In the minimal example below, I am trying to use the values of a character string vars in a regression formula. However, I am only able to pass the string of variable names ("v2+v3+v4") to the formula, not the real meaning of this string (e.g., "v2" is dat$v2).
I know there are better ways to run the regression (e.g., lm(v1 ~ v2 + v3 + v4, data=dat)). My situation is more complex, and I am trying to figure out how to use a character string in a formula. Any thoughts?
Updated below code
# minimal example
# create data frame
v1 <- rnorm(10)
v2 <- sample(c(0,1), 10, replace=TRUE)
v3 <- rnorm(10)
v4 <- rnorm(10)
dat <- cbind(v1, v2, v3, v4)
dat <- as.data.frame(dat)
# create objects of column names
c.2 <- colnames(dat)[2]
c.3 <- colnames(dat)[3]
c.4 <- colnames(dat)[4]
# shortcut to get to the type of object my full code produces
vars <- paste(c.2, c.3, c.4, sep="+")
### TRYING TO SOLVE FROM THIS POINT:
print(vars)
# [1] "v2+v3+v4"
# use vars in regression
regression <- paste0("v1", " ~ ", vars)
m1 <- lm(as.formula(regression), data=dat)
Update:
#Arun was correct about the missing "" on v1 in the first example. This fixed my example, but I was still having problems with my real code. In the code chunk below, I adapted my example to better reflect my actual code. I chose to create a simpler example at first thinking that the problem was the string vars.
Here's an example that does not work :) Uses the same data frame dat created above.
dv <- colnames(dat)[1]
r2 <- colnames(dat)[2]
# the following loop creates objects r3, r4, r5, and r6
# r5 and r6 are interaction terms
for (v in 3:4) {
r <- colnames(dat)[v]
assign(paste("r",v,sep=""),r)
r <- paste(colnames(dat)[2], colnames(dat)[v], sep="*")
assign(paste("r",v+2,sep=""),r)
}
# combine r3, r4, r5, and r6 then collapse and remove trailing +
vars2 <- sapply(3:6, function(i) {
paste0("r", i, "+")
})
vars2 <- paste(vars2, collapse = '')
vars2 <- substr(vars2, 1, nchar(vars2)-1)
# concatenate dv, r2 (as a factor), and vars into `eq`
eq <- paste0(dv, " ~ factor(",r2,") +", vars2)
Here is the issue:
print(eq)
# [1] "v1 ~ factor(v2) +r3+r4+r5+r6"
Unlike regression in the first example, eq does not bring in the column names (e.g., v3). The object names (e.g., r3) are retained. As such, the following lm() command does not work.
m2 <- lm(as.formula(eq), data=dat)
I see a couple issues going on here. First, and I don't think this is causing any trouble, but let's make your data frame in one step so you don't have v1 through v4 floating around both in the global environment as well as in the data frame. Second, let's just make v2 a factor here so that we won't have to deal with making it a factor later.
dat <- data.frame(v1 = rnorm(10),
v2 = factor(sample(c(0,1), 10, replace=TRUE)),
v3 = rnorm(10),
v4 = rnorm(10) )
Part One Now, for your first part, it looks like this is what you want:
lm(v1 ~ v2 + v3 + v4, data=dat)
Here's a simpler way to do that, though you still have to specify the response variable.
lm(v1 ~ ., data=dat)
Alternatively, you certainly can build up the function with paste and call lm on it.
f <- paste(names(dat)[1], "~", paste(names(dat)[-1], collapse=" + "))
# "v1 ~ v2 + v3 + v4"
lm(f, data=dat)
However, my preference in these situations is to use do.call, which evaluates expressions before passing them to the function; this makes the resulting object more suitable for calling functions like update on. Compare the call part of the output.
do.call("lm", list(as.formula(f), data=as.name("dat")))
Part Two About your second part, it looks like this is what you're going for:
lm(factor(v2) + v3 + v4 + v2*v3 + v2*v4, data=dat)
First, because v2 is a factor in the data frame, we don't need that part, and secondly, this can be simplified further by better using R's methods for using arithmetical operations to create interactions, like this.
lm(v1 ~ v2*(v3 + v4), data=dat)
I'd then simply create the function using paste; the loop with assign, even in the larger case, is probably not a good idea.
