Suppose I have a data such like this:
n=100
df<-data.frame(y=rnorm(n,2,3),
x1=rbinom(n,1,0.3),
x2=rbinom(n,10, 0.5),
x3=rnorm(n, 50, 20),
x4=rnorm(n, 3, 2))
and I have a basic model y=x1+x2:
mod0<-as.formula(y~x1+x2)
lm0<-lm(mod0, data=df)
what I want is to update the mod0 with an interaction term between x1 and each of x3 and x4 such that mod1<-y~x1+x2+x1*x3 and mod2<-y~x1+x2+x1*x4. Since I have a bunch of variables need to loop through, I am wondering what it may be a best way for that.
Rather than assigning loose variables in the global environment, it would be better to assemble the formulae in a list. This can be done with lapply(), and leveraging the useful trick that allows you to create a formula from a character string:
lapply(names(df)[-(1:3)],function(var) formula(paste0('y~x1+x2+x1*',var),env=globalenv()));
## [[1]]
## y ~ x1 + x2 + x1 * x3
##
## [[2]]
## y ~ x1 + x2 + x1 * x4
Related
I am new to R and I want to improve the following script with an *apply function (I have read about apply, but I couldn't manage to use it). I want to use lm function on multiple independent variables (which are columns in a data frame). I used
for (i in (1:3) {
assign(paste0('lm.',names(data[i])), lm(formula=formula(i),data=data))
}
Formula(i) is defined as
formula=function(x)
{
as.formula ( paste(names(data[x]),'~', paste0(names(data[-1:-3]), collapse = '+')), env=parent.frame() )
}
Thank you.
If I don't get you wrong, you are working with a dataset like this:
set.seed(0)
dat <- data.frame(y1 = rnorm(30), y2 = rnorm(30), y3 = rnorm(30),
x1 = rnorm(30), x2 = rnorm(30), x3 = rnorm(30))
x1, x2 and x3 are covariates, and y1, y2, y3 are three independent response. You are trying to fit three linear models:
y1 ~ x1 + x2 + x3
y2 ~ x1 + x2 + x3
y3 ~ x1 + x2 + x3
Currently you are using a loop through y1, y2, y3, fitting one model per time. You hope to speed the process up by replacing the for loop with lapply.
You are on the wrong track. lm() is an expensive operation. As long as your dataset is not small, the costs of for loop is negligible. Replacing for loop with lapply gives no performance gains.
Since you have the same RHS (right hand side of ~) for all three models, model matrix is the same for three models. Therefore, QR factorization for all models need only be done once. lm allows this, and you can use:
fit <- lm(cbind(y1, y2, y3) ~ x1 + x2 + x3, data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
If you check str(fit), you will see that this is not a list of three linear models; instead, it is a single linear model with a single $qr object, but with multiple LHS. So $coefficients, $residuals and $fitted.values are matrices. The resulting linear model has an additional "mlm" class besides the usual "lm" class. I created a special mlm tag collecting some questions on the theme, summarized by its tag wiki.
If you have a lot more covariates, you can avoid typing or pasting formula by using .:
fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
Caution: Do not write
y1 + y2 + y3 ~ x1 + x2 + x3
This will treat y = y1 + y2 + y3 as a single response. Use cbind().
Follow-up:
I am interested in a generalization. I have a data frame df, where first n columns are dependent variables (y1,y2,y3,....) and next m columns are independent variables (x1+x2+x3+....). For n = 3 and m = 3 it is fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)). But how to do this automatically, by using the structure of the df. I mean something like (for i in (1:n)) fit <- lm(cbind(df[something] ~ df[something], data = dat)). That "something" I have created it with paste and paste0. Thank you.
So you are programming your formula, or want to dynamically generate / construct model formulae in the loop. There are many ways to do this, and many Stack Overflow questions are about this. There are commonly two approaches:
use reformulate;
use paste / paste0 and formula / as.formula.
