How do I find equal columns in R? - r

Question

Given the following:
a <- c(1,2,3)
b <- c(1,2,3)
c <- c(4,5,6)
A <- cbind(a,b,c)
I want to find which columns in A are equal to for example my vector a.
My first attempt would be:
> which(a==A)
[1] 1 2 3 4 5 6
Which did not do that. (Too be honest I don't even understand what that did)
Second attempt was:
a==A
a b c
[1,] TRUE TRUE FALSE
[2,] TRUE TRUE FALSE
[3,] TRUE TRUE FALSE
which definitely is a step in the right direction but it seems extended into a matrix. What I would have preferred is something like just one of the rows. How do I compare a vector to columns and how do I find columns in a matrix that are equal to a vector?

Answer 1

Use identical. That is R's "scalar" comparison operator; it returns a single logical value, not a vector.
apply(A, 2, identical, a)
# a b c
# TRUE TRUE FALSE
If A is a data frame in your real case, you're better off using sapply or vapply because apply coerces it's input to a matrix.
d <- c("a", "b", "c")
B <- data.frame(a, b, c, d)
apply(B, 2, identical, a) # incorrect!
# a b c d
# FALSE FALSE FALSE FALSE
sapply(B, identical, a) # correct
# a b c d
# TRUE TRUE FALSE FALSE
But note that data.frame coerces character inputs to factors unless you ask otherwise:
sapply(B, identical, d) # incorrect
# a b c d
# FALSE FALSE FALSE FALSE
C <- data.frame(a, b, c, d, stringsAsFactors = FALSE)
sapply(C, identical, d) # correct
# a b c d
# FALSE FALSE FALSE TRUE
Identical is also considerably faster than using all + ==:
library(microbenchmark)
a <- 1:1000
b <- c(1:999, 1001)
microbenchmark(
all(a == b),
identical(a, b))
# Unit: microseconds
# expr min lq median uq max
# 1 all(a == b) 8.053 8.149 8.2195 8.3295 17.355
# 2 identical(a, b) 1.082 1.182 1.2675 1.3435 3.635

Answer 2

If you add an extra row:
> A
a b c
[1,] 1 1 4 4
[2,] 2 2 5 2
[3,] 3 3 6 1
Then you can see that this function is correct:
> hasCol=function(A,a){colSums(a==A)==nrow(A)}
> A[,hasCol(A,a)]
a b
[1,] 1 1
[2,] 2 2
[3,] 3 3
But the earlier version accepted doesn't:
> oopsCol=function(A,a){colSums(a==A)>0}
> A[,oopsCol(A,a)]
a b
[1,] 1 1 4
[2,] 2 2 2
[3,] 3 3 1
It returns the 4,2,1 column because the 2 matches the 2 in 1,2,3.

Answer 3

Surely there's a better solution but the following works:
> a <- c(1,2,3)
> b <- c(1,2,3)
> c <- c(4,5,6)
> A <- cbind(a,b,c)
> sapply(1:ncol(A), function(i) all(a==A[,i]))
[1] TRUE TRUE FALSE
And to get the indices:
> which(sapply(1:ncol(A), function(i) all(a==A[,i])))
[1] 1 2

Answer 4

colSums(a==A)==nrow(A)
Recycling of == makes a effectively a matrix which has all columns equal to a and dimensions equal to those of A. colSums sums each column; while TRUE behaves like 1 and FALSE like 0, columns equal to a will have sum equal to the number of rows. We use this observation to finally reduce the answer to a logical vector.
EDIT:
library(microbenchmark)
A<-rep(1:14,1000);c(7,2000)->dim(A)
1:7->a
microbenchmark(
apply(A,2,function(b) identical(a,b)),
apply(A,2,function(b) all(a==b)),
colSums(A==a)==nrow(A))
# Unit: microseconds
# expr min lq median
# 1 apply(A, 2, function(b) all(a == b)) 9446.210 9825.6465 10278.335
# 2 apply(A, 2, function(b) identical(a, b)) 9324.203 9915.7935 10314.833
# 3 colSums(A == a) == nrow(A) 120.252 121.5885 140.185
# uq max
# 1 10648.7820 30588.765
# 2 10868.5970 13905.095
# 3 141.7035 162.858