Hi:
On Fri, Jul 22, 2011 at 2:28 AM, Marko <markho1984 at googlemail.com>
wrote:> Hello,
> i have following problem and I hope you can help me a little bit
>
>
> My dataframe looks like:
>
> df
> a ? ? ? m ?d ? typ ? ?value
> 1950 ?1 ?1 ? ?5 ? ? ?-4.1
> 1950 ?1 ?2 ? ?9 ? ? ? 2.7
> 1950 ?1 ?3 ? ?3 ? ? ?-1.3
> 1950 ?1 ?4 ? ?5 ? ? ?-1.9
> 1950 ?1 ?5 ? ?2 ? ? ? 0.2
> 1950 ?1 ?6 ? ?8 ? ? ? 0.5
> 1951 ?1 ?1 ? ?4 ? ? ? 1.3
> ....
>
> It consists by daily observations from 1950- 2009.
>
> Now, I get with....
>
> for (i in df$V5)
> neu <- tapply(df[,5],list(df$V4,df$V1),mean)
>
> the yearly means of the types (1-18) for every year.
Other ways to do this, which output data frames rather than matrices,
would include
# the formula interface below works with R-2.11.0 +
aggregate(value ~ a + typ, data = df, FUN = mean)
library(plyr)
ddply(df, .(a, typ), summarise, m = mean(value))
Both would give you the 'long form' of the data. One could use the
cast() function in the reshape[2] package or the reshape() function
from the base package to convert it to 'wide'
form.>
> The new df looks like:
>
> new
> typ ? ? ? 1950 ? ? ? 1951 ? ? ? 1952 ? ? ? ?1953 ? ? ? ?1954 ? ? ? ?1955
> 1956 ? ? ? ? ? ?... ? ? ? 2009
> 1 ? 0.40588235 -0.1714286 -1.8111111 ? 5.4000000 ?-0.9555556 ?2.65833333
> 2 ?-3.17777778 ?1.4130435 -0.9166667 ?-4.9000000 ? 0.2900000 ?3.54285714
> 3 ? 0.08888889 -2.0000000 -2.9666667 ? 2.2000000 ?-1.8600000 -0.50000000
> ...
> 18
>
>
> Now, i would like to generate a timeseries with the means, according to the
> different types.
> For example: For all days in the year 1950 with the typ 1, I would like to
> write the mean value for Typ 1 in year 1950. In year 1951 I would like to
> write the mean value for typ 1 in 1951 etc. (for all 18 types)
>
> The output should look like as following:
>
> erg
> a ? ? ? m ?d ? typ ? ?value ? mean_typ_year
> 1950 ?1 ?1 ? ?1 ? ? ?-4.1 ? ? ?0,4
> 1950 ?1 ?2 ? ?2 ? ? ? 2.7 ? ? ?Mean (Typ2 1950)
> 1950 ?1 ?3 ? ?1 ? ? ?-1.3 ? ? ?0,4
> 1950 ?1 ?4 ? ?5 ? ? ?-1.9 ? ? ?Mean (Typ5 1950)
> 1950 ?1 ?5 ? ?2 ? ? ? 0.2 ? ? ?...
> 1950 ?1 ?6 ? ?8 ? ? ? 0.5 ? ? ?...
> 1951 ?1 ?1 ? ?1 ? ? ? 1.3 ? ? ?-0,17
> 1951 ?1 ?2 ? ?2 ? ? ? 2.1 ? ? ?Mean (Typ2 1951)
> ....
>
> I hope you can help me by solving this problem
It sounds like you want something like ave(). One approach (untested)
might be as follows:
ddply(df, .(a, typ), transform, mean_typ_year = mean(value))
You may want to re-sort the data afterward because ddply() will sort
by a x typ combinations rather than a x m x d. You could also use the
transform() function, perhaps something like
transform(df, mean_typ_year = ave(value, list(a, typ), FUN = mean))
##------
Here's a toy example to illustrate:
df <- data.frame(a = factor(rep(LETTERS[1:3], each = 6)),
b = factor(rep(letters[1:3], each = 2)),
d = rep(1:6, 3),
val = rnorm(18, m = 40))
library('plyr')
ddply(df, .(a, b), transform, m = mean(val))
transform(df, m = ave(val, list(a, b), FUN = mean))
HTH,
Dennis
>
> Best regards,
> Marko
>
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