I prefer to reformulate for its neatness, however, it does not support multiple LHS in the formula. It also needs some special treatment if you want to transform the LHS. So In the following I would use paste solution.
For you data frame df, you may do
paste0("cbind(", paste(names(df)[1:n], collapse = ", "), ")", " ~ .")
A more nice-looking way is to use sprintf and toString to construct the LHS:
sprintf("cbind(%s) ~ .", toString(names(df)[1:n]))
Here is an example using iris dataset:
string_formula <- sprintf("cbind(%s) ~ .", toString(names(iris)[1:2]))
# "cbind(Sepal.Length, Sepal.Width) ~ ."
You can pass this string formula to lm, as lm will automatically coerce it into formula class. Or you may do the coercion yourself using formula (or as.formula):
formula(string_formula)
# cbind(Sepal.Length, Sepal.Width) ~ .
Remark:
This multiple LHS formula is also supported elsewhere in R core:
the formula method for function aggregate;
ANOVA analysis with aov.
I am new to R and I want to improve the following script with an *apply function (I have read about apply, but I couldn't manage to use it). I want to use lm function on multiple independent variables (which are columns in a data frame). I used
for (i in (1:3) {
assign(paste0('lm.',names(data[i])), lm(formula=formula(i),data=data))
}
Formula(i) is defined as
formula=function(x)
{
as.formula ( paste(names(data[x]),'~', paste0(names(data[-1:-3]), collapse = '+')), env=parent.frame() )
}
Thank you.
If I don't get you wrong, you are working with a dataset like this:
set.seed(0)
dat <- data.frame(y1 = rnorm(30), y2 = rnorm(30), y3 = rnorm(30),
x1 = rnorm(30), x2 = rnorm(30), x3 = rnorm(30))
x1, x2 and x3 are covariates, and y1, y2, y3 are three independent response. You are trying to fit three linear models:
y1 ~ x1 + x2 + x3
y2 ~ x1 + x2 + x3
y3 ~ x1 + x2 + x3
Currently you are using a loop through y1, y2, y3, fitting one model per time. You hope to speed the process up by replacing the for loop with lapply.
You are on the wrong track. lm() is an expensive operation. As long as your dataset is not small, the costs of for loop is negligible. Replacing for loop with lapply gives no performance gains.
Since you have the same RHS (right hand side of ~) for all three models, model matrix is the same for three models. Therefore, QR factorization for all models need only be done once. lm allows this, and you can use:
fit <- lm(cbind(y1, y2, y3) ~ x1 + x2 + x3, data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
If you check str(fit), you will see that this is not a list of three linear models; instead, it is a single linear model with a single $qr object, but with multiple LHS. So $coefficients, $residuals and $fitted.values are matrices. The resulting linear model has an additional "mlm" class besides the usual "lm" class. I created a special mlm tag collecting some questions on the theme, summarized by its tag wiki.
If you have a lot more covariates, you can avoid typing or pasting formula by using .:
fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
Caution: Do not write
y1 + y2 + y3 ~ x1 + x2 + x3
This will treat y = y1 + y2 + y3 as a single response. Use cbind().
Follow-up:
I am interested in a generalization. I have a data frame df, where first n columns are dependent variables (y1,y2,y3,....) and next m columns are independent variables (x1+x2+x3+....). For n = 3 and m = 3 it is fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)). But how to do this automatically, by using the structure of the df. I mean something like (for i in (1:n)) fit <- lm(cbind(df[something] ~ df[something], data = dat)). That "something" I have created it with paste and paste0. Thank you.
So you are programming your formula, or want to dynamically generate / construct model formulae in the loop. There are many ways to do this, and many Stack Overflow questions are about this. There are commonly two approaches:
use reformulate;
use paste / paste0 and formula / as.formula.
I prefer to reformulate for its neatness, however, it does not support multiple LHS in the formula. It also needs some special treatment if you want to transform the LHS. So In the following I would use paste solution.
For you data frame df, you may do
paste0("cbind(", paste(names(df)[1:n], collapse = ", "), ")", " ~ .")
