Reputation: 936
I have a data frame like below:
Group1 Group2 Group3 Group4
A B A B
A C B A
B B B B
A C B D
A D C A
I want to add a new column to the data frame which will have the count of unique elements in each row. Desired output:
Group1 Group2 Group3 Group4 Count
A B A B 2
A C B A 3
B B B B 1
A C B D 4
A D C A 3
I am able to find such a count for each row using
length(unique(c(df[,c(1,2,3,4)][1,])))
I want to do the same thing for all rows in the data frame. I tried apply() with var=1 but without success. Also, it would be great if you could provide a more elegant solution to this.
Upvotes: 6
Views: 3994
Reputation: 886938
We can use apply
with MARGIN =1
to loop over the rows
df1$Count <- apply(df1, 1, function(x) length(unique(x)))
df1$Count
#[1] 2 3 1 4 3
Or using tidyverse
library(dplyr)
df1 %>%
rowwise() %>%
do(data.frame(., Count = n_distinct(unlist(.))))
# A tibble: 5 × 5
# Group1 Group2 Group3 Group4 Count
#* <chr> <chr> <chr> <chr> <int>
#1 A B A B 2
#2 A C B A 3
#3 B B B B 1
#4 A C B D 4
#5 A D C A 3
We can also use regex
to do this in a faster way. It is based on the assumption that there is only a single character per each cell
nchar(gsub("(.)(?=.*?\\1)", "", do.call(paste0, df1), perl = TRUE))
#[1] 2 3 1 4 3
More detailed explanation is given here
Upvotes: 9
Reputation: 11128
Athough there are some pretty great solutions mentioned over here, You can also use, data.table
:
DATA:
df <- data.frame(g1 = c("A","A","B","A","A"),g2 = c("B", "C", "B","C","D"),g3 = c("A","B","B","B","C"),g4 = c("B","A","B","D","A"),stringsAsFactors = F)
Code:
EDIT: After the David Arenberg's comment,added (.I) instead of 1:nrow(df). Thanks for valuable comments
library(data.table)
setDT(df)[, id := .I ]
df[, count := uniqueN(c(g1, g2, g3, g4)), by=id ]
df
Output:
> df
g1 g2 g3 g4 id count
1: A B A B 1 2
2: A C B A 2 3
3: B B B B 3 1
4: A C B D 4 4
5: A D C A 5 3
Upvotes: 2
Reputation: 12937
duplicated
in base R:
df$Count <- apply(df,1,function(x) sum(!duplicated(x)))
# Group1 Group2 Group3 Group4 Count
#1 A B A B 2
#2 A C B A 3
#3 B B B B 1
#4 A C B D 4
#5 A D C A 3
Upvotes: 3