Deleting every n-th row in a dataframe
If you want to get the each of the nth columns from a data frame or vector etc use modulo subsetting...
Select the nth columns by repeating sets here as modulo of 3 (choose nth as you desire)
> x <- c(1,2,3,4,5,6)
> d <- rbind(x,x,x)
> df <- as.data.frame(d, row.names=T)
> c <- 1:ncol(df)
> c
[1] 1 2 3 4 5 6
c%%3 ### nth cycle, here every 3
[1] 1 2 0 1 2 0
#select the every 3rd column of every 3
> df[, c%%3==0]
V3 V6
1 3 6
2 3 6
3 3 6
#every first column of every 3
> df[, c%%3==1]
V1 V4
1 1 4
2 1 4
3 1 4
#every 2nd column of every 3
> df[, c%%3==2]
V2 V5
1 2 5
2 2 5
3 2 5
#drop the 3rd columns
> df[, !(c%%3==0)]
V1 V2 V4 V5
1 1 2 4 5
2 1 2 4 5
3 1 2 4 5
etc... swap c<-nrow(df)
for subsetting rows..
You could create a function as follows
Nth.delete<-function(dataframe, n)dataframe[-(seq(n,to=nrow(dataframe),by=n)),]
Let's test it out
DF<-data.frame(A=1:15, B=rnorm(15), C=sample(LETTERS,15))
Nth.delete(DF, 3)
I wish to add the tidyverse
style approach to this problem, using the %%
operator.
library(dplyr)
df <- data.frame(V1 = seq(26), V2 = letters)
df %>% filter(row_number() %% 2 != 0) ## Delete even-rows
df %>% filter(row_number() %% 2 != 1) ## Delete odd-rows
df %>% filter(row_number() %% 3 != 1) ## Delete every 3rd row starting from 1
You can use the same idea to select every n-th row, of course. See here