I have columns of data and would like to use the WD as an independent variable and E1-E14 as dependent variable and do a regression for each and write the output to a csv file. Please helpenter image description here
This what I did, however it outputs the same results for all columns. I think the mod variable is being incorrectly set.
mod <- function(y) lm(E1 ~ WD , data = data)
lapply(data[,5:16], mod)
This is probably what you want, but it may requires some changes to get to the full extent of what you need.
df <- data.frame(WD = c(1,1,0,0,0,1,1,1,0,0),
E1 = rnorm(10,0,1),
E2 = rnorm(10,0,1))
mod <- function(x){
lm(WD ~ x, data = df)
}
sapply(df[setdiff(names(df),"WD")],mod)
Related
I am trying to create multiple linear regression models from a list of variable combinations (I also have them separately as a data-frame if that is more useful!)
The list of variables looks like this:
Vars
x1+x2+x3
x1+x2+x4
x1+x2+x5
x1+x2+x6
x1+x2+x7
The loop I'm using looks like this:
for (i in 1:length(var_list)){
lm(independent_variable ~ var_list[i],data = training_data)
i+1
}
However it is not recognizing the string of var_list[i] which gives x1+x2+x3 etc. as a model input.
Does any-one know how to fix it?
Thanks for your help.
You don't even have to use loops. Apply should work nicely.
training_data <- as.data.frame(matrix(sample(1:64), nrow = 8))
colnames(training_data) <- c("independent_variable", paste0("x", 1:7))
Vars <- as.list(c("x1+x2+x3",
"x1+x2+x4",
"x1+x2+x5",
"x1+x2+x6",
"x1+x2+x7"))
allModelsList <- lapply(paste("independent_variable ~", Vars), as.formula)
allModelsResults <- lapply(allModelsList, function(x) lm(x, data = training_data))
If you need models summaries you can add :
allModelsSummaries = lapply(allModelsResults, summary)
For example you can access the coefficient R² of the model lm(independent_variable ~ x1+x2+x3) by doing this:
allModelsSummaries[[1]]$r.squared
I hope it helps.
We can create the formula with paste
out <- vector('list', length(var_list))
for (i in seq_along(var_list)){
out[[i]] <- lm(paste('independent_variable', '~', var_list[i]),
data = training_data)
}
Or otherwise, it can be done with reformulate
lm(reformulate(var_list[i], 'independent_variable'), data = training_data)
I have a dataframe with many variables. I want to apply a linear regression to explain the last one with the others. So as I had to much to write I thought about creating a string with the independent variables e.g. Var1 + Var2 +...+ VarK. I achieved it pasting "+" to all column names except for the last one with this code:
ExVar <- toString(paste(names(datos)[1:11], "+ ", collapse = ''))
I also had to remove the last "+":
ExVar <- substr(VarEx, 1, nchar(ExVar)-2)
So I copied and pasted the ExVar string within the lm() function and the result looked like this:
m1 <- lm(calidad ~ Var1 + Var 2 +...+ Var K)
The question is: Is there any way to use "ExVar" within the lm() function as a string, not as a variable, to have a cleaner code?
For better understanding:
If I use this code:
m1 <- lm(calidad ~ ExVar)
It is interpreting ExVar as a independent variable.
The following will all produce the same results. I am providing multiple methods because there is are simpler ways of doing what you are asking (see examples 2 and 3) instead of writing the expression as a string.
First, I will generate some example data:
n <- 100
p <- 11
dat <- array(rnorm(n*p),c(n,p))
dat <- as.data.frame(dat)
colnames(dat) <- paste0("X",1:p)
If you really want to specify the model as a string, this example code will help:
ExVar <- toString(paste(names(dat[2:11]), "+ ", collapse = ''))
ExVar <- substr(ExVar, 1, nchar(ExVar)-3)
model1 <- paste("X1 ~ ",ExVar)
fit1 <- lm(eval(parse(text = model1)),data = dat)
Otherwise, note that the 'dot' notation will specify all other variables in the model as predictors.
fit2 <- lm(X1 ~ ., data = dat)
Or, you can select the predictors and outcome variables by column, if your data is structured as a matrix.
dat <- as.matrix(dat)
fit3 <- lm(dat[,1] ~ dat[,-1])
All three of these fit objects have the same estimates:
fit1
fit2
fit3
if you have a dataframe, and you want to explain the last one using all the rest then you can use the code below:
lm(calidad~.,dat)
or you can use
lm(rev(dat))#Only if the last column is your response variable
Any of the two above will give you the results needed.
