# [R] Selecting maximums between different variables

Law, Jason Jason.Law at portlandoregon.gov
Thu Oct 17 22:21:05 CEST 2013

```See ?pmax for getting the max for each year.

do.call('pmax', oil[-1])

Or equivalently:

pmax(oil\$TX, oil\$CA, oil\$AL, oil\$ND)

apply and which.max will give you the index:

i <- apply(oil[-1], 1, which.max)

which you can use to extract the state:

names(oil[-1])[i]

Jason

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Tim Umbach
Sent: Thursday, October 17, 2013 9:49 AM
To: r-help at r-project.org
Subject: [R] Selecting maximums between different variables

Hi there,

another beginners question, I'm afraid. Basically i want to selct the maximum of values, that correspond to different variables. I have a table of oil production that looks somewhat like this:

oil <- data.frame( YEAR = c(2011, 2012),
TX = c(20000, 30000),
CA = c(40000, 25000),
AL = c(20000,
21000),

ND = c(21000,60000))

Now I want to find out, which state produced most oil in a given year. I tried this:

attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)

Which works, but it doesnt't give me the corresponding values (i.e. it just gives me the maximum output, not what state its from).
So I tried this:

oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]

None of them work, but they don't give error messages either, the output is just "NA". The problem is, in my eyes, that I'm comparing the values of different variables with each other. Because if i change the structure of the dataframe (which I can't do with the real data, at least not with out doing it by hand with a huge dataset), it looks like this and works
perfectly:

oil2 <- data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 20000, 40000, 20000, 21000),
oil_2012 = c(2012, 30000, 25000, 21000, 60000)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]

Any help is much appreciated.

Thanks, Tim Umbach

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