[R] Group several variables and apply a function to the group

[Pete Brecknock]

Aurélien PHILIPPOT wrote
> 
> Dear R-experts,
> I am struggling with the following problem, and I am looking for advice
> from more experienced R-users: I have a data frame with 2 identifying
> variables (comn and mi), and an output variable (x). comn is a variable
> for
> a company and mi is a variable for a month.
> 
> comn<-c("abc", "abc", "abc", "abc", "abc", "abc", "xyz", "xyz","xyz",
> "xyz")
> mi<- c("1", "1","1", "2", "2", "2", "1", "1", "3", "3")
> x<- c("-0.0031", "0.0009", "-0.007", "0.1929","0.0087", "0.099","-0.089",
> "0.005", "-0.0078", "0.67" )
> df<- data.frame(comn=comn, mi=mi, x=x)
> 
> 
> For each company, within a particular month, I would like to compute the
> standard deviation of x: for example, for abc, I would like to compute the
> sd of x for month1 (when mi=1) and for month2 (when mi=2).
> 
> In other languages (Stata for instance), I would create a grouping
> variable
> (group comnn and mi) and then, apply the sd function for each group.
> 
> However, I don't find an elegant way to do the same in R:
> 
> I was thinking about the following: I could subset my data frame by mi and
> create one file per month, and then make a loop and in each file, use a
> "by" operator for each comn. I am sure it would work, but I feel that it
> would be like killing an ant with a tank.
> 
> I was wondering if anyone knew a more straightforward way to implement
> that
> kind of operation?
> 
> Thanks a lot,
> 
> Best,
> Aurelien
> 
> 	[[alternative HTML version deleted]]
> 
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> 

One way would be to use the aggregate function. 

# Your Data ... 
# Note:  I have removed the quotes off the output variable x
comn<-c("abc", "abc", "abc", "abc", "abc", "abc", "xyz", "xyz","xyz", "xyz")
mi<- c("1", "1","1", "2", "2", "2", "1", "1", "3", "3")
x<- c(-0.0031, 0.0009, -0.007, 0.1929,0.0087, 0.099,-0.089, 0.005, -0.0078,
0.67)
df<- data.frame(comn=comn, mi=mi, x=x)

# Aggregate Function
aggregate(df$x, by=list(df$comn,df$mi),FUN=sd)

HTH

Pete

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