CODEWITHSUNDEEP
r
showkey
showkey

Reputation: 348

Cut and Table function in R

x<-sample(30:60,50,TRUE)
y<-cut(x,breaks=c(30,40,50,60))
y
[1] (30,40] (30,40] (50,60] (40,50] (40,50] (40,50] (40,50] (30,40] (30,40]
[10] (50,60] (30,40] (50,60] (30,40] (30,40] (50,60] (50,60] (50,60] (30,40]
[19] (50,60] (30,40] (40,50] (40,50] (30,40] (30,40] (30,40] (40,50] (30,40]
[28] (50,60] (40,50] (40,50] (30,40] (50,60] (40,50] (50,60] (50,60] (30,40]
[37] (50,60] (50,60] (30,40] (50,60] (30,40] (30,40] <NA> (40,50] (30,40]
[46] (40,50] (30,40] (30,40] (30,40] (30,40]
Levels: (30,40] (40,50] (50,60]
table(y)
y
(30,40] (40,50] (50,60]
23 12 14
table(y)[1]
(30,40]
23

Question1:

If right=FALSE is added ,the intervals are
(30,40] (40,50] (50,60]

If right=TRUE is added,the intervals are
[30,40) [40,50) [50,60)

how can i get the interval such as [30,40] (40,50] (50,60] or '[30,40) [40,50) [50,60]`?

Question2:

table(y)[1]  
(30,40]
23

I know there are 23 numbers in the interval (30,40], I can get them all with x[x<=40 & x>30].
Is there any better way to get the result?

Upvotes: 0

Views: 3515

Answers (2)

Greg Snow
Greg Snow

Reputation: 49640

For your question # 2 it depends on what you mean by "better".

Here is one option:

library(TeachingDemos)
x[ 30 %<% x %<=% 40 ]

Or, instead of using cut you can use findInterval:

y <- findInterval(x, c(30,40,50,60))
x[ y==1 ]

You could also look at the subset function.

If these don't match your definition of "better", then tell us more about what you want.

Upvotes: 0

Ricardo Saporta
Ricardo Saporta

Reputation: 55340

For question 2, simply use the output of your cut

x[y == "(30,40]"]

Upvotes: 1

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