在 R 中，如何根据来自其他两列的输入使用 CIr 函数的结果填充两列？

Question

Thanks for any help in advance. I have a dataset with correlation values in a column called 'exit' and corresponding sample sizes (n) in a column called 'samplesize' in a data frame called 'dataset'.

My task is to create an R script to populate two full columns (CIleft and CIright) with the confidence interval outputs using the CIr function within the "psychometric" package for each row of data. This CIr function operates as follows, outputting the left and right confidence interval values:

CIr(r = .9, n = 100, level = .95)  
[1] 0.8546667 0.9317133

Below is my unsuccessful script.

CI <- function(x)
{
  require(psychometric)
  library(psychometric)
  r <- x["dataset$exit"];
  n <- x["dataset$samplesize"];
  results <- CIr(r, n, level = .95);
  x["dataset$CIleft"] <- results[1];
  x["dataset$CIright"] <- results[2];
}

One complication (which I believe may be relevant) is that test runs of "CI(x)" in the console produce the following errors:

// Error in CIz(z, n, level) : (list) object cannot be coerced to type 'double' 

Then entering dataset2 <- as.matrix(dataset) and trying CI(x) again yields:

Error in dataset2$exit : $ operator is invalid for atomic vectors 

And for

dataset3 <- lapply(dataset$exit, as.numeric)
dataset4 <- lapply(dataset$samplesize, as.numeric)

trying CI(x) again yields:

Error in 1 + x : non-numeric argument to binary operator //

Can anyone assist in helping me populate each row of my data frame with the appropriate output for CIleft and CIright , given that r = 'exit' , and n = 'samplesize' ?

Answer 1

I don't think you need a function.

library("psychometric")

dataset$lwr = NULL
dataset$upr = NULL

for (row in 1:nrow(dataset)){
   dataset[["lwr"]][row] <- CIr(r = dataset[["exit"]][row], n = dataset[["samplesize"]][row], level = .95)[1]
   dataset[["upr"]][row] <- CIr(r = dataset[["exit"]][row], n = dataset[["samplesize"]][row], level = .95)[2]
}

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I will note though that it's generally advisable to avoid for loops in R because of its architecture (ie, they're slow). Perhaps someone else can provide a solution with something else, eg, apply . However, if you only have a small dataframe, the speed cost of using a for loop is unlikely to be noticeable.

---

Test Data:

set.seed(55); m = rnorm(26, 20, 40); dataset = data.frame( exit = seq(0, 1, 0.04), samplesize = abs(round(m)))
dataset$samplesize[dataset$samplesize == 0] = 5
dataset$exit[dataset$exit == 1] = 0.99