CODEWITHSUNDEEP
rdataframe
Vinterwoo
Vinterwoo

Reputation: 3941

Calculate row means on subset of columns

Given a sample data frame:

C1<-c(3,2,4,4,5)
C2<-c(3,7,3,4,5)
C3<-c(5,4,3,6,3)
DF<-data.frame(ID=c("A","B","C","D","E"),C1=C1,C2=C2,C3=C3)

DF
    ID C1 C2 C3
  1  A  3  3  5
  2  B  2  7  4
  3  C  4  3  3
  4  D  4  4  6
  5  E  5  5  3

What is the best way to create a second data frame that would contain the ID column and the mean of each row? Something like this:

ID  Mean
A    3.66
B    4.33
C    3.33
D    4.66
E    4.33

Something similar to: 

RM<-rowMeans(DF[,2:4])

I'd like to keep the means aligned with their ID's.

Upvotes: 77

Views: 207914

Answers (8)

AnilGoyal
AnilGoyal

Reputation: 26238

rowwise() in dplyr can be used in such situations

library(dplyr)
#> 

DF %>% 
  rowwise() %>% 
  summarise(ID,
            Mean = mean(c_across(C1:C3))) 
#> # A tibble: 5 × 2
#>   ID     Mean
#>   <chr> <dbl>
#> 1 A      3.67
#> 2 B      4.33
#> 3 C      3.33
#> 4 D      4.67
#> 5 E      4.33

Still, if you want to use rowMeans that can also be used in piped syntax

DF %>% 
  mutate(Mean = rowMeans(.[-1]))

#>   ID C1 C2 C3     Mean
#> 1  A  3  3  5 3.666667
#> 2  B  2  7  4 4.333333
#> 3  C  4  3  3 3.333333
#> 4  D  4  4  6 4.666667
#> 5  E  5  5  3 4.333333

. is actually a special argument which passes the result of previous piped syntax to next pipe operation.

Upvotes: 1

Baraliuh
Baraliuh

Reputation: 2141

Awnser adapted from: here for N different groups of columns

library(dplyr, warn.conflicts = FALSE)
library(purrr)
row_means <- DF %>% 
        dplyr::select(where(is.numeric)) %>% 
        split.default(stringr::str_remove(names(df), '[0-9]')) %>% 
        map(rowMeans) %>% 
        setNames(paste0("mean_", names(.)))
    DF %>% 
        mutate(
            !!!row_means
        )

Upvotes: 1

zx8754
zx8754

Reputation: 56229

Using dplyr:

library(dplyr)

DF %>%
  transmute(ID,
            Mean = rowMeans(across(C1:C3)))

Or

DF %>%
  transmute(ID,
            Mean = rowMeans(select(., C1:C3)))

#   ID     Mean
# 1  A 3.666667
# 2  B 4.333333
# 3  C 3.333333
# 4  D 4.666667
# 5  E 4.333333

Upvotes: 16

MadmanLee
MadmanLee

Reputation: 558

rowMeans is nice, but if you are still trying to wrap your head around the apply family of functions, this is a good opprotunity to begin understanding it.

DF['Mean'] <- apply(DF[,2:4], 1, mean)

Notice I'm doing a slightly different assignment than the first example. This approach makes it easier to incorporate it into for loops.

Upvotes: 3

user12059497
user12059497

Reputation:

(Another solution using pivot_longer & pivot_wider from latest Tidyr update)

You should try using pivot_longer to get your data from wide to long form Read latest tidyR update on pivot_longer & pivot_wider (https://tidyr.tidyverse.org/articles/pivot.html)

library(tidyverse)
C1<-c(3,2,4,4,5)
C2<-c(3,7,3,4,5)
C3<-c(5,4,3,6,3)
DF<-data.frame(ID=c("A","B","C","D","E"),C1=C1,C2=C2,C3=C3)

Output here 

  ID     mean
  <fct> <dbl>
1 A      3.67
2 B      4.33
3 C      3.33
4 D      4.67
5 E      4.33

Upvotes: 0

Jilber Urbina
Jilber Urbina

Reputation: 61214

Calculate row means on a subset of columns:

Create a new data.frame which specifies the first column from DF as an column called ID and calculates the mean of all the other fields on that row, and puts that into column entitled 'Means': 

data.frame(ID=DF[,1], Means=rowMeans(DF[,-1]))
  ID    Means
1  A 3.666667
2  B 4.333333
3  C 3.333333
4  D 4.666667
5  E 4.333333

Upvotes: 66

Nadegelia
Nadegelia

Reputation: 251

You can create a new row with $ in your data frame corresponding to the Means

DF$Mean <- rowMeans(DF[,2:4])

Upvotes: 25

BenBarnes
BenBarnes

Reputation: 19454

Starting with your data frame DF, you could use the data.table package:

library(data.table)

## EDIT: As suggested by @MichaelChirico, setDT converts a
## data.frame to a data.table by reference and is preferred
## if you don't mind losing the data.frame
setDT(DF)

# EDIT: To get the column name 'Mean':

DF[, .(Mean = rowMeans(.SD)), by = ID]

#      ID     Mean
# [1,]  A 3.666667
# [2,]  B 4.333333
# [3,]  C 3.333333
# [4,]  D 4.666667
# [5,]  E 4.333333

Upvotes: 33

Related Questions