Got it, I did not know of the 'recode' function in car.
So you would like to recode those specific columns then? Once again, we
can do it without a loop, this time with the help of a function called
lapply, which applies a function to each item in a list in turn.
Try:
reverse_me_varnames <- c("HEQUAL", "HREVDIS1", "HREVDIS2")
reversed_varnames <-paste("R", reverse_me_varnames, sep = "")
## See ?paste
mdf[reversed_varnames] <-
lapply(mdf[reverse_me_varnames],
function(x) recode(x, recodes = "5:7=NA; 1=4; 2=3; 3=2; 4=1;",
as.factor.result = FALSE))
Now what does this actually mean? To the left of '<-' is simply the new
columns of our data.frame. We want to then use lapply to do some
function to a list of objects. The first argument to lapply is that
list. In this case, it is simply the columns of the data.frame you want
reversed. A data.frame is a list in R. See ?list and ?data.frame.
Then, the next argument to lapply is a function that we want to perform
on each element in our list. So, we create a function that accepts as
input a variable I simply call 'x'. This 'x' is going to be an item
from the list we passed lapply, which is one of the columns of mdf in
'reverse_me_varnames'.
We then use the recode function in the car package to recode x, in a
similar way to what you tried before. This function of x we define will
get called three times in the above example, once for each of
reverse_me_varnames. It will then assign those three new columns to the
left-hand side of the <- operator, which are three newly-named columns.
To see why what you tried before did not work, with the for loop, try:
mdf$HEQUAL
contrasted with
t1 <- c("HEQUAL")
mdf$t1
From the help for ?Extract, $ does not allow 'computed' indices.
I hope this helps!
Erik
Donald Braman wrote:
Erik,
Your example was just what I needed to generate the data -- many, many
thanks! The names() function was something I had not grasped fully. I
now have this and it works very nicely:
var_list <- c("HEQUAL", "EWEALTH", "ERADEQ", "HREVDIS1", "EDISCRIM",
"HREVDIS2")
mdf <- data.frame(replicate(length(var_list), sample(7,100, replace =
TRUE))) ## generate random data
names(mdf) ## default names
names(mdf) <- var_list ## use our names
mdf
I'm still trying to figure out how to recode (using the car package)
data into new variables using a similar loop. Basically, I'm not sure
how to call the variable name and append it to the dataframe name in a
loop. In Stata I'd do this using single quotes, but clearly that's not
how R works. I tried several variations on this:
reverse_me_varnames <- c("HEQUAL", "HREVDIS1", "HREVDIS2")
reversed_varnames <- c("RHEQUAL", "RHREVDIS1", "RHREVDIS2")
for(i in 1:length(reverse_me_varnames))
{mdf$reversed_varnames[i] <- recode(mdf$reverse_me_varnames[i],
'5:7=NA; 1=4; 2=3; 3=2; 4=1;', as.factor.result=FALSE)
While I don't get an error message, the data don't change. Any advice
on reverse coding non-continguous variables?
On Mon, May 19, 2008 at 4:12 PM, Donald Braman <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]>> wrote:
Many thanks --
You are right; I had rnorm() and sample() mixed up in my code. I'll
work on generating a normal ordinal sample next.
Cheers, Don
On Mon, May 19, 2008 at 4:07 PM, Erik Iverson
<[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>> wrote:
Hello -
Donald Braman wrote:
# I'm new to R and am trying to get the hang of how it handles
# dataframes & loops. If anyone can help me with some simple
tasks,
# I'd be much obliged.
# First, i'd like to generate some random data in a dataframe
# to efficiently illustrate what I'm up to.
# let's say I have six variables as listed below (I really
# have hundreds, but a few will illustrate the point).
# I want to generate my dataframe (mdf)
# with the 6 variables X 100 values with rnorm(7).
# How do I do this? I tried many variations on the following:
var_list <- c("HEQUAL", "EWEALTH", "ERADEQ", "HREVDIS1",
"EDISCRIM",
"HREVDIS2")
for(i in 1:length(var_list)) {var_list[1] <- rnorm(100)}
mdf <- data.frame(cbind(varlist[1:length(var_list)])
mdf
There are many ways to do this. Do you mean that you want 6
columns, 100 observations in each column, each a sample from a
normal distribution with mean = 7 and sd = 1? You can do this
without looping in one of several ways. If you are coming from
a SAS environment (my guess since you talk of looping over
data.frames), you may be used to looping through a data object.
In R, you can usually avoid this since many functions are
vectorized, or take a 'whole object' approach.
var_list <- c("HEQUAL", "EWEALTH", "ERADEQ", "HREVDIS1",
"EDISCRIM", "HREVDIS2")
mdf <- data.frame(replicate(6, rnorm(100, 7))) ## generate
random data
names(mdf) ## default names
names(mdf) <- var_list ## use our names
# Then, I'd like to recode the variables that begin with the
letter "H".
# I've tried many variations of the following, but to no avail:
reverse_list <- c("HEQUAL", "HREVDIS1", "HREVDIS2")
reversed_list <- c("RHEQUAL", "RHREVDIS1", "RHREVDIS2")
for(i in 1:length(reverse_list))
{mdf[ ,e_reversed_list][[i]] <- recode(mdf[
,e_reverse_list][[i]],
'5:99=NA; 1=4; 2=3; 3=2; 4=1; ', as.factor.result=FALSE)
I'm not quite sure what you are after here. What do you mean by
recode? What package is your 'recode' function located in?
It appears that you may be under the impression that the
data.frame contains integers, but certainly it will not since it
was generated with rnorm? sample can generate a samples of the
type you may be after, for example,
> sample(7, 100, replace = TRUE)
Best,
Erik Iverson
--
Donald Braman
http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
http://research.yale.edu/culturalcognition
http://ssrn.com/author=286206
--
Donald Braman
http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
http://research.yale.edu/culturalcognition
http://ssrn.com/author=286206
______________________________________________
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.