f <- paste(names(dat)[1], "~", names(dat)[2], "* (",
paste(names(dat)[-c(1:2)], collapse=" + "), ")")
# "v1 ~ v2 * ( v3 + v4 )"
It can then be called using either lm directly or with do.call.
lm(f, data=dat)
do.call("lm", list(as.formula(f), data=as.name("dat")))
About your code The problem you had with trying to use r3 etc was that you wanted the contents of the variable r3, not the value r3. To get the value, you need get, like this, and then you'd collapse the values together with paste.
vars <- sapply(paste0("r", 3:6), get)
paste(vars, collapse=" + ")
However, a better way would be to avoid assign and just build a vector of the terms you want, like this.
vars <- NULL
for (v in 3:4) {
vars <- c(vars, colnames(dat)[v], paste(colnames(dat)[2],
colnames(dat)[v], sep="*"))
}
paste(vars, collapse=" + ")
A more R-like solution would be to use lapply:
vars <- unlist(lapply(colnames(dat)[3:4],
function(x) c(x, paste(colnames(dat)[2], x, sep="*"))))
TL;DR: use paste.
create_ctree <- function(col){
myFormula <- paste(col, "~.", collapse="")
ctree(myFormula, data)
}
create_ctree("class")
Related
I am comparing two lists of formulas to see if some previously computed models can be reused. Right now I'm doing this like this:
set.seed(123)
# create some random formulas
l1 <- l2 <- list()
for (i in 1:10) {
l1[[i]] <- as.formula(paste("z ~", paste(sample(letters, 3), collapse = " + ")))
l2[[i]] <- as.formula(paste("z ~", paste(sample(letters, 3), collapse = " + ")))
}
# at least one appears in the other list
l1[[5]] <- l2[[7]]
# helper function to convert formulas to character strings
as.formulaCharacter <- function(x) paste(deparse(x))
# convert both lists to strings
s1 <- sapply(l1, as.formulaCharacter)
s2 <- sapply(l2, as.formulaCharacter)
# look up elements of one vector in the other
idx <- match(s1, s2, nomatch = 0L) # 7
s1[idx] # found matching elements
However, I noticed that some formulas are not retrieved although they are practically equivalent.
f1 <- z ~ b + c + b:c
f2 <- z ~ c + b + c:b
match(as.formulaCharacter(f1), as.formulaCharacter(f2)) # no match
I get why this result is different, the strings just aren't the same, but I'm struggling with how to extend this approach method to also work for formulas with reordered elements. I could use strsplit to first sort all formula components independently, but that sounds horribly inefficient to me.
Any ideas?
If the formulas are restricted to a sum of terms which contain colon separated variables then we can create a standardized string by extracting the term labels, exploding those with colons, sorting them, pasting the exploded terms back together, sorting this and turning that into a formula string.
stdize <- function(fo) {
s <- strsplit(attr(terms(f2), "term.labels"), ":")
terms <- sort(sapply(lapply(s, sort), paste, collapse = ":"))
format(reformulate(terms, all.vars(fo)[1]))
}
stdize(f1) == stdize(f2)
## [1] TRUE
In the minimal example below, I am trying to use the values of a character string vars in a regression formula. However, I am only able to pass the string of variable names ("v2+v3+v4") to the formula, not the real meaning of this string (e.g., "v2" is dat$v2).
I know there are better ways to run the regression (e.g., lm(v1 ~ v2 + v3 + v4, data=dat)). My situation is more complex, and I am trying to figure out how to use a character string in a formula. Any thoughts?
Updated below code
# minimal example
# create data frame
v1 <- rnorm(10)
v2 <- sample(c(0,1), 10, replace=TRUE)
v3 <- rnorm(10)
v4 <- rnorm(10)
dat <- cbind(v1, v2, v3, v4)
dat <- as.data.frame(dat)
# create objects of column names
c.2 <- colnames(dat)[2]
c.3 <- colnames(dat)[3]
c.4 <- colnames(dat)[4]
# shortcut to get to the type of object my full code produces
vars <- paste(c.2, c.3, c.4, sep="+")
### TRYING TO SOLVE FROM THIS POINT:
print(vars)
# [1] "v2+v3+v4"
# use vars in regression
regression <- paste0("v1", " ~ ", vars)
m1 <- lm(as.formula(regression), data=dat)
Update:
#Arun was correct about the missing "" on v1 in the first example. This fixed my example, but I was still having problems with my real code. In the code chunk below, I adapted my example to better reflect my actual code. I chose to create a simpler example at first thinking that the problem was the string vars.
