问题
I have asked about and receive great help for computing Euclidean distance in R before. Now, I need to compute the Euclidean distance from the first point relative to all the other points within the track data. Here is how my data looks like:
dput(head(t1))
structure(list(A = c(0L, 0L, 0L, 0L, 0L, 0L), T = 0:5, X = c(668L,
668L, 668L, 668L, 668L, 668L), Y = c(259L, 259L, 259L, 259L,
259L, 259L), V = c(NA, 0, 0, 0, 0, 0)), .Names = c("A", "T",
"X", "Y", "V"), row.names = c(NA, 6L), class = "data.frame")
And SimonO101 was so great in giving me a code that will compute the Euclidean distance from the starting position to the final position for each track:
## Split the data
dfs <- split(t1,t1$A)
## Find hypotenuse between first and last rows for each A
lapply( dfs , function(x){
j <- nrow(x)
str <- x[1,c("X","Y")]
end <- x[j,c("X","Y")]
dist <- sqrt( sum( (end - str)^2 ) )
return( dist )
} )
How do I edit the code, so that it will not just have the Euclidean distance from start to end, but from every X,Y position? Thanks again!
EDIT: And also: How to visualize the results as a matrix. Thank you
回答1:
I would use the dist
function in the stats
package. You can apply it to your data easily enough:
lapply( dfs , function(x) dist( x[,c("X","Y")] , diag = TRUE )[1:nrow(x)] )
The idea being that we operate on each dataframe, applying the dist
function to the "X"
and "Y"
columns of each data frame. The subsetting at the end ( [1:nrow(x)]
)is used to return only the distances between the first point and all the other points. Remove this subsetting if you want a full distance matrix for each track.
回答2:
A possible solution :
f <- function(x) {
j <- nrow(x)
end <- as.numeric(x[j,c("X","Y")])
x <- x[-j,c("X", "Y")]
dist <- colSums((t(x)-end)^2)
dist
}
lapply(dfs, f)
来源:https://stackoverflow.com/questions/16192483/computing-net-distance-euclidean-distance-in-r