使用R将单个列拆分为多个观测

Question

我正在处理HCUP数据，这在一个列中有一个值的范围，需要分割成多个列。以下是可供参考的HCUP数据框架：

code            label
61000-61003     excision of CNS
0169T-0169T     ventricular shunt

预期的产出应是：

code            label
61000           excision of CNS
61001           excision of CNS
61002           excision of CNS
61003           excision of CNS
0169T           ventricular shunt

我解决这个问题的方法是使用包splitstackshape并使用以下代码

library(data.table)
library(splitstackshape)

cSplit(hcup, "code", "-")[, list(code = code_1:code_2, by = label)]

这种方法会导致内存问题。是否有更好的方法来解决这个问题？

一些评论：

• 除了"T“之外，数据还有许多字母。
• 这封信可以在前面，也可以在最后面，但不能介于两个数字之间。
• 字母在一个范围内从"T“到"U”没有变化

Answer 1

下面是一个使用dplyr和Hmisc的all.is.numeric的解决方案

library(dplyr)
library(Hmisc)
library(tidyr)
dat %>% separate(code, into=c("code1", "code2")) %>%
        rowwise %>%
        mutate(lists = ifelse(all.is.numeric(c(code1, code2)),
                         list(as.character(seq(from = as.numeric(code1), to = as.numeric(code2)))),
                         list(code1))) %>%
        unnest(lists) %>%
        select(code = lists, label)

Source: local data frame [5 x 2]

   code             label
  (chr)            (fctr)
1 61000   excision of CNS
2 61001   excision of CNS
3 61002   excision of CNS
4 61003   excision of CNS
5 0169T ventricular shunt

要用字符值修正范围的编辑。把简单说得更简单一点：

dff %>% mutate(row = row_number()) %>%
        separate(code, into=c("code1", "code2")) %>%
        group_by(row) %>%
        summarise(lists = if(all.is.numeric(c(code1, code2)))
                              {list(str_pad(as.character(
                                   seq(from = as.numeric(code1), to = as.numeric(code2))),
                                         nchar(code1), pad="0"))}
                          else if(grepl("^[0-9]", code1))
                              {list(str_pad(paste0(as.character(
                                   seq(from = extract_numeric(code1), to = extract_numeric(code2))),
                                      strsplit(code1, "[0-9]+")[[1]][2]),
                                         nchar(code1), pad = "0"))}
                          else
                              {list(paste0(
                                      strsplit(code1, "[0-9]+")[[1]],
                                      str_pad(as.character(
                                    seq(from = extract_numeric(code1), to = extract_numeric(code2))),
                                         nchar(gsub("[^0-9]", "", code1)), pad="0")))},
                   label = first(label)) %>%
        unnest(lists) %>%
        select(-row)
Source: local data frame [15 x 2]

               label lists
               (chr) (chr)
1    excision of CNS 61000
2    excision of CNS 61001
3    excision of CNS 61002
4  ventricular shunt 0169T
5  ventricular shunt 0170T
6  ventricular shunt 0171T
7    excision of CNS 01000
8    excision of CNS 01001
9    excision of CNS 01002
10    some procedure A2543
11    some procedure A2544
12    some procedure A2545
13    some procedure A0543
14    some procedure A0544
15    some procedure A0545

数据：

dff <- structure(list(code = c("61000-61002", "0169T-0171T", "01000-01002", 
"A2543-A2545", "A0543-A0545"), label = c("excision of CNS", "ventricular shunt", 
"excision of CNS", "some procedure", "some procedure")), .Names = c("code", 
"label"), row.names = c(NA, 5L), class = "data.frame")

Answer 2

原始答案：查看下面的更新。

首先，通过将第一行添加到底部，使示例数据更具挑战性。

dff <- structure(list(code = c("61000-61003", "0169T-0169T", "61000-61003"
), label = c("excision of CNS", "ventricular shunt", "excision of CNS"
)), .Names = c("code", "label"), row.names = c(NA, 3L), class = "data.frame")

dff
#          code             label
# 1 61000-61003   excision of CNS
# 2 0169T-0169T ventricular shunt
# 3 61000-61003   excision of CNS

我们可以使用序列运算符:来获取code列的序列，用tryCatch()包装以避免出现错误，并保存无法排序的值。首先，我们用破折号-来分割值，然后通过lapply()运行它。

xx <- lapply(
    strsplit(dff$code, "-", fixed = TRUE), 
    function(x) tryCatch(x[1]:x[2], warning = function(w) x)
)
data.frame(code = unlist(xx), label = rep(dff$label, lengths(xx)))
#     code             label
# 1  61000   excision of CNS
# 2  61001   excision of CNS
# 3  61002   excision of CNS
# 4  61003   excision of CNS
# 5  0169T ventricular shunt
# 6  0169T ventricular shunt
# 7  61000   excision of CNS
# 8  61001   excision of CNS
# 9  61002   excision of CNS
# 10 61003   excision of CNS