Here's an example that does not work :) Uses the same data frame dat created above.
dv <- colnames(dat)[1]
r2 <- colnames(dat)[2]
# the following loop creates objects r3, r4, r5, and r6
# r5 and r6 are interaction terms
for (v in 3:4) {
r <- colnames(dat)[v]
assign(paste("r",v,sep=""),r)
r <- paste(colnames(dat)[2], colnames(dat)[v], sep="*")
assign(paste("r",v+2,sep=""),r)
}
# combine r3, r4, r5, and r6 then collapse and remove trailing +
vars2 <- sapply(3:6, function(i) {
paste0("r", i, "+")
})
vars2 <- paste(vars2, collapse = '')
vars2 <- substr(vars2, 1, nchar(vars2)-1)
# concatenate dv, r2 (as a factor), and vars into `eq`
eq <- paste0(dv, " ~ factor(",r2,") +", vars2)
Here is the issue:
print(eq)
# [1] "v1 ~ factor(v2) +r3+r4+r5+r6"
Unlike regression in the first example, eq does not bring in the column names (e.g., v3). The object names (e.g., r3) are retained. As such, the following lm() command does not work.
m2 <- lm(as.formula(eq), data=dat)
I see a couple issues going on here. First, and I don't think this is causing any trouble, but let's make your data frame in one step so you don't have v1 through v4 floating around both in the global environment as well as in the data frame. Second, let's just make v2 a factor here so that we won't have to deal with making it a factor later.
dat <- data.frame(v1 = rnorm(10),
v2 = factor(sample(c(0,1), 10, replace=TRUE)),
v3 = rnorm(10),
v4 = rnorm(10) )
Part One Now, for your first part, it looks like this is what you want:
lm(v1 ~ v2 + v3 + v4, data=dat)
Here's a simpler way to do that, though you still have to specify the response variable.
lm(v1 ~ ., data=dat)
Alternatively, you certainly can build up the function with paste and call lm on it.
f <- paste(names(dat)[1], "~", paste(names(dat)[-1], collapse=" + "))
# "v1 ~ v2 + v3 + v4"
lm(f, data=dat)
However, my preference in these situations is to use do.call, which evaluates expressions before passing them to the function; this makes the resulting object more suitable for calling functions like update on. Compare the call part of the output.
do.call("lm", list(as.formula(f), data=as.name("dat")))
Part Two About your second part, it looks like this is what you're going for:
lm(factor(v2) + v3 + v4 + v2*v3 + v2*v4, data=dat)
First, because v2 is a factor in the data frame, we don't need that part, and secondly, this can be simplified further by better using R's methods for using arithmetical operations to create interactions, like this.
lm(v1 ~ v2*(v3 + v4), data=dat)
I'd then simply create the function using paste; the loop with assign, even in the larger case, is probably not a good idea.
f <- paste(names(dat)[1], "~", names(dat)[2], "* (",
paste(names(dat)[-c(1:2)], collapse=" + "), ")")
# "v1 ~ v2 * ( v3 + v4 )"
It can then be called using either lm directly or with do.call.
lm(f, data=dat)
do.call("lm", list(as.formula(f), data=as.name("dat")))
About your code The problem you had with trying to use r3 etc was that you wanted the contents of the variable r3, not the value r3. To get the value, you need get, like this, and then you'd collapse the values together with paste.
vars <- sapply(paste0("r", 3:6), get)
paste(vars, collapse=" + ")
However, a better way would be to avoid assign and just build a vector of the terms you want, like this.
vars <- NULL
for (v in 3:4) {
vars <- c(vars, colnames(dat)[v], paste(colnames(dat)[2],
colnames(dat)[v], sep="*"))
}
paste(vars, collapse=" + ")
A more R-like solution would be to use lapply:
vars <- unlist(lapply(colnames(dat)[3:4],
function(x) c(x, paste(colnames(dat)[2], x, sep="*"))))
TL;DR: use paste.
create_ctree <- function(col){
myFormula <- paste(col, "~.", collapse="")
ctree(myFormula, data)
}
create_ctree("class")
In the minimal example below, I am trying to use the values of a character string vars in a regression formula. However, I am only able to pass the string of variable names ("v2+v3+v4") to the formula, not the real meaning of this string (e.g., "v2" is dat$v2).
