[R] loop for comparing two or more groups using bootstrapping

[Kristi Glover]

Dear Jim,

Thank you very much for the code. I run it but it gave me row names like "year224", "year142".

are these the difference between columns? If we want to get bootstrapping means of difference between years (year2-year1; year3-year1), its CI and exact p value, how can we get it?

thanks

KG

----

head(daT)

colname.mat<-combn(paste0("year",1:4),2)

samplenames<-apply(colname.mat,2,paste,collapse="")

k<-10

for(column in 1:ncol(colname.mat)) {

 assign(samplenames[column],replicate(k,sample(unlist(daT[,colname.mat[,column]]),3,TRUE)))

}


> get(samplenames[1])
         [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
year224 0.556 0.667 0.571 0.526 0.629 0.696 0.323 0.526 0.256 0.667
year142 0.324 0.324 0.706 0.638 0.600 0.294 0.612 0.688 0.432 0.387
year237 0.571 0.696 0.629 0.471 0.462 0.471 0.452 0.595 0.333 0.435




________________________________
From: Jim Lemon <drjimlemon using gmail.com>
Sent: September 11, 2018 1:44 AM
To: Kristi Glover
Cc: r-help mailing list
Subject: Re: [R] loop for comparing two or more groups using bootstrapping

Hi Kristy,
Try this:

colname.mat<-combn(paste0("year",1:4),2)
samplenames<-apply(colname.mat,2,paste,collapse="")
k<-10000
for(column in 1:ncol(colname.mat)) {
 assign(samplenames[column],replicate(k,sample(unlist(daT[,colname.mat[,column]]),3,TRUE)))
}

Then use get(samplenames[1]) and so on to access the values.

Jim
On Tue, Sep 11, 2018 at 4:52 PM Kristi Glover <kristi.glover using hotmail.com> wrote:
>
> Hi R users,
>
> I was trying to test a null hypothesis of difference between two groups was 0. I have many years data, such as year1, year2,, year3, year4 and I was trying to compare between year1 and year2, year1 and year3, year2 and year3 and so on and have used following code with an example data.
>
>
> I tried to make a loop but did not work to compare between many years, and also want to obtain the exact p value. Would you mind to help me to make a loop?
>
> Thanks for your help.
>
>
> KG
>
>
> daT<-structure(list(year1 = c(0.417, 0.538, 0.69, 0.688, 0.688, 0.606,
> 0.667, 0.7, 0.545, 0.462, 0.711, 0.642, 0.744, 0.604, 0.612,
> 0.667, 0.533, 0.556, 0.444, 0.526, 0.323, 0.308, 0.195, 0.333,
> 0.323, 0.256, 0.345, 0.205, 0.286, 0.706, 0.7, 0.6, 0.571, 0.364,
> 0.429, 0.326, 0.571, 0.424, 0.341, 0.387, 0.341, 0.324, 0.696,
> 0.696, 0.583, 0.556, 0.645, 0.435, 0.471, 0.556), year2 = c(0.385,
> 0.552, 0.645, 0.516, 0.629, 0.595, 0.72, 0.638, 0.557, 0.588,
> 0.63, 0.744, 0.773, 0.571, 0.723, 0.769, 0.667, 0.667, 0.526,
> 0.476, 0.294, 0.323, 0.222, 0.556, 0.263, 0.37, 0.357, 0.25,
> 0.323, 0.778, 0.667, 0.636, 0.583, 0.432, 0.412, 0.333, 0.571,
> 0.39, 0.4, 0.452, 0.326, 0.471, 0.7, 0.75, 0.615, 0.462, 0.556,
> 0.4, 0.696, 0.465), year3 = c(0.435, 0.759, 0.759, 0.759, 0.714,
> 0.593, 0.651, 0.683, 0.513, 0.643, 0.652, 0.757, 0.791, 0.649,
> 0.78, 0.5, 0.5, 0.5, 0.533, 0.429, 0.333, 0.286, 0.231, 0.533,
> 0.303, 0.417, 0.333, 0.333, 0.357, 0.909, 1, 0.952, 0.8, 0.556,
> 0.529, 0.562, 0.762, 0.513, 0.733, 0.611, 0.733, 0.647, 0.909,
> 0.857, 0.8, 0.556, 0.588, 0.562, 0.857, 0.513), year4 = c(0.333,
> 0.533, 0.6, 0.483, 0.743, 0.5, 0.691, 0.619, 0.583, 0.385, 0.653,
> 0.762, 0.844, 0.64, 0.667, 0.571, 0.571, 0.615, 0.421, 0.5, 0.205,
> 0.308, 0.25, 0.6, 0.242, 0.308, 0.276, 0.235, 0.211, 0.9, 0.632,
> 0.72, 0.727, 0.356, 0.5, 0.368, 0.5, 0.41, 0.562, 0.514, 0.4,
> 0.409, 0.632, 0.72, 0.727, 0.4, 0.5, 0.421, 0.5, 0.462)), .Names = c("year1",
> "year2", "year3", "year4"), row.names = c(NA, -50L), class = "data.frame")
>
> head(daT)
>
> # null hypothesis; difference is equal to zero
>
> dif1.2<-daT$year2-daT$year1
>
> k=10000
>
> mysamples1.2=replicate(k, sample(dif1.2, replace=T))
>
> mymeans1.2=apply(mysamples1.2, 2, mean)
>
> quantile(mymeans1.2, c(0.025, 0.975))
>
> hist(mysamples1.2)
>
> mean(mymeans1.2)
>
> #what is p value?
>
>
> #similarly Now I want to compare between year 1 and year3,
>
> dif1.3<-daT$year3-daT$year1
>
> mysamples1.3=replicate(k, sample(dif1.3, replace=T))
>
> mymeans1.3=apply(mysamples1.3, 2, mean)
>
> quantile(mymeans1.3, c(0.025, 0.975))
>
>
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