R help - Jul 2009 - Testing for membership in an array of strings

As an R beginner, I feel brain dead today as I can not find the answer  
to a relatively simple question.

Given a array of string values, for example lets say "mary",
"bob",
"danny", "sue", and "jane".

I am trying to determine how to perform a logical test to determine if  
a variable is an exact match for one of the string values in the array  
when the number of strings in the array is variable and without using  
a for loop and without comparing each value.  Considering the power of  
R, I thought this would be easy, but its not obvious to me.

Now I may not yet be one with the R fu so a bit more context.

I have a data frame that contains a column with text values. What I am  
trying to do is use the subset function on the data frame to select  
only data for "sue" or "jane" (for example.)  But maybe I
have not
taken the correct approach?

So obviously I could do something like the following.

subset( data_frame, name = "sue" | name == "jane", select =
c(name,
age, birthdate))

However, my subset needs to be much more than 2 and being lazy I do  
not want to type "| name == "some text" for each one.

Is there an other way?

Neil

Try this:

c("mary", "sue") %in% c("mary", "bob",
"danny", "sue","jane")

On Thu, Jul 2, 2009 at 11:13 AM, Neil Tiffin <neilt@neiltiffin.com> wrote:
> As an R beginner, I feel brain dead today as I can not find the answer to a
> relatively simple question.
>
> Given a array of string values, for example lets say "mary",
"bob",
> "danny", "sue", and "jane".
>
> I am trying to determine how to perform a logical test to determine if a
> variable is an exact match for one of the string values in the array when
> the number of strings in the array is variable and without using a for loop
> and without comparing each value.  Considering the power of R, I thought
> this would be easy, but its not obvious to me.
>
> Now I may not yet be one with the R fu so a bit more context.
>
> I have a data frame that contains a column with text values. What I am
> trying to do is use the subset function on the data frame to select only
> data for "sue" or "jane" (for example.)  But maybe I
have not taken the
> correct approach?
>
> So obviously I could do something like the following.
>
> subset( data_frame, name = "sue" | name == "jane",
select = c(name, age,
> birthdate))
>
> However, my subset needs to be much more than 2 and being lazy I do not
> want to type "| name == "some text" for each one.
>
> Is there an other way?
>
> Neil
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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On Jul 2, 2009, at 9:13 AM, Neil Tiffin wrote:
> As an R beginner, I feel brain dead today as I can not find the  
> answer to a relatively simple question.
>
> Given a array of string values, for example lets say "mary",
"bob",
> "danny", "sue", and "jane".
>
> I am trying to determine how to perform a logical test to determine  
> if a variable is an exact match for one of the string values in the  
> array when the number of strings in the array is variable and  
> without using a for loop and without comparing each value.   
> Considering the power of R, I thought this would be easy, but its  
> not obvious to me.
>
> Now I may not yet be one with the R fu so a bit more context.
>
> I have a data frame that contains a column with text values. What I  
> am trying to do is use the subset function on the data frame to  
> select only data for "sue" or "jane" (for example.) 
But maybe I
> have not taken the correct approach?
>
> So obviously I could do something like the following.
>
> subset( data_frame, name = "sue" | name == "jane",
select = c(name,
> age, birthdate))
>
> However, my subset needs to be much more than 2 and being lazy I do  
> not want to type "| name == "some text" for each one.
>
> Is there an other way?
>
> Neil
Try this:

   subset(data_frame, name %in% c("sue", "jane"), select =
c(name,
age, birthdate))

or if your vector of names is:

   Names <-  c("mary", "bob", "danny",
"sue", "jane")

   subset(data_frame, name %in% Names, select = c(name, age, birthdate))


See ?"%in%" for more information.

HTH,

Marc Schwartz

On 2 July 2009 at 09:13, Neil Tiffin wrote:
| Given a array of string values, for example lets say "mary",
"bob",
| "danny", "sue", and "jane".
| 
| I am trying to determine how to perform a logical test to determine if  
| a variable is an exact match for one of the string values in the array  
| when the number of strings in the array is variable and without using  
| a for loop and without comparing each value.  Considering the power of  
| R, I thought this would be easy, but its not obvious to me.
| 
| Now I may not yet be one with the R fu so a bit more context.
| 
| I have a data frame that contains a column with text values. What I am  
| trying to do is use the subset function on the data frame to select  
| only data for "sue" or "jane" (for example.)  But maybe I
have not
| taken the correct approach?
| 
| So obviously I could do something like the following.
| 
| subset( data_frame, name = "sue" | name == "jane", select
= c(name,
| age, birthdate))
| 
| However, my subset needs to be much more than 2 and being lazy I do  
| not want to type "| name == "some text" for each one.
| 
| Is there an other way?

Yup, e.g. using the %in% operator:
> set.seed(42)  # fix rng so that you get the same data.frame
> neil <- data.frame(rownb=1:20, name=sample(c("mary",
"bob", "danny", "sue",
> "jane"), 20, replace=TRUE))
> head(neil)    # quick sanity check
  rownb  name
1     1  jane
2     2  jane
3     3   bob
4     4  jane
5     5   sue
6     6 danny
> neil[ neil$name %in% c("sue", "jane"), ]
   rownb name
1      1 jane
2      2 jane
4      4 jane
5      5  sue
7      7  sue
9      9  sue
10    10  sue
12    12  sue
13    13 jane
16    16 jane
17    17 jane
> 
Cheers, Dirk

-- 
Three out of two people have difficulties with fractions.