How to programatically compare an entire row in R?

Question

I have the following dataframe in R:
data= Time X1 X2 X3 1 1 0 0 2 1 1 1 3 0 0 1 4 1 1 1 5 0 0 0 6 0 1 1 7 1 1 1 8 0 0

Answer 1

Here's a couple of options using a merge:

merge(list(X1=0,X2=1,X3=1), dat)
#or
merge(setNames(list(0,1,1),c("X1","X2","X3")), dat)

Or even using positional indexes based on what columns you want matched up:

L <- list(0,1,1)
merge(L, dat, by.x=seq_along(L), by.y=2:4)

All of which return:

#  X1 X2 X3 Time
#1  0  1  1    6

If your matching variables are all of the same type, you could also safely do it via matrix comparison like:

dat[colSums(t(dat[c("X1","X2","X3")]) == c(0,1,1)) == 3,]

Answer 2

apply(data, 1, function(x) all(x==c(0,1,1)))

This will go down each row of the frame and return TRUE for each row where the row is equal to c(0,1,1).

Answer 3

this is your data

mydf <- structure(list(Time = 1:10, X1 = c(1L, 1L, 0L, 1L, 0L, 0L, 1L, 
0L, 1L, 0L), X2 = c(0L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L), 
    X3 = c(0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 0L)), .Names = c("Time", 
"X1", "X2", "X3"), class = "data.frame", row.names = c(NA, -10L
))

Using subset

subset(mydf, X1 == 0 & X2==1 & X3==1)
#  Time X1 X2 X3
#6    6  0  1  1

another way

mydf[mydf$X1 ==0 & mydf$X2 ==1 & mydf$X3 ==1, ] 
#   Time X1 X2 X3
#6    6  0  1  1

or like this

mydf[mydf$X1 ==0 & mydf$X2 & mydf$X3 %in% c(1,1), ]
#  Time X1 X2 X3
#6    6  0  1  1

you can also do that by

library(dplyr)
filter(mydf, X1==0 & X2==1 & X3==1)
#   Time X1 X2 X3
#1    6  0  1  1