide again.
> This is the code, that can be seen by running the function name
> at a
> command line.
>
>
> sd
> #function (x, na.rm = FALSE)
> #sqrt(var(if (is.vector(x) || is.factor(x)) x else as.double(x),
> # na.rm = na.rm))
> #<bytecode: 0x55f3ce900848>
> #<environment: namespace:stats>
>
>
>
> >
> > bootprop <- function(data, index){
> > d <- data[index, ]
> > sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]],
> na.rm = TRUE)
> > }
> >
> > R <- 1e3
> > set.seed(2020)
> > b <- boot(daT, bootprop, R)
> > b
> > b$t0 # original
> > sd(b$t) # bootstrapped estimate of the SE of the sample prop.
> > sd(b$t)/sqrt(1000)
> > pandit*(1-pandit)
> >
> > hist(b$t, freq = FALSE)
>
>
> Try plotting the normal densities for both cases, the red line is
> clearly wrong.
>
>
> f <- function(x, xbar, s){
> dnorm(x, mean = xbar, sd = s)
> }
>
> hist(b$t, freq = FALSE)
> curve(f(x, xbar = b$t0, s = sd(b$t)), from = 0, to = 1, col =
> "blue",
> add = TRUE)
> curve(f(x, xbar = b$t0, s = sd(b$t)/sqrt(R)), from = 0, to = 1,
> col =
> "red", add = TRUE)
>
>
> Hope this helps,
>
> Rui Barradas
>
> >
> >
> >
> >
> > On Fri, Jan 22, 2021 at 3:07 PM Rui Barradas
> <ruipbarradas using sapo.pt <mailto:ruipbarradas using sapo.pt>
> > <mailto:ruipbarradas using sapo.pt <mailto:ruipbarradas using sapo.pt>>>
> wrote:
> >
> > Hello,
> >
> > Something like this, using base package boot?
> >
> >
> > library(boot)
> >
> > bootprop <- function(data, index){
> > d <- data[index, ]
> > sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]],
> na.rm = TRUE)
> > }
> >
> > R <- 1e3
> > set.seed(2020)
> > b <- boot(daT, bootprop, R)
> > b
> > b$t0 # original
> > sd(b$t) # bootstrapped estimate of the SE of the sample
> prop.
> > hist(b$t, freq = FALSE)
> >
> >
> > Hope this helps,
> >
> > Rui Barradas
> >
> > Às 21:57 de 22/01/21, Marna Wagley escreveu:
> > > Hi All,
> > > I was trying to estimate standard error (SE) for the
> proportion
> > value using
> > > some kind of randomization process (bootstrapping or
> jackknifing)
> > in R, but
> > > I could not figure it out.
> > >
> > > Is there any way to generate SE for the proportion?
> > >
> > > The example of the data and the code I am using is
> attached for your
> > > reference. I would like to generate the value of
> proportion with
> > a SE using
> > > a 1000 times randomization.
> > >
> > > dat<-structure(list(Sample = structure(c(1L, 12L, 13L,
> 14L, 15L, 16L,
> > > 17L, 18L, 19L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
> 11L), .Label
> > = c("id1",
> > > "id10", "id11", "id12", "id13", "id14", "id15",
> "id16", "id17",
> > > "id18", "id19", "Id2", "id3", "id4", "id5", "id6",
> "id7", "id8",
> > > "id9"), class = "factor"), Time1 = c(0L, 1L, 1L, 1L,
> 0L, 0L,
> > > 1L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 1L, 0L, 1L, 0L),
> Time2 = c(1L,
> > > 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 1L,
> 0L, 1L, 0L,
> > > 1L, 1L)), .Names = c("Sample", "Time1", "Time2"), class =
> > "data.frame",
> > > row.names = c(NA,
> > > -19L))
> > > daT<-data.frame(dat %>%
> > > mutate(Time1.but.not.in.Time2 = case_when(
> > > Time1 %in% "1" & Time2 %in% "0" ~ "1"),
> > > Time2.but.not.in.Time1 = case_when(
> > > Time1 %in% "0" & Time2 %in% "1" ~ "1"),
> > > BothTimes = case_when(
> > > Time1 %in% "1" & Time2 %in% "1" ~ "1")))
> > > daT
> > > summary(daT)
> > >
> > > cols.num <-
> c("Time1.but.not.in.Time2","Time2.but.not.in.Time1",
> > > "BothTimes")
> > > daT[cols.num] <- sapply(daT[cols.num],as.numeric)
> > > summary(daT)
> > > ProportionValue<-sum(daT$BothTimes,
> na.rm=T)/sum(daT$Time1, na.rm=T)
> > > ProportionValue
> > > standard error??
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
> > > R-help using r-project.org <mailto:R-help using r-project.org>
> <mailto:R-help using r-project.org <mailto:R-help using r-project.org>>
> 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.
> > >
> >
>
More information about the R-help
mailing list