Dipu
September 20, 2023, 4:32am
1
set.seed(11)
for(i in 1:4){
dat1<- rnorm(10, 0, 1)
dat2<- sample(dat1, 6, replace = T)
print(dat1)
}
If dat2 is run in the loop, then the result for dat1 is different from the output of dat1 if we don't run dat2. How can I fix dat1 while taking a sample within a loop?
FJCC
September 20, 2023, 1:16pm
2
Does this do what you want?
set.seed(11)
dat1<- rnorm(10, 0, 1)
dat2 <- vector(mode = "list", length = 4)
for(i in 1:4){
dat2[[i]] <- sample(dat1, 6, replace = T)
}
dat2
#> [[1]]
#> [1] 1.32360565 0.02659437 -0.59103110 0.62491779 1.32360565 -0.93415132
#>
#> [[2]]
#> [1] -1.51655310 -1.51655310 0.02659437 0.02659437 -1.51655310 -1.00412058
#>
#> [[3]]
#> [1] -1.00412058 0.02659437 -0.04572296 0.62491779 1.32360565 0.62491779
#>
#> [[4]]
#> [1] 0.62491779 -0.04572296 1.32360565 -1.36265335 -0.59103110 1.17848916
Created on 2023-09-20 with reprex v2.0.2
system
Closed
October 11, 2023, 1:16pm
3
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