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Hi Christian,
Use the fact that 0 * lat[i] == 0 and 1 * lat[i] == lat[i], where lat[i]
is one of your lat values. Also use the fact that R is vectorized.
So we have:
grid.id<-c(1:4)
lat<-c(10,12,13,15)
species1<-c(0,0,0,1)
species2<-c(1,1,0,0)
species3<-c(1,1,1,1)
## don't need cbind here &
## data is not a good name, use something else
df <- data.frame(grid.id, lat, species1, species2, species3)
So to get what you want:
df[,-(1:2)] * df$lat
and if you want to assign this into df to make the change permanent:
df[,-(1:2)] <- df[,-(1:2)] * df$lat
We drop the first two columns so you don't need to know how many species
you have, just that there are two columns we don't want to mess with.
HTH
G
On Wed, 2008-01-30 at 19:47 +0100, Christian Hof wrote:
> Dear all,
> maybe somebody can provide some help for this problem:
>
> Example:
> I've got the following dataframe "data":
>
> grid.id<-c(1:4)
> lat<-c(10,12,13,15)
> species1<-c(0,0,0,1)
> species2<-c(1,1,0,0)
> species3<-c(1,1,1,1)
> data<-data.frame(cbind(grid.id,lat,species1,species2,species3))
>
> How can I, out of "data" make a new dataframe, where the cells of value
> "1" in the species columns ("species1" to "species3") are replaced by
> the respective "lat" values automatically - so that the final dataframe
> looks like this:
>
> specieslat1<-c(0,0,0,15)
> specieslat2<-c(10,12,0,0)
> specieslat3<-c(10,12,13,15)
> data.frame(cbind(grid.id,lat,specieslat1,specieslat2,specieslat3))
>
> ?
>
> Thanks a lot,
> cheers
> Christian
>
--
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Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
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