我有一个包含char vectrors的数据结构（见下文）。它来自json来源，有点乱。
我需要组合/连接到一个大字符串，其中lat / long对由|分隔以逗号分隔的lat和long值以及删除的名称。
即 “53.193418，-2881248 | 53.1905138631287，-2.89043889005541 |等。”。
我试过
piped.data<-unname(paste(b, sep="|", collapse=","))
这使得我可以将值与逗号配对并删除名称。
我只需要将管道添加到各个对
有什么想法吗？
dput(b)
structure(c("53.193418", "-2.881248", "53.1905138631287", "-2.89043889005541",
"53.186744", "-2.890165", "53.189836", "-2.893896", "53.1884117",
"-2.88802", "53.1902965", "-2.8919373", "53.1940384", "-2.8972299",
"53.1934748", "-2.8814698", "53.1894004", "-2.8886692", "53.1916771",
"-2.8846099"), .Names = c("location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude",
"location.coordinate.latitude", "location.coordinate.longitude"
))
答案 0 :(得分：7)
我会将您的“b”转换为2列matrix并粘贴：
apply(matrix(b, ncol = 2, byrow = TRUE), 1, paste, collapse = "|")
# [1] "53.193418|-2.881248" "53.1905138631287|-2.89043889005541"
# [3] "53.186744|-2.890165" "53.189836|-2.893896"
# [5] "53.1884117|-2.88802" "53.1902965|-2.8919373"
# [7] "53.1940384|-2.8972299" "53.1934748|-2.8814698"
# [9] "53.1894004|-2.8886692" "53.1916771|-2.8846099"
我想我误解了你的问题。
如果它是你想要的一个长字符串，首先用逗号分隔，然后用管道分隔，你需要粘贴两次：
paste(apply(matrix(b, ncol = 2, byrow = TRUE), 1, paste, collapse = ","),
collapse = "|")
答案 1 :(得分：6)
另一种选择是将矢量重塑为data.frame，
do.call(paste, c(data.frame(matrix(b, ncol=2, byrow=TRUE)),
list(sep=",", collapse="|")))
答案 2 :(得分：5)
另一种选择是
paste(tapply(b, gl(length(b)/2, 2), toString), collapse = "|")
# [1] "53.193418, -2.881248|53.1905138631287, -2.89043889005541|53.186744, -2.890165|53.189836,
# -2.893896|53.1884117, -2.88802|53.1902965, -2.8919373|53.1940384, -2.8972299|53.1934748,
# -2.8814698|53.1894004, -2.8886692|53.1916771, -2.8846099"
如果您不想在逗号后面留空格，请执行
paste(tapply(b, gl(length(b)/2, 2), paste, collapse = ","), collapse = "|")
修改 所以@akrun和@SvenHohenstein能够对他们的解决方案进行矢量化，所以这里有一些插图的基准
b <- rep(b, 1e3)
library(microbenchmark)
microbenchmark(
SH = paste(paste(b[c(TRUE, FALSE)], b[c(FALSE, TRUE)], sep = ","), collapse = "|"),
akrun1 = paste(c(rbind(b,rep(c(',','|'), length.out = length(b))))[-length(b)*2], collapse = ""),
akrun2 = paste(vapply(split(b,cumsum(grepl('latitude',names(b)))), paste, collapse=",", character(1L)), collapse="|"),
akrun3 = as.data.table(matrix(b, ncol=2, byrow=TRUE))[, paste(V1, V2, sep=',',collapse="|")],
AM = paste(apply(matrix(b, ncol = 2, byrow = TRUE), 1, paste, collapse = ","), collapse = "|"),
DA = paste(tapply(b, gl(length(b)/2, 2), paste, collapse = ","), collapse = "|"),
BA = do.call(paste, c(data.frame(matrix(b, ncol=2, byrow=TRUE)), list(sep=",", collapse="|")))
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# SH 6.207338 6.275886 6.633830 6.472943 6.915140 10.556983 100
#akrun1 8.738792 8.790045 9.301718 9.049665 9.611671 11.899290 100
#akrun2 40.676819 42.329860 45.361688 43.887247 46.427638 109.963421 100
#akrun3 4.648384 4.831599 5.019834 4.901934 5.217579 5.798325 100
# AM 38.322320 40.905073 43.108411 42.457375 44.875023 56.236726 100
# DA 47.102466 49.679579 52.092028 51.237417 53.694154 68.123738 100
# BA 5.227204 5.366769 6.147758 5.494207 5.806313 55.938247 100
答案 3 :(得分：4)
你可以尝试
paste(sapply(split(b,cumsum(grepl('latitude',names(b)))),
toString),collapse="|")
如果您不需要space
paste(sapply(split(b,cumsum(grepl('latitude',names(b)))),
paste, collapse=","), collapse="|")
或者使用vapply
faster
paste(vapply(split(b,cumsum(grepl('latitude',names(b)))),
paste, collapse=",", character(1L)), collapse="|")
或者
paste(c(rbind(b,rep(c(',','|'),length.out=length(b))))[
-length(b)*2],collapse="")
或
library(data.table)
as.data.table(matrix(b, ncol=2, byrow=TRUE))[,
paste(V1, V2, sep=',',collapse="|")]
答案 4 :(得分：4)
您可以使用逻辑索引和矢量回收：
paste(paste(b[c(TRUE, FALSE)], b[c(FALSE, TRUE)], sep = ","), collapse = "|")
答案 5 :(得分：2)
您可以这样做：
tmp <- apply(matrix(b, ncol = 2, byrow = TRUE), MARGIN = 1, FUN = paste, collapse = ",")
paste(tmp, collapse = "|")
# [1] "53.193418,-2.881248|53.1905138631287,-2.89043889005541|53.186744,-2.890165|53.189836,-2.893896|53.1884117,-2.88802|53.1902965,-2.8919373|53.1940384,-2.8972299|53.1934748,-2.8814698|53.1894004,-2.8886692|53.1916771,-2.8846099"