Help to count in apply(...)

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Dear Gianni,

I think you want this:

tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

E.g.:

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- 0.5
tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

Robert

On Fri, Mar 13, 2009 at 9:58 AM, gianni lavaredo
dear User

I'm quite new to R and trying to learn the basic and sorry for the email. I
don't wish abuse of mail-list. I had read manual and Help online about
tapply (...). I have a data frame and I need "count" the Fe value with sum
for every ID sample area

fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)

I need to count the Fe area with Fe$K <= ?fer.conc.max

# give me the summatory and not the count
fer.conc.max <- tapply(areaFe$K, areaFe$ID, sum, na.rm=TRUE)

I tried sever combination that I found in mail-list archive but alway I have
a error message

thanks Gianni

? ? ? ?[[alternative HTML version deleted]]

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I see, I did not understand that. This is a quick fix, please check if
it is OK. Using which.max might be more elegant.

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)
fermax <- rep (fer.conc.max, each=10)
tapply(areaFe$K > 0.9 * fermax, areaFe$ID, sum, na.rm=TRUE)

Robert

On Fri, Mar 13, 2009 at 5:02 PM, gianni lavaredo
Thanks Robert

the problem (on your example) this I need

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
#find the max value in each ID area
fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)
fer.conc.max
??????? 1???????? 2???????? 3???????? 4???????? 5???????? 6
7???????? 8
0.9339687 0.9413441 0.9605550 0.9968569 0.9647085 0.9563837 0.9731356
0.9278238
??????? 9??????? 10
0.9496833 0.9111127
#count the?value?below or equal the fe.comc.max
tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)
Error tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm = TRUE) :
? dims [product 10] do not match the length of object [100]

2009/3/13 Robert Hijmans <r.hijmans at gmail.com>
Dear Gianni,

I think you want this:

tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

E.g.:

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- 0.5
tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

Robert

On Fri, Mar 13, 2009 at 9:58 AM, gianni lavaredo
<gianni.lavaredo at gmail.com> wrote:
dear User

I'm quite new to R and trying to learn the basic and sorry for the
email. I
don't wish abuse of mail-list. I had read manual and Help online about
tapply (...). I have a data frame and I need "count" the Fe value with
sum
for every ID sample area

fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)

I need to count the Fe area with Fe$K <= ?fer.conc.max

# give me the summatory and not the count
fer.conc.max <- tapply(areaFe$K, areaFe$ID, sum, na.rm=TRUE)

I tried sever combination that I found in mail-list archive but alway I
have
a error message

thanks Gianni

? ? ? ?[[alternative HTML version deleted]]

_______________________________________________
R-sig-Geo mailing list
R-sig-Geo at stat.math.ethz.ch
https://stat.ethz.ch/mailman/listinfo/r-sig-geo

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I think the solution is using 'merge'

soil.fe <- data.frame(
  ID =c(1,1,1,1,2,2,3,3,4,5,5,5,5,5,9,9,9,9,9,9,10,10),
  K =c(6.25,12.50,25.00,50.00,80.00,12.00,6.25,12.50,25.00,50.00,76.00,67.00,67.00,56.00,98.00,32.00,33.00,43.00,54.00,76.00,85.00,67.00))

 fer.conc.max <- tapply(soil.fe$K, soil.fe$ID, max, na.rm=TRUE)
 fm <- matrix(as.numeric(cbind(names(fer.conc.max), fer.conc.max)), ncol=2)
 colnames(fm) <- c('ID', 'maxFe')
 merge(soil.fe, fm)

 R

On Fri, Mar 13, 2009 at 8:29 PM, gianni lavaredo
Thanks Robert,

I have a data frame with difefferent length of ID and I change the codes in
this mode

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)
# I am sure that all my value are under 100
tapply(areaFe$K?<= 100, areaFe$ID, sum, na.rm=TRUE)

