计算宽数据框中每对坐标之间的距离 [英] calculate distance between each pair of coordinates in wide dataframe
问题描述
我想计算两个链接的空间坐标集(在我的伪数据集中为program和admin)之间的距离.数据采用宽格式，因此两对坐标都在同一行.
I want to calculate the distance between two linked set of spatial coordinates (program and admin in my fake dataset). The data are in a wide format, so both pairs of coordinates are in the same row.
library(sp)
set.seed(1)
n <- 100
program.id <- seq(1, n)
c1 <- cbind(runif(n, -90, 90), runif(n, -180, 180))
c2 <- cbind(runif(n, -90, 90), runif(n, -180, 180))
dat <- data.frame(cbind(program.id, c1, c2))
names(dat) <- c("program.id", "program.lat", "program.long", "admin.lat", "admin.long")
head(dat)
# program.id program.lat program.long admin.lat admin.long
# 1 1 -42.20844 55.70061 -41.848523 62.536404
# 2 2 -23.01770 -52.84898 -50.643849 -145.851172
# 3 3 13.11361 -82.70635 3.023431 -2.665397
# 4 4 73.47740 177.36626 -41.588893 -13.841337
# 5 5 -53.69725 48.05758 -57.389701 -44.922049
# 6 6 71.71014 -103.24507 3.343705 176.795719
我知道如何使用sp包在program或admin之间创建距离矩阵:
I know how to create a matrix of distances among program or admin using the sp package:
ll <- c("program.lat", "program.long")
coords <- dat[ll]
dist <- apply(coords, 1,
function(eachPoint) spDistsN1(as.matrix(coords),
eachPoint, longlat=TRUE))
但是我要做的是在每对坐标之间创建一个距离(dist.km)的nx1向量，并将其添加到dat中.
But what I want to do is create a nx1 vector of distances (dist.km) between each pair of coordinates and add it to dat.
# program.id program.lat program.long admin.lat admin.long dist.km
# 1 1 -42.20844 55.70061 -41.848523 62.536404 567.35
# 2 2 -23.01770 -52.84898 -50.643849 -145.851172 8267.86
# ...
有什么建议吗?我花了一段时间来解决旧的SO问题，但是似乎没有什么是正确的.很高兴被证明是错误的.
Any suggestions? I've spent a while going through old SO questions, but nothing seems quite right. Happy to be proven wrong.
更新
@Amit的解决方案适用于我的玩具数据集:
@Amit's solution works for my toy dataset:
apply(dat,1,function(x) spDistsN1(matrix(x[2:3],nrow=1),x[3:4],longlat=TRUE))
但是我想我需要交换lat的顺序，把lat long列的顺序长到lat之前.来自?spDistsN1:
But I think I need to swap the order of the lat, long the order of the lat long columns so long comes before lat. From ?spDistsN1:
pts: A matrix of 2D points, first column x/longitude, second column y/latitude, or a SpatialPoints or SpatialPointsDataFrame object
此外，除非我误解了逻辑，否则我认为Amit的解决方案应该使用cols [2:3]和[4:5]，而不是[2:3]和[3:4].
Also, unless I've misunderstood the logic, I think Amit's solution should grab cols [2:3] and [4:5], not [2:3] and [3:4].
我现在的挑战是将其应用于我的实际数据.我在下面转载了一部分.
My challenge now is applying this to my actual data. I've reproduced a portion below.
library(sp)
dat <- structure(list(ID = 1:4,
subcounty = c("a", "b", "c", "d"),
pro.long = c(33.47627919, 31.73605491, 31.54073482, 31.51748984),
pro.lat = c(2.73996953, 3.26530095, 3.21327597, 3.17784981),
sub.long = c(33.47552, 31.78307, 31.53083, 31.53083),
sub.lat = c(2.740362, 3.391209, 3.208736, 3.208736)),
.Names = c("ID", "subcounty", "pro.long", "pro.lat", "sub.long", "sub.lat"),
row.names = c(NA, 4L), class = "data.frame")
head(dat)
# ID subcounty pro.long pro.lat sub.long sub.lat
# 1 1 a 33.47628 2.739970 33.47552 2.740362
# 2 2 b 31.73605 3.265301 31.78307 3.391209
# 3 3 c 31.54073 3.213276 31.53083 3.208736
# 4 4 d 31.51749 3.177850 31.53083 3.208736
apply(dat, 1, function(x) spDistsN1(matrix(x[3:4], nrow=1),
x[5:6],
longlat=TRUE))
我收到错误:Error in spDistsN1(matrix(x[3:4], nrow = 1), x[5:6], longlat = TRUE) : pts must be numeric
我很困惑，因为这些列是数字:
I'm confused because these columns are numeric:
> is.numeric(dat$pro.long)
[1] TRUE
> is.numeric(dat$pro.lat)
[1] TRUE
> is.numeric(dat$sub.long)
[1] TRUE
> is.numeric(dat$sub.lat)
[1] TRUE
推荐答案
您遇到的问题是apply(...)将第一个参数强制转换为矩阵.根据定义，矩阵必须具有相同数据类型的所有元素.由于dat(dat$subcounty)中的一列是char，因此apply(...)将所有内容强制转换为char.在您的测试数据集中，所有内容都是数字，因此您没有这个问题.
The problem you're having is thatapply(...) coerces the first argument to a matrix. By definition, a matrix must have all elements of the same data type. Since one of the columns in dat (dat$subcounty) is char, apply(...) coerces everything to char. In your test dataset, everything was numeric, so you didn't have this problem.
这应该有效:
dat$dist.km <- sapply(1:nrow(dat),function(i)
spDistsN1(as.matrix(dat[i,3:4]),as.matrix(dat[i,5:6]),longlat=T))
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