Issue with pasting 5 columns groups in R

空扰寡人 提交于 2019-12-02 07:23:05

Using DF defined in the Note at the end create a sprintf formatting string fmt and then run it.

If there are NA's in DF then they will appear in the output as the string "NA". If you prefer to omit them completely then replace them with the empty string in DF before running the code below, i.e. run DF[is.na(DF)] <- "" first.

fmt <- paste(rep(strrep("%s", 5), ncol(DF)/5), collapse = "-") # %s%s%s%s%s-%s%s%s%s%s
Output <- do.call("sprintf", c(fmt, DF))
data.frame(DF, Output, stringsAsFactors = FALSE)

giving:

  V29 V30 V31 V32 V33 V34 V35 V36 V37 V38                  Output
1 044   N 005   E 026 044   N 006   E 011 044N005E026-044N006E011

or using DF2 from Note in place of DF we get:

  V29 V30 V31 V32 V33 V34 V35 V36 V37 V38                  Output
1 044   N 005   E 026 044   N 006   E 011 044N005E026-044N006E011
2 045   S 006   F 027 045   S 007   F 012 045S006F027-045S007F012

data.table

If, as per comment, you want to use data.table then use this (with fmt from above):

library(data.table)

DT <- data.table(DF)
DT[, Output:=do.call("sprintf", c(fmt, .SD))]

Note

Lines <- "
  V29  V30  V31  V32  V33  V34 V35 V36 V37 V38 
  044  N    005  E    026  044 N   006 E   011 "
DF <- read.table(text = Lines, header = TRUE, colClasses = "character")

Lines2 <- "
  V29 V30 V31 V32 V33 V34 V35 V36 V37 V38
1 044   N 005   E 026 044   N 006   E 011
2 045   S 006   F 027 045   S 007   F 012"
DF2 <- read.table(text = Lines2, header = TRUE, colClasses = "character")

Expanding your data a little bit:

x <- read.table(stringsAsFactors=FALSE, header=TRUE, as.is=TRUE, colClasses="character", text="
V29  V30  V31  V32  V33  V34 V35 V36 V37 V38    V29a V30a V31a V32a V33a V34a V35a V36a V37a V38a
044  N    005  E    026  044 N   006 E   011    044  N    005  E    026  044  N    006  E    011 
044  N    005  E    026  044 N   006 E   011    044  N    005  E    026  044  N    006  E    011 ")

The answer:

sapply(split.default(x, (seq_len(ncol(x))-1) %/% 5),
       function(s) paste(apply(s, 1, paste0, collapse = ""), collapse = "-"))
#                         0                         1                         2 
# "044N005E026-044N005E026" "044N006E011-044N006E011" "044N005E026-044N005E026" 
#                         3 
# "044N006E011-044N006E011" 

This can easily be assigned to a column of the same frame.

Explanation:

  • to break a frame up by 5 columns, split comes to mind, but the default use of split(...) will use split.data.frame which splits by row, not column, so we use split.default (which works by column). From there, you can see how we're grouping things:

    (seq_len(ncol(x))-1) %/% 5
    #  [1] 0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3
    
  • For each of these groups, we get a 5-column frame:

    split.default(x, (seq_len(ncol(x))-1) %/% 5)
    # $`0`
    #   V29 V30 V31 V32 V33
    # 1  44   N   5   E  26
    # 2  44   N   5   E  26
    # $`1`
    #   V34 V35 V36 V37 V38
    # 1  44   N   6   E  11
    # 2  44   N   6   E  11
    ### truncated for brevity
    

    So we use sapply to do something to each of these frames, returning it (in this case) simplified. (If we specify simplify=FALSE or if not all of them are the same length, then it will be returned unsimplified, as a list instead of a vector).

  • The function we apply to each frame is apply(., 1, paste0, collapse0) which will return a vector of the 5-column pastes, something like:

    apply(s, 1, paste0, collapse = "")
    # $`0`
    # [1] ""044N005E026" "044N005E026""
    

    Because we want them combined, we surround it as paste(apply(...), collapse = "-").

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!