Your R code looks correct. Because this is a straightforward calculation, I would be surprised if there were any differences with SPSS. It may be worthwhile to check if SPSS gives partial correlations or semipartial correlations. For example, if you take the correlation between
py <- resid(lm(y ~ z1 + z2,data=mydat2)) and x where mydat2 has missing values removed, you get 0.47. On Tue, Dec 1, 2009 at 8:24 PM, dadrivr <dadr...@gmail.com> wrote: > > I am trying to calculate a partial correlation and p-values. Unfortunately, > the results in R are different than what SPSS gives. > > Here is an example in R (calculating the partial correlation of x and y, > controlling for z1 and z2): > > x <- c(1,20,14,30,9,4,8) > y <- c(5,6,7,9,NA,10,6) > z1 <- c(13,8,16,14,26,13,20) > z2 <- c(12,NA,2,5,8,16,13) > fmx <- lm(x ~ z1 + z2, na.action = na.exclude) > fmy <- lm(y ~ z1 + z2, na.action = na.exclude) > yres <- resid(fmy) > xres <- resid(fmx) > cor(xres, yres, use = "p") > ct <- cor.test(xres, yres) > ct$estimate > ct$p.value > > R give me: > r = .65, p = .23 > > However, SPSS calculates: > r = .46, p = .70 > > I think something may be different with R's handling of missing data, as > when I replace the NA's with values, R and SPSS give the same r-values, > albeit different p-values still. I am doing pairwise case exclusion in both > R and SPSS. Any ideas why I'm getting different values? Is something wrong > with my formula in R? Any help would be greatly appreciated. Thanks! > ______________________________________________ R-help@r-project.org mailing list 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.
