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How to detect a binary column defined with 0 and 1 in an R data frame?
If a column in an R data frame has only two values 0 and 1 then we call it a binary column but it is not necessary that a binary column needs to be defined with 0 and 1 only but it is a general convention. To detect a binary column defined with 0 and 1 in an R data frame, we can use the apply function as shown in the below examples.
Example
Consider the below data frame −
x1<-sample(0:1,20,replace=TRUE) x2<-rnorm(20,1,0.57) x3<-sample(1:5,20,replace=TRUE) x4<-rpois(20,5) x5<-rpois(20,1) df1<-data.frame(x1,x2,x3,x4,x5) df1
Output
x1 x2 x3 x4 x5 1 1 0.6203256 4 4 0 2 1 0.6133840 3 6 0 3 1 2.2600124 2 5 1 4 1 0.9189756 4 0 0 5 0 0.6335537 1 4 2 6 1 0.6631676 2 3 1 7 1 1.2910049 1 3 1 8 1 1.6811408 5 5 1 9 0 0.9246393 5 8 0 10 1 1.7744186 2 5 1 11 0 1.5409120 1 8 0 12 1 0.5852456 2 9 0 13 1 0.6707639 1 2 1 14 1 1.4045163 3 7 3 15 0 1.1463694 5 4 3 16 0 1.4744266 4 3 2 17 1 1.7846723 2 6 0 18 1 1.2694807 2 4 0 19 1 1.2146714 1 6 0 20 0 1.2323528 4 6 0
Detecting which column in df1 is binary −
Example
apply(df1,2,function(x) {all(x %in% 0:1)})
Output
x1 x2 x3 x4 x5 TRUE FALSE FALSE FALSE FALSE
Example
y1<-rnorm(20) y2<-sample(0:1,20,replace=TRUE) y3<-sample(0:1,20,replace=TRUE) y4<-sample(0:1,20,replace=TRUE) y5<-rexp(20,1.38) df2<-data.frame(y1,y2,y3,y4,y5) df2
Output
y1 y2 y3 y4 y5 1 -0.41990195 1 1 0 0.17901907 2 -0.82665045 0 0 0 0.61638486 3 0.30680950 0 1 1 0.46992402 4 1.00525636 1 0 0 0.30043897 5 0.88557771 0 1 1 0.30998419 6 0.36112442 0 1 1 0.48023858 7 -1.13961239 1 1 1 1.12290153 8 -0.27722960 1 0 0 0.61928866 9 -0.42098660 0 0 1 0.12449119 10 0.17711381 0 0 0 0.07402737 11 -0.37249602 0 1 1 2.71841887 12 1.16715519 0 0 0 0.43469615 13 1.14925253 0 1 0 1.78815398 14 0.35758175 1 1 1 2.73568363 15 0.42990962 0 1 1 2.08840339 16 -0.57891804 1 1 1 1.45000159 17 -0.84741815 1 0 0 0.33356979 18 0.97735362 0 1 1 0.19783634 19 0.92523768 1 0 1 0.14692741 20 -0.07879863 0 0 0 0.49631861
Detecting which column in df1 is binary −
Example
apply(df2,2,function(x) {all(x %in% 0:1)})
Output
y1 y2 y3 y4 y5 FALSE TRUE TRUE TRUE FALSE
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