A more nice-looking way is to use sprintf and toString to construct the LHS:
sprintf("cbind(%s) ~ .", toString(names(df)[1:n]))
Here is an example using iris dataset:
string_formula <- sprintf("cbind(%s) ~ .", toString(names(iris)[1:2]))
# "cbind(Sepal.Length, Sepal.Width) ~ ."
You can pass this string formula to lm, as lm will automatically coerce it into formula class. Or you may do the coercion yourself using formula (or as.formula):
formula(string_formula)
# cbind(Sepal.Length, Sepal.Width) ~ .
Remark:
This multiple LHS formula is also supported elsewhere in R core:
the formula method for function aggregate;
ANOVA analysis with aov.
I am new to R and I want to improve the following script with an *apply function (I have read about apply, but I couldn't manage to use it). I want to use lm function on multiple independent variables (which are columns in a data frame). I used
for (i in (1:3) {
assign(paste0('lm.',names(data[i])), lm(formula=formula(i),data=data))
}
Formula(i) is defined as
formula=function(x)
{
as.formula ( paste(names(data[x]),'~', paste0(names(data[-1:-3]), collapse = '+')), env=parent.frame() )
}
Thank you.
If I don't get you wrong, you are working with a dataset like this:
set.seed(0)
dat <- data.frame(y1 = rnorm(30), y2 = rnorm(30), y3 = rnorm(30),
x1 = rnorm(30), x2 = rnorm(30), x3 = rnorm(30))
x1, x2 and x3 are covariates, and y1, y2, y3 are three independent response. You are trying to fit three linear models:
y1 ~ x1 + x2 + x3
y2 ~ x1 + x2 + x3
y3 ~ x1 + x2 + x3
Currently you are using a loop through y1, y2, y3, fitting one model per time. You hope to speed the process up by replacing the for loop with lapply.
You are on the wrong track. lm() is an expensive operation. As long as your dataset is not small, the costs of for loop is negligible. Replacing for loop with lapply gives no performance gains.
Since you have the same RHS (right hand side of ~) for all three models, model matrix is the same for three models. Therefore, QR factorization for all models need only be done once. lm allows this, and you can use:
fit <- lm(cbind(y1, y2, y3) ~ x1 + x2 + x3, data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
If you check str(fit), you will see that this is not a list of three linear models; instead, it is a single linear model with a single $qr object, but with multiple LHS. So $coefficients, $residuals and $fitted.values are matrices. The resulting linear model has an additional "mlm" class besides the usual "lm" class. I created a special mlm tag collecting some questions on the theme, summarized by its tag wiki.
If you have a lot more covariates, you can avoid typing or pasting formula by using .:
fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)
#Coefficients:
# y1 y2 y3
#(Intercept) -0.081155 0.042049 0.007261
#x1 -0.037556 0.181407 -0.070109
#x2 -0.334067 0.223742 0.015100
#x3 0.057861 -0.075975 -0.099762
Caution: Do not write
y1 + y2 + y3 ~ x1 + x2 + x3
This will treat y = y1 + y2 + y3 as a single response. Use cbind().
Follow-up:
I am interested in a generalization. I have a data frame df, where first n columns are dependent variables (y1,y2,y3,....) and next m columns are independent variables (x1+x2+x3+....). For n = 3 and m = 3 it is fit <- lm(cbind(y1, y2, y3) ~ ., data = dat)). But how to do this automatically, by using the structure of the df. I mean something like (for i in (1:n)) fit <- lm(cbind(df[something] ~ df[something], data = dat)). That "something" I have created it with paste and paste0. Thank you.
So you are programming your formula, or want to dynamically generate / construct model formulae in the loop. There are many ways to do this, and many Stack Overflow questions are about this. There are commonly two approaches:
use reformulate;
use paste / paste0 and formula / as.formula.
I prefer to reformulate for its neatness, however, it does not support multiple LHS in the formula. It also needs some special treatment if you want to transform the LHS. So In the following I would use paste solution.