1? 2? 3? 4? 5? 6? 7? 8? 9 10
10 10 10 10 10 10 10 10 10 10

It's not elegant mode, and I ask please if there an elegant mode to write
below (use "fer.conc.max")

soil.fe <- data.frame(
??ID =c(1,1,1,1,2,2,3,3,4,5,5,5,5,5,9,9,9,9,9,9,10,10),
??K
=c(6.25,12.50,25.00,50.00,80.00,12.00,6.25,12.50,25.00,50.00,76.00,67.00,67.00,56.00,98.00,32.00,33.00,43.00,54.00,76.00,85.00,67.00))

fer.conc.max <- tapply(soil.fe$K, soil.fe$ID, max, na.rm=TRUE)
#ee max value
fer.conc.max

1??? 2??? 3??? 4??? 5??? 9?? 10
50.0 80.0 12.5 25.0 76.0 98.0 85.0

# I am sure that all my value are under 100
tapply(soil.fe$K <= 100, soil.fe$ID, sum, na.rm=TRUE)

1? 2? 3? 4? 5? 9 10
?4? 2? 2? 1? 5? 6? 2

Thanks

2009/3/13 Robert Hijmans <r.hijmans at gmail.com>
I see, I did not understand that. This is a quick fix, please check if
it is OK. Using which.max might be more elegant.

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)
fermax <- rep (fer.conc.max, each=10)
tapply(areaFe$K > 0.9 * fermax, areaFe$ID, sum, na.rm=TRUE)

Robert

On Fri, Mar 13, 2009 at 5:02 PM, gianni lavaredo
<gianni.lavaredo at gmail.com> wrote:
Thanks Robert

the problem (on your example) this I need

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
#find the max value in each ID area
fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)
fer.conc.max
??????? 1???????? 2???????? 3???????? 4???????? 5???????? 6
7???????? 8
0.9339687 0.9413441 0.9605550 0.9968569 0.9647085 0.9563837 0.9731356
0.9278238
??????? 9??????? 10
0.9496833 0.9111127
#count the?value?below or equal the fe.comc.max
tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)
Error tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm = TRUE) :
? dims [product 10] do not match the length of object [100]

2009/3/13 Robert Hijmans <r.hijmans at gmail.com>
Dear Gianni,

I think you want this:

tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

E.g.:

ID <- rep(1:10, each=10)
K <- runif(100)
areaFe <- as.data.frame(cbind(ID, K))
fer.conc.max <- 0.5
tapply(areaFe$K <= fer.conc.max, areaFe$ID, sum, na.rm=TRUE)

Robert

On Fri, Mar 13, 2009 at 9:58 AM, gianni lavaredo
<gianni.lavaredo at gmail.com> wrote:
dear User

I'm quite new to R and trying to learn the basic and sorry for the
email. I
don't wish abuse of mail-list. I had read manual and Help online
about
tapply (...). I have a data frame and I need "count" the Fe value
with
sum
for every ID sample area

fer.conc.max <- tapply(areaFe$K, areaFe$ID, max, na.rm=TRUE)

I need to count the Fe area with Fe$K <= ?fer.conc.max

# give me the summatory and not the count
fer.conc.max <- tapply(areaFe$K, areaFe$ID, sum, na.rm=TRUE)

I tried sever combination that I found in mail-list archive but alway
I
have
a error message

thanks Gianni

? ? ? ?[[alternative HTML version deleted]]

_______________________________________________
R-sig-Geo mailing list
R-sig-Geo at stat.math.ethz.ch
https://stat.ethz.ch/mailman/listinfo/r-sig-geo

I think the solution is using 'merge'

soil.fe <- data.frame(
?ID =c(1,1,1,1,2,2,3,3,4,5,5,5,5,5,9,9,9,9,9,9,10,10),
?K =c(6.25,12.50,25.00,50.00,80.00,12.00,6.25,12.50,25.00,50.00,76.00,67.00,67.00,56.00,98.00,32.00,33.00,43.00,54.00,76.00,85.00,67.00))

?fer.conc.max <- tapply(soil.fe$K, soil.fe$ID, max, na.rm=TRUE)
?fm <- matrix(as.numeric(cbind(names(fer.conc.max), fer.conc.max)), ncol=2)
?colnames(fm) <- c('ID', 'maxFe')
?merge(soil.fe, fm)
Another alternative is to use ddply from the plyr package:

ddply(soil.fe, .(ID), transform, maxFe = max(K, na.rm = T))

You can find out more at http://had.co.nz/plyr.

Hadley
http://had.co.nz/