Hi lili,
You can extend it like this. I checked this with two values each for
pop and mood, and it looked okay. Obviously I didn't check the result
with 365 values for each, but it ran okay.
# these values are the centers of the black cells
lat<-rep(28:38,11)
lon<-rep(98:108,each=11)
pop<-matrix(s
Hi Jim,
Thanks. It works. I now have more complex problems. If at each blackcell,
there are two variables such as pop and mood. For each variable, there are
daily records in one year, so 365 records for pop and 365 records for mood.
The averaged values for the redcells should be daily records too.
Hi lily,
It's not too hard to do it using dataframes. Getting the indexing
right is usually that hardest part:
# these values are the centers of the black cells
lat<-rep(28:38,11)
lon<-rep(98:108,each=11)
pop<-sample(80:200,121)
# just use the data.frame function
blackcells<-data.frame(lat=lat,lon
Hi lily,
You could also create "blackcells" as a dataframe (which is itself a
type of list). I used a list as I thought it would be a more general
solution if there were different numbers of values for different grid
cells. The use of 1 for the comparison was due to the grid increments
being 1. If
Hi Jim,
Thanks. Yes, the two assumptions are correct, and they reflect the
datasets. I have an uncertainty about the code below. Why do you use
abs(blackcells[[i]]$lat - redcell$lat) <1 rather than a different number
than 1? Second, why to construct blackcells as a list, rather than a
dataframe. B
Hi lily,
There are one or two assumptions to be made here. First is that the
latitude and longitude values of the "black" cells are equally spaced
as in your illustration. Second, that all latitude and longitude
values for the "red" cells fall at the corners of four "black" cells.
You can get the
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