按组添加ID列

我想根据两列纬度和经度在R中创build一个唯一的ID，以便重复的位置具有相同的簇ID。

例如：

LAT LONG Cluster_ID 13.5330 -15.4180 1 13.5330 -15.4180 1 13.5330 -15.4180 1 13.5330 -15.4180 1 13.5330 -15.4170 2 13.5330 -15.4170 2 13.5330 -15.4170 2 13.5340 -14.9350 3 13.5340 -14.9350 3 13.5340 -15.9170 4 13.3670 -14.6190 5 

Answer 1

这是使用interaction的一种方式。

 d <- read.table(text='LAT LONG 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4170 13.5330 -15.4170 13.5330 -15.4170 13.5340 -14.9350 13.5340 -14.9350 13.5340 -15.9170 13.3670 -14.6190', header=TRUE) d <- transform(d, Cluster_ID = as.numeric(interaction(LAT, LONG, drop=TRUE))) # LAT LONG Cluster_ID # 1 13.533 -15.418 2 # 2 13.533 -15.418 2 # 3 13.533 -15.418 2 # 4 13.533 -15.418 2 # 5 13.533 -15.417 3 # 6 13.533 -15.417 3 # 7 13.533 -15.417 3 # 8 13.534 -14.935 4 # 9 13.534 -14.935 4 # 10 13.534 -15.917 1 # 11 13.367 -14.619 5 

编辑：结合@ Spacedman的build议提供drop=TRUE interaction 。

Answer 2

数据：

 dat <- read.table(text=" LAT LONG 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4170 13.5330 -15.4170 13.5330 -15.4170 13.5340 -14.9350 13.5340 -14.9350 13.5340 -15.9170 13.3670 -14.6190", header = TRUE) 

这些命令创build一个以1开头的idvariables：

 comb <- with(dat, paste(LAT, LONG)) within(dat, Cluster_ID <- match(comb, unique(comb))) 

输出：

  LAT LONG Cluster_ID 1 13.533 -15.418 1 2 13.533 -15.418 1 3 13.533 -15.418 1 4 13.533 -15.418 1 5 13.533 -15.417 2 6 13.533 -15.417 2 7 13.533 -15.417 2 8 13.534 -14.935 3 9 13.534 -14.935 3 10 13.534 -15.917 4 11 13.367 -14.619 5 

Answer 3

.GRP被添加到data.table 1.8.3中，允许您执行以下操作：

 # Your data, as a data.frame dat <- read.table(text='LAT LONG 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4170 13.5330 -15.4170 13.5330 -15.4170 13.5340 -14.9350 13.5340 -14.9350 13.5340 -15.9170 13.3670 -14.6190', header=TRUE) # Convert it to a data.table # with keys as the combination of LAT and LONG library(data.table) DT <- data.table(dat, key="LAT,LONG") DT[, Cluster_ID:=.GRP, by=key(DT)] DT # LAT LONG Cluster_ID # 1: 13.367 -14.619 1 # 2: 13.533 -15.418 2 # 3: 13.533 -15.418 2 # 4: 13.533 -15.418 2 # 5: 13.533 -15.418 2 # 6: 13.533 -15.417 3 # 7: 13.533 -15.417 3 # 8: 13.533 -15.417 3 # 9: 13.534 -15.917 4 # 10: 13.534 -14.935 5 # 11: 13.534 -14.935 5 

Answer 4

比较性能提示解决scheme

 df <- read.table(text='LAT LONG 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4180 13.5330 -15.4170 13.5330 -15.4170 13.5330 -15.4170 13.5340 -14.9350 13.5340 -14.9350 13.5340 -15.9170 13.3670 -14.6190', header=TRUE) f1 <- function(df, cols) { df$id <- as.numeric(interaction(df[cols], drop = TRUE)) df } f2 <- function(df, cols) { comb <- do.call(paste, c(as.list(df[cols]), sep = ".")) df$id <- match(comb, unique(comb)) df } f2(df, 1:2) #> LAT LONG id #> 1 13.533 -15.418 1 #> 2 13.533 -15.418 1 #> 3 13.533 -15.418 1 #> 4 13.533 -15.418 1 #> 5 13.533 -15.417 2 #> 6 13.533 -15.417 2 #> 7 13.533 -15.417 2 #> 8 13.534 -14.935 3 #> 9 13.534 -14.935 3 #> 10 13.534 -15.917 4 #> 11 13.367 -14.619 5 microbenchmark::microbenchmark(f1(df, 1:2), f2(df, 1:2)) #> Unit: microseconds #> expr min lq mean median uq max neval cld #> f1(df, 1:2) 486.400 510.422 575.26659 573.3945 594.1165 1622.243 100 b #> f2(df, 1:2) 72.952 79.208 86.09265 83.5275 89.7195 159.740 100 a