Re: [R] two newbie questions

From: Donald Braman <dbraman_at_law.gwu.edu>
Date: Sun, 22 Jun 2008 19:10:53 -0400

Wow -- many thanks for the mind-*expanding* help! I'm really impressed by R's ability to handle this so concisely It's going to take me a while to get used to applying things to vectors, but the more I understand, the nicer R looks.

On Sun, Jun 22, 2008 at 6:59 PM, jim holtman <jholtman_at_gmail.com> wrote:

> This does away with the 'for' loops and uses 'expand.grid' to create
> the combinations. I think I got the right variables substituted:
>
> my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1))))
> var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
> "intv1", "intv2")
> names(my.df) <- var.list
>
> # I have some are DVs:
> dvs <- c("dv1", "dv2", "dv3")
>
> # some IVs:
> ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")
>
> # and some binary interaction variables:
> intvs <- c("intv1", "intv2")
> library(car)
> my.df[intvs] <- lapply(my.df[intvs], function(x)
> recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))
>
> # now I loop through a series of interactions using the vector numbers:
> # create a dataframe of values to check
> xpnd <- expand.grid(dvs, ivs, intvs) # create combinations
> invisible(apply(xpnd, 1, function(.row) {
> jpeg(paste(paste(.row, collapse="_"),".jpg", sep=''))
>
> my.fit <- lm( my.df[[.row[1]]] ~ my.df[[.row[2]]] + my.df[[.row[3]]] +
> my.df[[.row[2]]]:my.df[[.row[3]]])
> colors <- ifelse (my.df[[.row[3]]] == 1, "black", "grey")
> plot(my.df[[.row[2]]], my.df[[.row[1]]], xlab=.row[2],
> ylab=.row[1], col=colors, pch=".")
> curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
> curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
>
> dev.off()
> }
> ))
>
>
> On Sun, Jun 22, 2008 at 6:26 PM, Donald Braman <dbraman_at_law.gwu.edu>
> wrote:
> > # I've tried to make this easy to paste into R, though it's probably
> > so simple you won't need to.
> > # I have some data (there are many more variables, but this is a
> > reasonable approximation of it)
> >
> > # here's a fabricated data frame that is similar in form to mine:
> > my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1))))
> > var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
> > "intv1", "intv2")
> > names(my.df) <- var.list
> >
> > # I have some are DVs:
> > dvs <- c("dv1", "dv2", "dv3")
> >
> > # some IVs:
> > ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")
> >
> > # and some binary interaction variables:
> > intvs <- c("intv1", "intv2")
> > library(car)
> > my.df[intvs] <- lapply(my.df[intvs], function(x)
> > recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))
> >
> > # now I loop through a series of interactions using the vector numbers:
> > for(dv in 1:3) {
> > for(iv in 4:8) {
> > for (intv in 9:10) {
> > jpeg(paste(names(my.df[iv]), names(my.df[dv]), names(my.df[intv]),
> > ".jpg", sep="_"))
> > with(data.frame(my.df), {
> > my.fit <- lm( my.df[[dv]] ~ my.df[[iv]] + my.df[[intv]] +
> > my.df[[iv]]:my.df[[intv]])
> > colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
> > plot(my.df[[iv]], my.df[[dv]], xlab=names(my.df[iv]),
> > ylab=names(my.df[dv]), col=colors, pch=".")
> > curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
> > curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
> > })
> > dev.off()
> > }
> > }
> > }
> >
> >
> > # Question1: Works fine, but using the vector numbers feels kludgy --
> > especially if the variables in question aren't consecutive.
> > # Is there a more elegant way of doing this with my lists of variable
> > names? Something like this, for example:
> > for(dv in dvs) {
> > for(iv in ivs) {
> > for (intv in intvs) {
> > jpeg(paste(dv, iv, intv, ".jpg", sep="_"))
> > with(data.frame(my.df), {
> > my.fit <- lm(my.df[dv] ~ my.df[iv] + my.df[intv] +
> my.df[iv]:my.df[intv])
> > colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
> > plot(my.df[iv], my.df[dv], xlab=iv, ylab=names(dv), col=colors,
> pch=".")
> > curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
> > curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
> > })
> > dev.off()
> > }
> > }
> > }
> >
> > # Clearly that's wrong -- why it's wrong is obscure to me, though!
> > Please educate me!
> >
> > # Question2: Could this could be done by using "apply" rather than a
> loop?
> > # Or is looping better here bc there are several actions performed at
> > each iteration?
> > # I'm still trying to get my head around all the ways to ditch looping in
> R.
> >
> >
> > Donald Braman
> > http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
> > http://research.yale.edu/culturalcognition
> > http://ssrn.com/author=286206
> >
> > ______________________________________________
> > R-help_at_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.
> >
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>

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