I know there are better ways to run the regression (e.g., lm(v1 ~ v2 + v3 + v4, data=dat)). My situation is more complex, and I am trying to figure out how to use a character string in a formula. Any thoughts?
Updated below code
# minimal example
# create data frame
v1 <- rnorm(10)
v2 <- sample(c(0,1), 10, replace=TRUE)
v3 <- rnorm(10)
v4 <- rnorm(10)
dat <- cbind(v1, v2, v3, v4)
dat <- as.data.frame(dat)
# create objects of column names
c.2 <- colnames(dat)[2]
c.3 <- colnames(dat)[3]
c.4 <- colnames(dat)[4]
# shortcut to get to the type of object my full code produces
vars <- paste(c.2, c.3, c.4, sep="+")
### TRYING TO SOLVE FROM THIS POINT:
print(vars)
# [1] "v2+v3+v4"
# use vars in regression
regression <- paste0("v1", " ~ ", vars)
m1 <- lm(as.formula(regression), data=dat)
Update:
#Arun was correct about the missing "" on v1 in the first example. This fixed my example, but I was still having problems with my real code. In the code chunk below, I adapted my example to better reflect my actual code. I chose to create a simpler example at first thinking that the problem was the string vars.
Here's an example that does not work :) Uses the same data frame dat created above.
dv <- colnames(dat)[1]
r2 <- colnames(dat)[2]
# the following loop creates objects r3, r4, r5, and r6
# r5 and r6 are interaction terms
for (v in 3:4) {
r <- colnames(dat)[v]
assign(paste("r",v,sep=""),r)
r <- paste(colnames(dat)[2], colnames(dat)[v], sep="*")
assign(paste("r",v+2,sep=""),r)
}
# combine r3, r4, r5, and r6 then collapse and remove trailing +
vars2 <- sapply(3:6, function(i) {
paste0("r", i, "+")
})
vars2 <- paste(vars2, collapse = '')
vars2 <- substr(vars2, 1, nchar(vars2)-1)
# concatenate dv, r2 (as a factor), and vars into `eq`
eq <- paste0(dv, " ~ factor(",r2,") +", vars2)
Here is the issue:
print(eq)
# [1] "v1 ~ factor(v2) +r3+r4+r5+r6"
Unlike regression in the first example, eq does not bring in the column names (e.g., v3). The object names (e.g., r3) are retained. As such, the following lm() command does not work.
m2 <- lm(as.formula(eq), data=dat)
I see a couple issues going on here. First, and I don't think this is causing any trouble, but let's make your data frame in one step so you don't have v1 through v4 floating around both in the global environment as well as in the data frame. Second, let's just make v2 a factor here so that we won't have to deal with making it a factor later.
dat <- data.frame(v1 = rnorm(10),
v2 = factor(sample(c(0,1), 10, replace=TRUE)),
v3 = rnorm(10),
v4 = rnorm(10) )
Part One Now, for your first part, it looks like this is what you want:
lm(v1 ~ v2 + v3 + v4, data=dat)
Here's a simpler way to do that, though you still have to specify the response variable.
lm(v1 ~ ., data=dat)
Alternatively, you certainly can build up the function with paste and call lm on it.
f <- paste(names(dat)[1], "~", paste(names(dat)[-1], collapse=" + "))
# "v1 ~ v2 + v3 + v4"
lm(f, data=dat)
However, my preference in these situations is to use do.call, which evaluates expressions before passing them to the function; this makes the resulting object more suitable for calling functions like update on. Compare the call part of the output.
do.call("lm", list(as.formula(f), data=as.name("dat")))
Part Two About your second part, it looks like this is what you're going for:
lm(factor(v2) + v3 + v4 + v2*v3 + v2*v4, data=dat)
First, because v2 is a factor in the data frame, we don't need that part, and secondly, this can be simplified further by better using R's methods for using arithmetical operations to create interactions, like this.
lm(v1 ~ v2*(v3 + v4), data=dat)
I'd then simply create the function using paste; the loop with assign, even in the larger case, is probably not a good idea.
