Logical Operation in R

Question

I have three columns named i1,i2 and i3. Each of these columns is a vector of equal length. Each of them contain 1 and -1. I want to have an output column where the value should be

1 if all the entires in i1,i2 and i3 is 1
-1 if all the entries in i1,i2 and i3 is -1
0 for any other values Please suggest the way forward.

three columns of a data frame

Thanks for the response jogo. I am not sure it would work for three inputs simultaneously. I tried your suggestion in R. jj= (i1 | i2 | i3) & ( !(i1 & i2 & i3)) # Didn't work — Abhishek Kulkarni, Apr 11 '16 at 06:37
Yeah, sorry about that. Edited the question. Thanks a million! — Abhishek Kulkarni, Apr 11 '16 at 07:49

score 3 · Answer 1 · answered Apr 11 '16 at 06:40

3

Just try (x is your matrix):

(rowSums(x==1)==ncol(x)) - (rowSums(x==-1)==ncol(x))

answered Apr 11 '16 at 06:40

nicola

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I am yet to understand how this works but this is the solution I was looking for. – Abhishek Kulkarni Apr 11 '16 at 07:50

score 1 · Accepted Answer · answered Apr 11 '16 at 06:37

You can try the following

set.seed(1)
(x <- matrix(sample(c(-1,1), 60, TRUE), ncol = 3))

#       [,1] [,2] [,3]
#  [1,]   -1    1    1
#  [2,]   -1   -1    1
#  [3,]    1    1    1
#  [4,]    1   -1    1
#  [5,]   -1   -1    1
#  [6,]    1   -1    1
#  [7,]    1   -1   -1
#  [8,]    1   -1   -1
#  [9,]    1    1    1
# [10,]   -1   -1    1
# [11,]   -1   -1   -1
# [12,]   -1    1    1
# [13,]    1   -1   -1
# [14,]   -1   -1   -1
# [15,]    1    1   -1
# [16,]   -1    1   -1
# [17,]    1    1   -1
# [18,]    1   -1    1
# [19,]   -1    1    1
# [20,]    1   -1   -1

When the row sums are 3 or -3, multiply by the sign

rs <- rowSums(x)
cbind(x, ind = (abs(rs) == 3) * sign(rs))

#                ind
#  [1,] -1  1  1   0
#  [2,] -1 -1  1   0
#  [3,]  1  1  1   1
#  [4,]  1 -1  1   0
#  [5,] -1 -1  1   0
#  [6,]  1 -1  1   0
#  [7,]  1 -1 -1   0
#  [8,]  1 -1 -1   0
#  [9,]  1  1  1   1
# [10,] -1 -1  1   0
# [11,] -1 -1 -1  -1
# [12,] -1  1  1   0
# [13,]  1 -1 -1   0
# [14,] -1 -1 -1  -1
# [15,]  1  1 -1   0
# [16,] -1  1 -1   0
# [17,]  1  1 -1   0
# [18,]  1 -1  1   0
# [19,] -1  1  1   0
# [20,]  1 -1 -1   0

score 0 · Answer 3 · answered Apr 11 '16 at 06:40

0

You can do

f <- function(x) ifelse(all(x==1), 1, ifelse(all(x==-1), -1, 0))
apply(df, 1, f) # or:
df$result <- apply(df, 1, f)

answered Apr 11 '16 at 06:40

jogo

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Logical Operation in R

3 Answers3