我有两个数据帧：df1包含具有纬度坐标的观测值；df2的名称具有纬度坐标。我想创建一个新的变量df1$names，它为每个观测都有df2的名称，这些名称位于距离该观测指定距离内。
df1:的一些样本数据
df1 <- structure(list(lat = c(52.768, 53.155, 53.238, 53.253, 53.312, 53.21, 53.21, 53.109, 53.376, 53.317, 52.972, 53.337, 53.208, 53.278, 53.316, 53.288, 53.341, 52.945, 53.317, 53.249), lon = c(6.873, 6.82, 6.81, 6.82, 6.84, 6.748, 6.743, 6.855, 6.742, 6.808, 6.588, 6.743, 6.752, 6.845, 6.638, 6.872, 6.713, 6.57, 6.735, 6.917), cat = c(2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 3L, 2L, 2L, 2L, 2L, 2L), diff = c(6.97305555555555, 3.39815972222222, 14.2874305555556, -0.759791666666667, 34.448275462963, 4.38783564814815, 0.142430555555556, 0.698599537037037, 1.22914351851852, 7.0008912037037, 1.3349537037037, 8.67978009259259, 1.6090162037037, 25.9466782407407, 9.45068287037037, 4.76284722222222, 1.79163194444444, 16.8280787037037, 1.01336805555556, 3.51240740740741)), .Names = c("lat", "lon", "cat", "diff"), row.names = c(125L, 705L, 435L, 682L, 186L, 783L, 250L, 517L, 547L, 369L, 618L, 280L, 839L, 614L, 371L, 786L, 542L, 100L, 667L, 785L), class = "data.frame")
df2:的一些样本数据
df2 <- structure(list(latlonloc = structure(c(6L, 3L, 4L, 2L, 5L, 1L), .Label = c("Boelenslaan", "Borgercompagnie", "Froombosch", "Garrelsweer", "Stitswerd", "Tinallinge"), class = "factor"), lat = c(53.356789, 53.193886, 53.311237, 53.111339, 53.360848, 53.162031), lon = c(6.53493, 6.780792, 6.768608, 6.82354, 6.599604, 6.143804)), .Names = c("latlonloc", "lat", "lon"), class = "data.frame", row.names = c(NA, -6L))
使用geosphere包创建距离矩阵：
library(geosphere)
mat <- distm(df1[,c('lon','lat')], df2[,c('lon','lat')], fun=distHaversine)
得到的距离以米为单位(至少我认为是这样，否则距离矩阵有问题)。
使用(df1$cat)^2)*1000计算指定距离。我尝试了df1$names <- df2$latlonloc[apply(distmat, 1, which(distmat < ((df1$cat)^2)*1000 ))]，但收到一条错误消息：
Error in match.fun(FUN) :
'which(distmat < ((df1$cat)^2) * 1000)' is not a function, character or symbol
这可能不是正确的方法，但我需要的是：
df1$names <- #code or function which gives me a string of names which are within a specified distance of the observation
如何创建一个名称在观测指定距离内的字符串？
您需要对df1(或mat)的每一行进行操作，才能计算出df2中每一个对象的距离。从中，您可以选择符合距离标准的对象。
我想你对apply和which的使用有点困惑。要使which真正适用于您，您需要将其应用于mat的每一行，而您当前的代码将其应用到整个mat矩阵。还要注意，这里很难使用apply，因为您要将mat的每一行与((df1$cat)^2)*1000)定义的向量的相应元素进行比较。因此，我将向您展示使用sapply和lapply的示例。您也可以在这里使用mapply，但我认为sapply/mapply语法更清晰。
为了解决您想要的输出，我展示了两个例子。对于df1中的每一行，返回一个列表，其中包含df2中在距离阈值内的项目的名称。这不会很容易作为变量返回到原始的df1中，因为列表中的每个元素都可以包含多个名称。第二个示例将这些名称粘贴为一个逗号分隔的字符串，以便创建您要查找的新变量。
示例1:
out1 <- lapply(1:nrow(df1), function(x) {
df2[which(mat[x,] < (((df1$cat)^2)*1000)[x]),'latlonloc']
})
结果：
> str(out1)
List of 20
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..: 2
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..: 4
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..: 6 4 5
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..:
$ : Factor w/ 6 levels "Boelenslaan",..: 4
$ : Factor w/ 6 levels "Boelenslaan",..:
示例2:
out2 <- sapply(1:nrow(df1), function(x) {
paste(df2[which(mat[x,] < (((df1$cat)^2)*1000)[x]),'latlonloc'], collapse=',')
})
结果：
> out2
[1] "" ""
[3] "" ""
[5] "" ""
[7] "" "Borgercompagnie"
[9] "" "Garrelsweer"
[11] "" ""
[13] "" ""
[15] "Tinallinge,Garrelsweer,Stitswerd" ""
[17] "" ""
[19] "Garrelsweer" ""
我认为第二个可能最接近你想要的。