On Jun 17, 2008, at 8:06 AM, Chuck Cleland wrote:
On 6/17/2008 6:59 AM, Steve Murray wrote:
Dear all,
I have used 'read.table' to create a data frame of 720 columns and
360 rows (and assigned this to 'Jan'). The row and column names
are numeric:
columnnames <- sprintf("%.2f", seq(from = -179.75, to = 179.75,
length = 720)). rnames <- sprintf("%.2f", seq(from = -89.75, to =
89.75, length = 360))
colnames(Jan) <- columnnames
rownames(Jan) <- rnames
A sample of the data looks like this:
head(Jan)
-179.75 -179.25 -178.75 -178.25 -177.75 -177.25 -176.75
-176.25 -175.75
-89.75 -56.9 -64.2 56.2 -90.0 56.9 -29.0 -91.0
34.0 -9.1
-89.25 37.9 19.3 -0.4 -12.3 -11.8 -92.1 9.2
-23.5 -0.2
-88.75 47.4 3.1 -47.4 46.4 34.2 6.1
-41.3 44.7 -10.3
-88.25 -20.3 34.5 -67.3 -99.9 37.9 -9.3 17.7
-17.2 63.4
-87.75 -46.4 47.4 12.4 -48.3 9.3 -33.8 38.1
10.8 -34.1
-87.25 -48.4 10.3 -89.3 -33.0 -1.1 -33.1 81.2
-8.3 -47.2
I'm hoping to get the whole dataset into the form of columns, so
that, for example, the first row (as shown above) would look like
this:
Latitude Longitude Value
-89.75 -179.75 -56.9
-89.75 -179.25 -64.2
-89.75 -178.75 56.2
-89.75 -178.25 -90.0
-89.75 -177.75 56.9
-89.75 -177.25 -29.0
-89.75 -176.75 -91.0
-89.75 -176.25 34.0
-89.75 -175.75 -9.1
As you can see, this would require the repeated printing of the
the row and column names (in this case '-89.75') - so it's not
just a case of rearranging the data, but creating 'more' data too.
I've tried to achieve this using 'reshape' and 'stack' (their help
files and after looking through the mailing archives), but I'm
obviously doing something wrong. For reshape, I'm getting errors
relating to the commands I enter, and for stack, I can only
produce two columns from my data (with the additional 3rd column
being a row count). In any case, these two columns refer to the
wrong values (it's producing output in the form of: row count
number, Longitude, Value).
I'd be very grateful if anyone could help me out with the commands
I need to enter in order to achieve the results I'm hoping for.
Here is an approach with reshape() on a much smaller example:
columnnames <- sprintf("%.2f", seq(from = -179.75, to = 179.75,
length = 5))
rnames <- sprintf("%.2f", seq(from = - 89.75, to = 89.75, length =
3))
Jan <- as.data.frame(matrix(runif(3*5), ncol=5))
colnames(Jan) <- columnnames
rownames(Jan) <- rnames
Jan$Latitude <- rownames(Jan)
Jan.long <- reshape(Jan, idvar="Latitude", direction="long",
varying = list(columnnames),
v.names="Value",
timevar="Longitude",
times=columnnames)
Jan.long[] <- sapply(Jan.long, as.numeric)
Here's another approach, using Chuck's example. I have two methods,
one produces a data frame, the other produces a matrix. It's up to
you. In the data frame example the first two columns are actually
factors, in the matrix they are numeric vectors. The other key
difference is that I start from a matrix, and I simply use the fact
that a matrix is just a vector with a dim attribute (and I use
as.numeric to drop the dim argument).
Jan <- matrix(runif(3*5), ncol=5)
Jan.long <- data.frame(Latitude=rep(rownames(Jan), ncol(Jan)),
Longitute=rep(colnames(Jan), each=nrow(Jan)), Value=as.numeric(Jan))
Jan.long <- cbind(Latitude=rep(as.numeric(rownames(Jan)), ncol(Jan)),
Longitute=rep(as.numeric(colnames(Jan)), each=nrow(Jan)),
Value=as.numeric(Jan))
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
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