For you data frame df, you may do
paste0("cbind(", paste(names(df)[1:n], collapse = ", "), ")", " ~ .")
A more nice-looking way is to use sprintf and toString to construct the LHS:
sprintf("cbind(%s) ~ .", toString(names(df)[1:n]))
Here is an example using iris dataset:
string_formula <- sprintf("cbind(%s) ~ .", toString(names(iris)[1:2]))
# "cbind(Sepal.Length, Sepal.Width) ~ ."
You can pass this string formula to lm, as lm will automatically coerce it into formula class. Or you may do the coercion yourself using formula (or as.formula):
formula(string_formula)
# cbind(Sepal.Length, Sepal.Width) ~ .
Remark:
This multiple LHS formula is also supported elsewhere in R core:
the formula method for function aggregate;
ANOVA analysis with aov.
Suppose I have one dependent variable, and 4 independent variables. I suspect only 3 of the independent variables are significant, so I use the glm(y~ x1 + x2 + x3...) function. Then I get some coefficients for these variables. Now I want to run glm(y ~ x1 + x2 + x3 + x4), but I want to specify that the x1, x2, x3 coefficients remain the same. How could I accomplish this?
Thanks!
I don't think you can fit a model where some of the independent variables have fixed parameters. What you can do is create a new variable y2 that equals the predicted value of your first model with x1+x2+x3. Then, you can fit a second model y~y2+x4 to include it as an independent variable along with x4.
So basically, something like this:
m1 <- glm(y~x1+x2+x3...)
data$y2 <- predict(glm, newdata=data)
m2 <- glm(y~y2+x4...)
I'm new to using R and I am trying to create a matrix of correlations. I have three independent variables (x1,x2,x3) and one dependent varaible (y).
I've been trying to use cor to make a matrix of the correlations, but so far I have bene unable to find a formula for doing this.
x1=rnorm(20)
x2=rnorm(20)
x3=rnorm(20)
y=rnorm(20)
data=cbind(y,x1,x2,x3)
cor(data)
If I have correctly understood, you have a matrix of 3 columns (say x1 to x3) and many rows (as y values). You may act as follows:
foo = matrix(runif(30), ncol=3) # creating a matrix of 3 columns
cor(foo)
If you have already your values in 3 vectors x1 to x3, you can make foo like this: foo=data.frame(x1,x2,x3)
Correct me if I'm wrong, but assuming this is related to a regression problem, this might be what you're looking for:
#Set the number of data points and build 3 independent variables
set.seed(0)
numdatpoi <- 7
x1 <- runif(numdatpoi)
x2 <- runif(numdatpoi)
x3 <- runif(numdatpoi)
#Build the dependent variable with some added noise
noisig <- 10
yact <- 2 + (3 * x1) + (5 * x2) + (10 * x3)
y <- yact + rnorm(n=numdatpoi, mean=0, sd=noisig)
#Fit a linear model
rmod <- lm(y ~ x1 + x2 + x3)
#Build the variance-covariance matrix. This matrix is typically what is wanted.
(vcv <- vcov(rmod))
#If needed, convert the variance-covariance matrix to a correlation matrix
(cm <- cov2cor(vcv))
From the above, here's the variance-covariance matrix:
(Intercept) x1 x2 x3
(Intercept) 466.5773 14.3368 -251.1715 -506.1587
x1 14.3368 452.9569 -170.5603 -307.7007
x2 -251.1715 -170.5603 387.2546 255.9756
x3 -506.1587 -307.7007 255.9756 873.6784
And, here's the associated correlation matrix:
(Intercept) x1 x2 x3
(Intercept) 1.00000000 0.03118617 -0.5908950 -0.7927735
x1 0.03118617 1.00000000 -0.4072406 -0.4891299
x2 -0.59089496 -0.40724064 1.0000000 0.4400728
x3 -0.79277352 -0.48912986 0.4400728 1.0000000