f <- paste(names(dat)[1], "~", names(dat)[2], "* (",
paste(names(dat)[-c(1:2)], collapse=" + "), ")")
# "v1 ~ v2 * ( v3 + v4 )"
It can then be called using either lm directly or with do.call.
lm(f, data=dat)
do.call("lm", list(as.formula(f), data=as.name("dat")))
About your code The problem you had with trying to use r3 etc was that you wanted the contents of the variable r3, not the value r3. To get the value, you need get, like this, and then you'd collapse the values together with paste.
vars <- sapply(paste0("r", 3:6), get)
paste(vars, collapse=" + ")
However, a better way would be to avoid assign and just build a vector of the terms you want, like this.
vars <- NULL
for (v in 3:4) {
vars <- c(vars, colnames(dat)[v], paste(colnames(dat)[2],
colnames(dat)[v], sep="*"))
}
paste(vars, collapse=" + ")
A more R-like solution would be to use lapply:
vars <- unlist(lapply(colnames(dat)[3:4],
function(x) c(x, paste(colnames(dat)[2], x, sep="*"))))
TL;DR: use paste.
create_ctree <- function(col){
myFormula <- paste(col, "~.", collapse="")
ctree(myFormula, data)
}
create_ctree("class")
I tried to predict the t121 columns using the "lm" command below like this,
Model<-lm(t121 ~ t1 + t2 + ..... +t120, mydata)
In my data dependent variables are more than 100, So it's difficult for predicting each columns using "lm" command that's why i want to write the program for my data like this given below i written,
for(j in 120:179){
model[[j+1]]<-lm(t[j+1] ~ add1(t1:t[j]),mydata)
}
Instead of add1 place i used add.bigq,sum commands but these three commands are not correct please tell me what is the command suitable for that place?
From what I understand, you want to write a loop that allows you to use lm with different formulas. The nice thing about lm is that it can take objects of the class formula as its first argument. Lets see how that works.
# Create a data set
df <- data.frame(col1=(1:10+rnorm(10)), col2 = 1:10, col3 = rnorm(10), col4 = rnorm(10))
If we want to run lm on col1 as the dependent and col2 as the independent variable, then we can do this:
model_a <- lm(col1 ~ col2, data = df)
form_b <- as.formula("col1 ~ col2")
model_b <- lm(form_b, data = df)
all.equal(model_a,model_b)
# [1] "Component “call”: target, current do not match when deparsed"
So the only thing that differed between the two models is that the function call was different (in model_b we used form_b, not col1 ~ col2). Other than that, the models are identical.
So now you know how to use the formula class to run lm. You can easily construct formulas with paste, by setting collapse to +
ind_vars <- paste(names(df)[-1],collapse = " + ")
form_lm <- paste(names(df)[1], "~", ind_vars)
form_lm
# [1] "col1 ~ col2 + col3 + col4"
If we want three different models, we can do a couple of things, for example:
lis <- list()
for (i in 2:length(names(df))) {
ind_vars <- paste(names(df)[2:i], collapse="+")
form_lm <- paste(names(df)[1], "~", ind_vars)
lis[[i-1]] <- lm(form_lm,data=df)
}
In the minimal example below, I am trying to use the values of a character string vars in a regression formula. However, I am only able to pass the string of variable names ("v2+v3+v4") to the formula, not the real meaning of this string (e.g., "v2" is dat$v2).
I know there are better ways to run the regression (e.g., lm(v1 ~ v2 + v3 + v4, data=dat)). My situation is more complex, and I am trying to figure out how to use a character string in a formula. Any thoughts?
Updated below code
# minimal example
# create data frame
v1 <- rnorm(10)
v2 <- sample(c(0,1), 10, replace=TRUE)
v3 <- rnorm(10)
v4 <- rnorm(10)
dat <- cbind(v1, v2, v3, v4)
dat <- as.data.frame(dat)
# create objects of column names
c.2 <- colnames(dat)[2]
c.3 <- colnames(dat)[3]
c.4 <- colnames(dat)[4]
# shortcut to get to the type of object my full code produces
vars <- paste(c.2, c.3, c.4, sep="+")
### TRYING TO SOLVE FROM THIS POINT:
print(vars)
# [1] "v2+v3+v4"
# use vars in regression
regression <- paste0("v1", " ~ ", vars)
m1 <- lm(as.formula(regression), data=dat)
Update:
#Arun was correct about the missing "" on v1 in the first example. This fixed my example, but I was still having problems with my real code. In the code chunk below, I adapted my example to better reflect my actual code. I chose to create a simpler example at first thinking that the problem was the string vars.
