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I have the following dataframe:

Index <- seq.int(1:10)
A <- c(5, 5, 3, 4, 3, 3, 2, 2, 4, 3)
B <- c(10, 11, 12, 12, 12, 11, 13, 13, 14, 13)
C <- c(7, 6, 7, 7, 6, 5, 6, 5, 5, 4)
df <- data.frame(Index, A, B, C)
> df
      Index A  B C
 [1,]     1 5 10 7
 [2,]     2 5 11 6
 [3,]     3 3 12 7
 [4,]     4 4 12 7
 [5,]     5 3 12 6
 [6,]     6 3 11 5
 [7,]     7 2 13 6
 [8,]     8 2 13 5
 [9,]     9 4 14 5
[10,]    10 3 13 4

I would like to generate the following three linear models:

lm(df$A ~ df$Index)
lm(df$B ~ df$Index)
lm(df$C ~ df$Index)

Is there a way to do this quickly and efficiently in one step (possibly using the lapply function)? My actual data frame has many more rows and columns. Thanks!

David Moore
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  • Both links give you automatic solution on this. The first link tells you how to use `sprintf` to construct formula and use "mlm" support of `lm`; the second link is specially for simple linear regression, using my function `general_paired_simpleLM`. – Zheyuan Li Oct 23 '18 at 22:39

2 Answers2

0

Try

df <- data.frame(Index, A, B, C) 
lm(cbind(A, B, C) ~ Index, data = df)
#Call:
#lm(formula = cbind(A, B, C) ~ Index, data = df)

#Coefficients:
#             A        B        C      
#(Intercept)   4.6000  10.2667   7.4000
#Index        -0.2182   0.3333  -0.2909

Note that df must be a dataframe.

markus
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  • I actually have over 50 columns I want to perform this on. Instead of using `cbind(A, B, C)`, is there a way I could perform this on `df[, 2:4]`? – David Moore Oct 23 '18 at 22:35
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    @DavidMoore You could do `lm(as.matrix(df[, 2:50]) ~ Index, data = df)`. Sorry for the late reply. – markus Oct 26 '18 at 07:07
0

You can also use a tidyverse approach like this:

library(tidyverse)
df %>% 
  gather(Variables, Value, A:C) %>% 
  split(.$Variables) %>% 
  map(~ lm(Value~Index, data = .)) %>% 
  map(summary)
Jilber Urbina
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