CODEWITHSUNDEEP
rdataframeregression
Sheldon
Sheldon

Reputation: 315

how to run lm regression for every column in R

I have data frame as:

df=data.frame(x=rnorm(100),y1=rnorm(100),y2=rnorm(100),y3=...)

I want to run a loop which regresses each column starting from the second column on the first column:

for(i in names(df[,-1])){
    model = lm(i~x, data=df)
}

But I failed. The point is that I want to do a loop of regression for each column and some column names is just a number (e.g. 404.1). I cannot find a way to run a loop for each column using the above command.

Upvotes: 4

Views: 13297

Answers (4)

Qbik
Qbik

Reputation: 6147

Generic R solution:

lapply(df[, -1], function(y) {
  lm(y ~ df$x)
})

Upvotes: 0

Wei_Du
Wei_Du

Reputation: 19

library(tidyverse)
df <- data.frame(x=rnorm(100),y1=rnorm(100),y2=rnorm(100))

head(df) you will see 

       x          y1          y2
1 -0.8955473  0.96571502 -0.16232461
2  0.5054406 -2.74246178 -0.18120499
3  0.1680144 -0.06316372 -0.53614623
4  0.2956123  0.94223922  0.38358329
5  1.1425223  0.43150919 -0.32185672
6 -0.3457060 -1.16637706 -0.06561134 

models <- df %>% 
  pivot_longer(
    cols = starts_with("y"),
    names_to = "y_name",
    values_to = "y_value"
  )

after this, head(models), you will get 

       x y_name y_value
   <dbl> <chr>    <dbl>
1 -0.896 y1      0.966 
2 -0.896 y2     -0.162 
3  0.505 y1     -2.74  
4  0.505 y2     -0.181 
5  0.168 y1     -0.0632
6  0.168 y2     -0.536

split(.$y_name) will split all data by different levels of y_name, and for each part of data, they will do the same function split(map(~lm(y_value ~ x, data = .))

After this, and head(models) you will get 

$y1

Call:
lm(formula = y_value ~ x, data = .)

Coefficients:
(Intercept)            x  
    0.14924      0.08237  


$y2

Call:
lm(formula = y_value ~ x, data = .)

Coefficients:
(Intercept)            x  
    0.11183      0.03141

If you want to tidy your results, you could do the following thing:

  tibble(
    dvsub = names(.),
    untidied = .
    ) %>%
  mutate(tidy = map(untidied, broom::tidy)) %>%
  unnest(tidy)

Then you will get View(models) like this:

  dvsub untidied     term        estimate std.error statistic p.value
  <chr> <named list> <chr>          <dbl>     <dbl>     <dbl>   <dbl>
1 y1    <lm>         (Intercept)   0.0367    0.0939     0.391   0.697
2 y1    <lm>         x             0.0399    0.0965     0.413   0.680
3 y2    <lm>         (Intercept)   0.0604    0.109      0.553   0.582
4 y2    <lm>         x            -0.0630    0.112     -0.561   0.576

So the whole code is as follows: 

models <- df %>% 
  pivot_longer(
    cols = starts_with("y"),
    names_to = "y_name",
    values_to = "y_value"
  ) %>%
  split(.$y_name) %>%
  map(~lm(y_value ~ x, data = .)) %>%
  tibble(
    dvsub = names(.),
    untidied = .
    ) %>%
  mutate(tidy = map(untidied, broom::tidy)) %>%
  unnest(tidy)

Upvotes: 1

Anastasia Vishnyakova
Anastasia Vishnyakova

Reputation: 179

Another solution with broom and tidyverse: 

library(tidyverse)
library(broom)
df <- data.frame(x=rnorm(100),y1=rnorm(100),y2=rnorm(100))

result <- df %>% 
  gather(measure, value, -x) %>%
  nest(-measure) %>%
  mutate(fit = map(data, ~ lm(value ~ x, data = .x)),
         tidied = map(fit, tidy)) %>%
  unnest(tidied)

Upvotes: 3

Gin_Salmon
Gin_Salmon

Reputation: 847

Your code looks fine except when you call i within lm, R will read i as a string, which you can't regress things against. Using get will allow you to pull the column corresponding to i.

df=data.frame(x=rnorm(100),y1=rnorm(100),y2=rnorm(100),y3=rnorm(100))

storage <- list()
for(i in names(df)[-1]){
  storage[[i]] <- lm(get(i) ~ x, df)
}

I create an empty list storage, which I'm going to fill up with each iteration of the loop. It's just a personal preference but I'd also advise against how you've written your current loop:

 for(i in names(df[,-1])){
    model = lm(i~x, data=df)
}

You will overwrite model, thus returning only the last iteration results. I suggest you change it to a list, or a matrix where you can iteratively store results.

Hope that helps

Upvotes: 5

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