我们试图将序列运算符:应用于strsplit()中的每个元素，如果不可能采用x[1]:x[2]，则只返回这些元素的值，否则继续执行序列x[1]:x[2]。然后，我们只根据label列的长度复制xx列的值，以获得新的label列。

更新:这里是我针对您的编辑提出的。将上面的xx替换为

xx <- lapply(strsplit(dff$code, "-", TRUE), function(x) {
    s <- stringi::stri_locate_first_regex(x, "[A-Z]")
    nc <- nchar(x)[1L]
    fmt <- function(n) paste0("%0", n, "d")
    if(!all(is.na(s))) {
        ss <- s[1,1]
        fmt <- fmt(nc-1)
        if(ss == 1L) {
            xx <- substr(x, 2, nc)
            paste0(substr(x, 1, 1), sprintf(fmt, xx[1]:xx[2]))
        } else {
            xx <- substr(x, 1, ss-1)
            paste0(sprintf(fmt, xx[1]:xx[2]), substr(x, nc, nc))
        }
    } else {
        sprintf(fmt(nc), x[1]:x[2])
    }
})

是的，很复杂。现在，如果我们以下面的数据框架df2作为测试用例

df2 <- structure(list(code = c("61000-61003", "0169T-0174T", "61000-61003", 
"T0169-T0174"), label = c("excision of CNS", "ventricular shunt", 
"excision of CNS", "ventricular shunt")), .Names = c("code", 
"label"), row.names = c(NA, 4L), class = "data.frame") 

并从上面运行xx代码，我们可以得到以下结果。

data.frame(code = unlist(xx), label = rep(df2$label, lengths(xx)))
#     code             label
# 1  61000   excision of CNS
# 2  61001   excision of CNS
# 3  61002   excision of CNS
# 4  61003   excision of CNS
# 5  0169T ventricular shunt
# 6  0170T ventricular shunt
# 7  0171T ventricular shunt
# 8  0172T ventricular shunt
# 9  0173T ventricular shunt
# 10 0174T ventricular shunt
# 11 61000   excision of CNS
# 12 61001   excision of CNS
# 13 61002   excision of CNS
# 14 61003   excision of CNS
# 15 T0169 ventricular shunt
# 16 T0170 ventricular shunt
# 17 T0171 ventricular shunt
# 18 T0172 ventricular shunt
# 19 T0173 ventricular shunt
# 20 T0174 ventricular shunt

Answer 3

为这类代码创建一个排序规则：

seq_code <- function(from,to){

    ext = function(x, part) gsub("([^0-9]?)([0-9]*)([^0-9]?)", paste0("\\",part), x)

    pre = unique(sapply(list(from,to), ext, part = 1 ))
    suf = unique(sapply(list(from,to), ext, part = 3 ))

    if (length(pre) > 1 | length(suf) > 1){
        return("NO!")
    }

    num = do.call(seq, lapply(list(from,to), function(x) as.integer(ext(x, part = 2))))
    len = nchar(from)-nchar(pre)-nchar(suf)

    paste0(pre, sprintf(paste0("%0",len,"d"), num), suf)

}

使用@jeremycg的示例：

setDT(dff)[,.(
  label = label[1], 
  code  = do.call(seq_code, tstrsplit(code,'-'))
), by=.(row=seq(nrow(dff)))]

这给

    row             label  code
 1:   1   excision of CNS 61000
 2:   1   excision of CNS 61001
 3:   1   excision of CNS 61002
 4:   2 ventricular shunt 0169T
 5:   2 ventricular shunt 0170T
 6:   2 ventricular shunt 0171T
 7:   3   excision of CNS 01000
 8:   3   excision of CNS 01001
 9:   3   excision of CNS 01002
10:   4    some procedure A2543
11:   4    some procedure A2544
12:   4    some procedure A2545
13:   5    some procedure A0543
14:   5    some procedure A0544
15:   5    some procedure A0545

从@jeremycg的答复中复制的数据：

dff <- structure(list(code = c("61000-61002", "0169T-0171T", "01000-01002", 
"A2543-A2545", "A0543-A0545"), label = c("excision of CNS", "ventricular shunt", 
"excision of CNS", "some procedure", "some procedure")), .Names = c("code", 
"label"), row.names = c(NA, 5L), class = "data.frame")