问题
Stata's inlist allows us to refer to the real or string values of a variable. I was wondering whether R
has such a function.
Examples:
I want to choose eight states from the variable state
(you can think this as column state
in any dataframe where state
takes 50 string values (states of United States)).
inlist(state,"NC","AZ","TX","NY","MA","CA","NJ")
I want to choose nine values of age from the variable age
(you can think this as column age
in any dataframe where age
takes numerical values from 0 to 90).
inlist(age,16, 24, 45, 54, 67,74, 78, 79, 85)
Question:
age<-c(0:10) # for this problem age takes values from 0 to 10 only
data<-as.data.frame(age) # age is a variable of data frame data
data$m<-ifelse(c(1,7,9)%in%data$age,0,1) # generate a variable m which takes value 0 if age is 1, 7, and 8 and 1, otherwise
Expected output:
age m
1 0 1
2 1 0
3 2 1
4 3 1
5 4 1
6 5 1
7 6 1
8 7 0
9 8 1
10 9 0
11 10 1
回答1:
I think you want %in%
:
statevec <- c("NC","AZ","TX","NY","MA","CA","NJ")
state <- c("AZ","VT")
state %in% statevec ## TRUE FALSE
agevec <- c(16, 24, 45, 54, 67,74, 78, 79, 85)
age <- c(34,45)
age %in% agevec ## FALSE TRUE
edit: working on updated question.
Copying from @NickCox's link:
inlist(z,a,b,...)
Domain: all reals or all strings
Range: 0 or 1
Description: returns 1 if z is a member of the remaining arguments;
otherwise, returns 0. All arguments must be reals
or all must be strings. The number of arguments is
between 2 and 255 for reals and between 2 and 10 for
strings.
However, I'm not quite sure how this matches up with the original question. I don't know Stata well enough to know if z
can be a vector or not: it doesn't sound that way, in which case the original question (considering z=state
as a vector) doesn't make sense. If we consider that it can be a vector then the answer would be as.numeric(state %in% statevec)
-- I think.
Edit: Update by Ananda
Using your updated data, here's one approach, again using %in%
:
data <- data.frame(age=0:10)
within(data, {
m <- as.numeric(!age %in% c(1, 7, 9))
})
age m
1 0 1
2 1 0
3 2 1
4 3 1
5 4 1
6 5 1
7 6 1
8 7 0
9 8 1
10 9 0
11 10 1
This matches your expected output, by using !
(NOT) to invert the sense of %in%
. It seems to be a little backwards from the way I would think about it (normally, 0=FALSE
="is not in the list" and 1=TRUE
="is in the list") and my reading of Stata's definition, but if it's what you want ...
Or one can use ifelse
for more potential flexibility (i.e. values other than 0/1): substitute within(data, { m <- ifelse(age %in% c(1, 7, 9),0,1)})
in the code above.
来源:https://stackoverflow.com/questions/14295236/what-is-the-equivalent-of-stata-function-inlist-in-r