Here's an example that does not work :) Uses the same data frame dat created above.
dv <- colnames(dat)[1]
r2 <- colnames(dat)[2]
# the following loop creates objects r3, r4, r5, and r6
# r5 and r6 are interaction terms
for (v in 3:4) {
r <- colnames(dat)[v]
assign(paste("r",v,sep=""),r)
r <- paste(colnames(dat)[2], colnames(dat)[v], sep="*")
assign(paste("r",v+2,sep=""),r)
}
# combine r3, r4, r5, and r6 then collapse and remove trailing +
vars2 <- sapply(3:6, function(i) {
paste0("r", i, "+")
})
vars2 <- paste(vars2, collapse = '')
vars2 <- substr(vars2, 1, nchar(vars2)-1)
# concatenate dv, r2 (as a factor), and vars into `eq`
eq <- paste0(dv, " ~ factor(",r2,") +", vars2)
Here is the issue:
print(eq)
# [1] "v1 ~ factor(v2) +r3+r4+r5+r6"
Unlike regression in the first example, eq does not bring in the column names (e.g., v3). The object names (e.g., r3) are retained. As such, the following lm() command does not work.
m2 <- lm(as.formula(eq), data=dat)
I see a couple issues going on here. First, and I don't think this is causing any trouble, but let's make your data frame in one step so you don't have v1 through v4 floating around both in the global environment as well as in the data frame. Second, let's just make v2 a factor here so that we won't have to deal with making it a factor later.
dat <- data.frame(v1 = rnorm(10),
v2 = factor(sample(c(0,1), 10, replace=TRUE)),
v3 = rnorm(10),
v4 = rnorm(10) )
Part One Now, for your first part, it looks like this is what you want:
lm(v1 ~ v2 + v3 + v4, data=dat)
Here's a simpler way to do that, though you still have to specify the response variable.
lm(v1 ~ ., data=dat)
Alternatively, you certainly can build up the function with paste and call lm on it.
f <- paste(names(dat)[1], "~", paste(names(dat)[-1], collapse=" + "))
# "v1 ~ v2 + v3 + v4"
lm(f, data=dat)
However, my preference in these situations is to use do.call, which evaluates expressions before passing them to the function; this makes the resulting object more suitable for calling functions like update on. Compare the call part of the output.
do.call("lm", list(as.formula(f), data=as.name("dat")))
Part Two About your second part, it looks like this is what you're going for:
lm(factor(v2) + v3 + v4 + v2*v3 + v2*v4, data=dat)
First, because v2 is a factor in the data frame, we don't need that part, and secondly, this can be simplified further by better using R's methods for using arithmetical operations to create interactions, like this.
lm(v1 ~ v2*(v3 + v4), data=dat)
I'd then simply create the function using paste; the loop with assign, even in the larger case, is probably not a good idea.
f <- paste(names(dat)[1], "~", names(dat)[2], "* (",
paste(names(dat)[-c(1:2)], collapse=" + "), ")")
# "v1 ~ v2 * ( v3 + v4 )"
It can then be called using either lm directly or with do.call.
lm(f, data=dat)
do.call("lm", list(as.formula(f), data=as.name("dat")))
About your code The problem you had with trying to use r3 etc was that you wanted the contents of the variable r3, not the value r3. To get the value, you need get, like this, and then you'd collapse the values together with paste.
vars <- sapply(paste0("r", 3:6), get)
paste(vars, collapse=" + ")
However, a better way would be to avoid assign and just build a vector of the terms you want, like this.
vars <- NULL
for (v in 3:4) {
vars <- c(vars, colnames(dat)[v], paste(colnames(dat)[2],
colnames(dat)[v], sep="*"))
}
paste(vars, collapse=" + ")
A more R-like solution would be to use lapply:
vars <- unlist(lapply(colnames(dat)[3:4],
function(x) c(x, paste(colnames(dat)[2], x, sep="*"))))
TL;DR: use paste.
create_ctree <- function(col){
myFormula <- paste(col, "~.", collapse="")
ctree(myFormula, data)
}
create_ctree("class")