To do it your way:
EXV=as.formula(paste0("calidad~",paste0(names(datos)[-12],collapse = '+')))
lm(EXV,dat)
There is no need to do it this way since the lm function itself will do this by using the first code above.
I need to reproduce this code using all of these variables.
composite <- read.csv("file.csv", header = T, stringsAsFactors = FALSE)
composite <- subset(composite, select = -Date)
model1 <- lm(indepvariable ~., data = composite, na.action = na.exclude)
composite is a data frame with 82 variables.
UPDATE:
What I have done is found a way to create an object that contains only the significantly correlated variables, to narrow the number of independent variables down.
I have a variable now: sigvars, which is the names of an object that sorted a correlation matrix and picked out only the variables with correlation coefficients >0.5 and <-0.5. Here is the code:
sortedcor <- sort(cor(composite)[,1])
regvar = NULL
k = 1
for(i in 1:length(sortedcor)){
if(sortedcor[i] > .5 | sortedcor[i] < -.5){
regvar[k] = i
k = k+1
}
}
regvar
sigvars <- names(sortedcor[regvar])
However, it is not working in my lm() function:
model1 <- lm(data.matrix(composite[1]) ~ sigvars, data = composite)
Error: Error in model.frame.default(formula = data.matrix(composite[1]) ~ sigvars, : variable lengths differ (found for 'sigvars')
Think about what sigvars is for a minute...?
After sigvars <- names(sortedcor[regvar]), sigvars is a character vector of column names. Say your data have 100 rows and 5 variables come out as significant using the method you've chosen (which doesn't sound overly defensible to be). The model formula you are using will result in composite[, 1] being a vector of length 100 (100 rows) and sigvars being a character vector of length 5.
Assuming you have the variables you want to include in the model, then you could do:
form <- reformulate(sigvars, response = names(composite)[1])
model1 <- lm(form, data = composite)
or
model1 <- lm(composite[,1] ~ ., data = composite[, sigvars])
In the latter case, do yourself a favour and write the name of the dependent variable into the formula instead of composite[,1].
Also, you don't seem to have appreciated the difference between [i] and [i,j] for data frames, hence you are doing data.matrix(composite[1]) which is taking the first component of composite, leaving it as a data frame, then converting that to a matrix via the data.matrix() function. All you really need is just the name of the dependent variable on the LHS of the formula.
The error is here:
model1 <- lm(data.matrix(composite[1]) ~ sigvars, data = composite)
The sigvars is names(data). The equation is usually of the form lm(var1 ~ var2+var3+var4), you however have it as lm(var1 ~ var2 var3 var4).
Hopefully that helps.
I want to loop through the vars in a dataframe, calling lm() on each one, and so I wrote this:
findvars <- function(x = samsungData, dv = 'activity', id = 'subject') {
# Loops through the possible predictor vars, does an lm() predicting the dv
# from each, and returns a data.frame of coefficients, one row per IV.
r <- data.frame()
# All varnames apart from the dependent var, and the case identifier
ivs <- setdiff(names(x), c(dv, id))
for (iv in ivs) {
print(paste("trying", iv))
m <- lm(dv ~ iv, data = x, na.rm = TRUE)
# Take the absolute value of the coefficient, then transpose.
c <- t(as.data.frame(sapply(m$coefficients, abs)))
c$iv <- iv # which IV produced this row?
r <- c(r, c)
}
return(r)
}
This doesn't work, I believe b/c the formula in the lm() call consists of function-local variables that hold strings naming vars in the passed-in dataframe (e.g., "my_dependant_var" and "this_iv") as opposed to pointers to the actual variable objects.
I tried wrapping that formula in eval(parse(text = )), but could not get that to work.
If I'm right about the problem, can someone explain to me how to get R to resolve the contents of those vars iv & dv into the pointers I need? Or if I'm wrong, can someone explain what else is going on?
Many thanks!
Here is some repro code:
library(datasets)
data(USJudgeRatings)
findvars(x = USJudgeRatings, dv = 'CONT', id = 'DILG')
So there's enough bad stuff happening in your function besides your trouble with the formula, that I think someone should walk you through it all. Here are some annotations, followed by a better version:
#For small examples, "growing" objects isn't a huge deal,
# but you will regret it very, very quickly. It's a bad
# habit. Learn to ditch it now. So don't inititalize
# empty lists and data frames.
r <- data.frame()
ivs <- setdiff(names(x), c(dv, id))
for (iv in ivs) {
print(paste("trying", iv))
#There is no na.rm argument to lm, only na.action
m <- lm(dv ~ iv, data = x, na.rm = TRUE)
#Best not to name variables c, its a common function, see two lines from now!
