Dummy (factor) based on a pair of variables
df <- read.table(textConnection("y,i,j
+ 1,AUT,BEL
+ 2,AUT,GER
+ 3,BEL,GER"), header=T,sep=",", as.is=T)
> df
y i j
1 1 AUT BEL
2 2 AUT GER
3 3 BEL GER
> countries <- unique(c(df$i,df$j))
> countries
[1] "AUT" "BEL" "GER"
> df[countries] <- sapply(countries, function(x) df[x] <<- df$i == x
| df$j == x)
> df
y i j AUT BEL GER
1 1 AUT BEL TRUE TRUE FALSE
2 2 AUT GER TRUE FALSE TRUE
3 3 BEL GER FALSE TRUE TRUE
Obviously it would not be possible to test this arrangement with lm.
So I tried scaling it up and testing on:
dft <- data.frame(y=rnorm(100), i = sample(countries, 100,
replace=T), j= sample(countries, 100, replace=T))
#Removed all the duplicates with:
dft <- dft(dft$i != dft$j, ]
#and it did not give proper answers.
This seems to give correct answers
dft[countries] <- sapply(countries, function(y) apply(dft, 1,
function(x) x[2] == y | x[3] == y))
And application of those variables is handles in a reasonable manner
by the R formula parser:
> lm(y ~ AUT + BEL + GER, data=dft)
Call:
lm(formula = y ~ AUT + BEL + GER, data = dft)
Coefficients:
(Intercept) AUTTRUE BELTRUE GERTRUE
0.09192 0.15130 -0.29274 NA
-
David Winsemius
On Apr 18, 2009, at 4:09 PM, Jason Morgan wrote:
On 2009.04.18 15:58:30, Jason Morgan wrote:
On 2009.04.18 13:52:35, Serguei Kaniovski wrote:
I can generate the above dummies but can this design be imputed in a reg. model directly?
Oops, I apologize for not reading the whole question. Can you do the
following:
lm(y ~ I(ifelse(df$i=="AUT"|df$j=="AUT", 1, 0)) +
I(ifelse(df$i=="BEL"|df$j=="BEL", 1, 0)) +
I(ifelse(df$i=="GER"|df$j=="GER", 1, 0)), data=df)
If you exclude the ifelse(), you will get a vector of TRUE/FALSE,
which may or may not work.
~Jason
Hello Serguei,
I am sure there is a better way to do this, but the following seems
to
work:
# Create sample data.frame()
i <- c("AUT", "AUT", "BEL")
j <- c("BEL", "GER", "GER")
df <- data.frame(i=i, j=j)
# Create dummy vectors
df$d.aut <- ifelse(df$i=="AUT"|df$j=="AUT", 1, 0)
df$d.bel <- ifelse(df$i=="BEL"|df$j=="BEL", 1, 0)
df$d.ger <- ifelse(df$i=="GER"|df$j=="GER", 1, 0)
# Print results
df
HTH,
~Jason
-- Jason W. Morgan Graduate Student, Political Science *The Ohio State University*
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David Winsemius, MD Heritage Laboratories West Hartford, CT