R 我可以每隔一个逗号分隔一个字符吗？

假设我有dd，我想拆分cc列，最后得到一个targetdd

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假设我有dd，我想拆分cc列，最后得到一个targetdd。

一个简单的方法是用逗号拆分列，每两行分组，并使用toString汇总值

使用我们可以在一个字母后面拆分

library(dplyr)
dd %>% tidyr::separate_rows(cc, sep = ",(?=[a-z]+)")
# A tibble: 6 x 2
  aa    cc    
  <chr> <chr> 
1 we    de,34 
2 we    ff,55u
3 we    gf,55 
4 dd    de,34 
5 qw    de,34 
6 qw    ff,55u

基本R解决方案：

# Split the string into groups: 
cc_uld <- lapply(strsplit(dd$cc, ","), function(x){
      x <- unlist(strsplit(paste0(ifelse(grepl("\\d+", x), 
                                  paste0(x, ","), x), collapse = " "), ","))
    }
  )

# Replicate aa vector by the length of cc_uld list elements:
rolled_out_df <- data.frame(aa = rep(as.character(dd$aa), sapply(cc_uld, length)), 
                            cc = unlist(cc_uld), stringsAsFactors = FALSE)

# Prep the string to mirror the target:
rolled_out_df$cc <- gsub("\\s+", ",", trimws(rolled_out_df$cc, "both"))

使用strsplit。正则表达式从@A.Suliman处借用

do.call(rbind.data.frame, Map(cbind, dd$aa, strsplit(dd$cc, ",(?=[a-z]+)", perl=T)))
#      V1     V2
# we.1 we  de,34
# we.2 we ff,55u
# we.3 we  gf,55
# dd   dd  de,34
# qw.1 qw  de,34
# qw.2 qw ff,55u

字符串中sep每出现一次拆分的广义解：

---

您可以使用strsplit按逗号拆分，然后将它们逐个分组。

dd %>% 
 mutate(res = purrr::map(cc, ~ tibble(cc_new = unlist(stringr::str_split(.x, ",(?=[a-z]+)"))))) %>% 
 unnest(res)

# A tibble: 6 x 3
  aa    cc                 cc_new
  <chr> <chr>              <chr> 
1 we    de,34,ff,55u,gf,55 de,34 
2 we    de,34,ff,55u,gf,55 ff,55u
3 we    de,34,ff,55u,gf,55 gf,55 
4 dd    de,34              de,34 
5 qw    de,34,ff,55u       de,34 
6 qw    de,34,ff,55u       ff,55u

# Split the string into groups: 
cc_uld <- lapply(strsplit(dd$cc, ","), function(x){
      x <- unlist(strsplit(paste0(ifelse(grepl("\\d+", x), 
                                  paste0(x, ","), x), collapse = " "), ","))
    }
  )

# Replicate aa vector by the length of cc_uld list elements:
rolled_out_df <- data.frame(aa = rep(as.character(dd$aa), sapply(cc_uld, length)), 
                            cc = unlist(cc_uld), stringsAsFactors = FALSE)

# Prep the string to mirror the target:
rolled_out_df$cc <- gsub("\\s+", ",", trimws(rolled_out_df$cc, "both"))

do.call(rbind.data.frame, Map(cbind, dd$aa, strsplit(dd$cc, ",(?=[a-z]+)", perl=T)))
#      V1     V2
# we.1 we  de,34
# we.2 we ff,55u
# we.3 we  gf,55
# dd   dd  de,34
# qw.1 qw  de,34
# qw.2 qw ff,55u

nGroup_indexes <- function(vec, n) {
  vec_len <- length(vec)
  index_groups <- list()
  for (i in seq(n)) {
    index_groups[[i]] <- seq(from=i, to=vec_len, by=n)
  }
  index_groups
  suppressWarnings(index_mat <- Reduce(cbind, index_groups))
  colnames(index_mat) <- 1:n
  divisor <- vec_len %% n
  if (!(divisor * n == vec_len)) {
    index_mat[vec_len %/% n + 1, (vec_len %% n + 1):n] <- NA
  }
  index_mat
}

nGroups <- function(vec, n) {
  index_mat <- nGroup_indexes(vec, n)
  res <- lapply(1:nrow(index_mat), function(row_idx) vec[index_mat[row_idx, ]])
  res[[nrow(index_mat)]] <- na.omit(res[[nrow(index_mat)]])
  res
}

split_every_kth <- function(str, sep, k) {
  i_res <- strsplit(str, sep)[[1]]
  grouped_i_res <- nGroups(i_res, k)
  groups_joined <- lapply(grouped_i_res, function(vec) paste(vec, collapse=sep))
  unlist(groups_joined) # make back to vector
}

# > split_every_kth("a,b,c,d,e,f,g", ",", 2)
# [1] "a,b" "c,d" "e,f" "g"
# > split_every_kth("a,b,c,d,e,f,g", ",", 3)
# [1] "a,b,c" "d,e,f" "g"