# Also, use the coef() extractor functions, not $. That way, if/when
# authors change the object structure your code won't break.
#Finally, abs is vectorized, no need for sapply
c <- t(as.data.frame(sapply(m$coefficients, abs)))
#This is probably best stored in the name
c$iv <- iv # which IV produced this row?
#Growing objects == bad! Also, are you sure you know what happens when
# you concatenate two data frames?
r <- c(r, c)
}
return(r)
}
Try something like this instead:
findvars <- function(x,dv,id){
ivs <- setdiff(names(x),c(dv,id))
#initialize result list of the appropriate length
result <- setNames(vector("list",length(ivs)),ivs)
for (i in seq_along(ivs)){
result[[i]] <- abs(coef(lm(paste(dv,ivs[i],sep = "~"),data = x,na.action = na.omit)))
}
result
}
I've done a fair amount of reading here on SO and learned that I should generally avoid manipulation of formula objects as strings, but I haven't quite found how to do this in a safe manner:
tf <- function(formula = NULL, data = NULL, groups = NULL, ...) {
# Arguments are unquoted and in the typical form for lm etc
# Do some plotting with lattice using formula & groups (works, not shown)
# Append 'groups' to 'formula':
# Change y ~ x as passed in argument 'formula' to
# y ~ x * gr where gr is the argument 'groups' with
# scoping so it will be understood by aov
new_formula <- y ~ x * gr
# Now do some anova (could do if formula were right)
model <- aov(formula = new_formula, data = data)
# And print the aov table on the plot (can do)
print(summary(model)) # this will do for testing
}
Perhaps the closest I came was to use reformulate but that only gives + on the RHS, not *. I want to use the function like this:
p <- tf(carat ~ color, groups = clarity, data = diamonds)
and have the aov results for carat ~ color * clarity. Thanks in Advance.
Solution
Here is a working version based on #Aaron's comment which demonstrates what's happening:
tf <- function(formula = NULL, data = NULL, groups = NULL, ...) {
print(deparse(substitute(groups)))
f <- paste(".~.*", deparse(substitute(groups)))
new_formula <- update.formula(formula, f)
print(new_formula)
model <- aov(formula = new_formula, data = data)
print(summary(model))
}
I think update.formula can solve your problem, but I've had trouble with update within function calls. It will work as I've coded it below, but note that I'm passing the column to group, not the variable name. You then add that column to the function dataset, then update works.
I also don't know if it's doing exactly what you want in the second equation, but take a look at the help file for update.formula and mess around with it a bit.
http://stat.ethz.ch/R-manual/R-devel/library/stats/html/update.formula.html
tf <- function(formula,groups,d){
d$groups=groups
newForm = update(formula,~.*groups)
mod = lm(newForm,data=d)
}
dat = data.frame(carat=rnorm(10,0,1),color=rnorm(10,0,1),color2=rnorm(10,0,1),clarity=rnorm(10,0,1))
m = tf(carat~color,dat$clarity,d=dat)
m2 = tf(carat~color+color2,dat$clarity,d=dat)
tf2 <- function(formula, group, d) {
f <- paste(".~.*", deparse(substitute(group)))
newForm <- update.formula(formula, f)
lm(newForm, data=d)
}
mA = tf2(carat~color,clarity,d=dat)
m2A = tf2(carat~color+color2,clarity,d=dat)
EDIT:
As #Aaron pointed out, it's deparse and substitute that solve my problem: I've added tf2 as the better option to the code example so you can see how both work.
One technique I use when I have trouble with scoping and calling functions within functions is to pass the parameters as strings and then construct the call within the function from those strings. Here's what that would look like here.
tf <- function(formula, data, groups) {
f <- paste(".~.*", groups)
m <- eval(call("aov", update.formula(as.formula(formula), f), data = as.name(data)))
summary(m)
}
tf("mpg~vs", "mtcars", "am")
See this answer to one of my previous questions for another example of this: https://stackoverflow.com/a/7668846/210673.
Also see this answer to the sister question of this one, where I suggest something similar for use with xyplot: https://stackoverflow.com/a/14858661/210673