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I have a dataframe that has 1000 rows and 10 columns. First column is my y variable and rest of the columns are x variables. I would like to fit 10 different linear regression on 10 different subsets of data. For example: row1:100 first subset, row101:200 second subset etc...I would like to store output of each linear model (slope values) in a row of a new dataset. Is there an easy way to do this? I tried below:

for (i in 1:10 ) {
  model_var[i] = lm(y[(100*(i-1)+1:100*i]~.,var) 
  # var is my dataframe that has all the data
  #model_var[i] will store linear models
}

But I got an error. It seems that R doesn't allow to fit linear model to subset of a data.

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3 Answers 3

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2

A slightly more elegant solution based on @nograpes's answer:

Make up some data:

set.seed(101)
var <- data.frame(y=1:1000,matrix(runif(10000),nrow=1000))

Create a splitting variable (alternately see ggplot2::cut_number)

cutvar <- (seq(nrow(var))-1) %/% 100

Split the data and use lapply:

mList <- lapply(split(var,cutvar),lm,formula=y~.)

If you just want the coefficients then

t(sapply(mList,coef))

should extract them for you.

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1 Comment

If you used by it would do the splitting work for you: by(var,rep(1:10,each=10),lm,formula=y~.) or using your cutvar by(var,cutvar,lm,formula=y~.)
2

Another way to do is using rollapply from zoo package.

Using slightly different data to those of Ben Bolker and applying rollapply you can get it.

set.seed(1)
var <- data.frame(matrix(runif(10000),nrow=1000))
colnames(var) <- c("y", paste0("x", 1:9))

Coef <- rollapply(var, 
          width = 100, by=100, 
          FUN = function(z) {
            coef(lm(y~., data=as.data.frame(z)))
          },
          by.column = FALSE, align = "right") 

round(Coef, 3) # and here's the coefficients corresponding to the 10 regressions
      (Intercept)     x1     x2     x3     x4     x5     x6     x7     x8     x9
 [1,]       0.416 -0.253  0.093 -0.047  0.039  0.081  0.053 -0.022  0.084  0.006
 [2,]       0.656  0.144 -0.209 -0.150 -0.066  0.084  0.018 -0.114 -0.016  0.073
 [3,]       0.311 -0.134  0.006  0.047  0.036  0.020  0.082  0.172  0.211 -0.090
 [4,]       0.720 -0.110  0.094 -0.058 -0.018 -0.256 -0.058  0.074 -0.042  0.010
 [5,]       0.510  0.052  0.019 -0.193 -0.045  0.114 -0.093  0.044  0.059  0.051
 [6,]       1.044 -0.037 -0.300 -0.180  0.148  0.018 -0.187 -0.128 -0.182 -0.154
 [7,]       0.558  0.027 -0.231 -0.074  0.065  0.192 -0.022 -0.105 -0.002  0.046
 [8,]       0.496  0.156 -0.129 -0.061  0.025  0.028 -0.010  0.097 -0.031 -0.090
 [9,]       0.435  0.140  0.138 -0.170 -0.085 -0.069 -0.077 -0.056  0.190  0.105
[10,]       0.282  0.078  0.014 -0.005  0.110  0.149  0.001  0.175 -0.017 -0.033
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Comments

2

You need to subset both the y and the x variables. A simple way to do this would be to subset the var data.frame directly:

model_var<-list()
for (i in 1:10 ) 
  model_var[[i]] = lm(y~.,var[(100*(i-1)+1:100*i,])
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3 Comments

I am getting below error "object 'model_var' not found" . how to store lm models in an array object?
model_var <- list(); for (i in 1:10) { model_var[[i]] = ... } (this is a list, which is probably the best way to store lm models. What do you mean exactly by "an array object"?
In my original question I am referring to model_var[i] as an array object because i will be storing my LM models in them...that is what I meant by "an array object". I am using an array object because that enables me to use a for loop

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