CODEWITHSUNDEEP
rdataframe
zhlee
zhlee

Reputation: 53

check if every row in two columns has the same sign

Sorry that i didn't specify: V1 and V2 are factor values. When I

sign(as.numeric(df$V1)) == sign(as.numeric(df$V2))

Every case returns true.

Check if every row in two columns has the same sign. Return false if one is 0, but true when both are; an example is as follows:

V1  V2
-1  2.3
3.6 2
-2  -4
0   4
0   0

V1  V2   comparison
-1  2.3    false
3.6 2      true
-2  -4     true
0   4      false
0   0      true

I came up with this which always returns true:

output.df$comparison = (((as.numeric(output.df$V1) > 0) & (as.numeric(output.df$V2) > 0)) | ((as.numeric(output.df$V1) < 0) & (as.numeric(output.df$V2) < 0)))

Upvotes: 1

Views: 1047

Answers (3)

Peter Hahn
Peter Hahn

Reputation: 158

dplyr way:

df <- df %>% mutate(comparison = (sign(V1) == sign(V2))

Upvotes: 0

tmfmnk
tmfmnk

Reputation: 39858

Another base R idea could be:

apply(df, 1, function(x) sd(sign(x))) == 0

[1] FALSE  TRUE  TRUE FALSE  TRUE

Upvotes: 1

Ronak Shah
Ronak Shah

Reputation: 388982

You can compare their sign which will handle case for 0 as well. 

df$comparison <- sign(df$V1) == sign(df$V2)
df

#    V1   V2 comparison
#1 -1.0  2.3      FALSE
#2  3.6  2.0       TRUE
#3 -2.0 -4.0       TRUE
#4  0.0  4.0      FALSE
#5  0.0  0.0       TRUE

data

df <- structure(list(V1 = c(-1, 3.6, -2, 0, 0), V2 = c(2.3, 2, -4, 
4, 0)), class = "data.frame", row.names = c(NA, -5L))

Upvotes: 4

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