R 用|分隔符连接字符向量

我有一个包含字符向量的数据结构（见下文）。它有点混乱，因为它来自json源代码

我需要组合/连接到一个大字符串，其中lat/long对由|分隔，lat/long值由逗号分隔，名称已删除

i、 e.“53.193418、-2881248 | 53.1905138631287、-2.89043889005541等”

我试过了

piped.data<-unname(paste(b, sep="|", collapse=","))

你可以试试

 paste(sapply(split(b,cumsum(grepl('latitude',names(b)))),
             toString),collapse="|")

如果您不需要

空间

 paste(sapply(split(b,cumsum(grepl('latitude',names(b)))),
                             paste, collapse=","), collapse="|")

或者使用

vapply

，这样会更快一些

 paste(vapply(split(b,cumsum(grepl('latitude',names(b)))),
             paste, collapse=",", character(1L)), collapse="|")

或

或

我会将您的“b”转换为2列

矩阵

，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" 

编辑 我想我误解了你的问题

如果它是您想要的单个长字符串，首先用逗号分隔，然后用管道分隔，则需要粘贴两次：

paste(apply(matrix(b, ncol = 2, byrow = TRUE), 1, paste, collapse = ","), 
      collapse = "|")

你可以：

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"

---

tmp另一个选项是

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"

如果不希望逗号后面有空格，请执行以下操作

paste(tapply(b, gl(length(b)/2, 2), paste, collapse = ","), collapse = "|")

编辑：
所以@akrun和@SvenHohenstein能够矢量化他们的解决方案，下面是一些用于说明的基准

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

b您可以使用逻辑索引和向量回收：

paste(paste(b[c(TRUE, FALSE)], b[c(FALSE, TRUE)], sep = ","), collapse = "|")

另一个选项是将向量重塑为data.frame

do.call(paste, c(data.frame(matrix(b, ncol=2, byrow=TRUE)), 
        list(sep=",", collapse="|")))

@LeeJH如果你不需要
空间
，请使用
粘贴，collapse=“，”
你总能找到一种将所有内容矢量化的方法，不是吗：）@davidernburg我只是在探索一种只使用
粘贴
一次的方法。@davidernburg我不知道这是否比任何解决方案都快。只是一次尝试：-）我意识到我的解决方案基本上是您的扩展，但它的格式看起来更糟糕comment@baptiste，不用担心。我也考虑过这个选择，但在我的平板电脑上发布，不想尝试太多。不管怎么说，你的选择会变得更快。每个人所做的贡献都是值得赞赏的。这给了我们很多思考的东西。感谢@Davidernburg花时间运行基准测试。
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

paste(paste(b[c(TRUE, FALSE)], b[c(FALSE, TRUE)], sep = ","), collapse = "|")

do.call(paste, c(data.frame(matrix(b, ncol=2, byrow=TRUE)), 
        list(sep=",", collapse="|")))