8

I am working on HCUP data and this has range of values in one single column that needs to be split into multiple columns. Below is the HCUP data frame for reference :

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

The desired output should be :

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

My approach to this problem is using the package splitstackshape and using this code

library(data.table)
library(splitstackshape)

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

This approach leads to memory issues. Is there a better approach to this problem?

Some comments :

  • The data has many letters apart from "T".
  • The letter can be either in the front or at the very end but not in between two numbers.
  • There is no change of letter from "T" to "U" in one single range
4

5 回答 5

8

这是使用dplyrand all.is.numericfrom的解决方案Hmisc：

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

使用字符值修复范围的编辑。稍微降低了简单性：

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")
于 2015-10-13T23:57:37.403 回答
6

原始答案：请参阅下面的更新。

首先，通过将第一行添加到底部，我使您的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

我们可以使用序列操作符:来获取code列的序列，用 包裹，tryCatch()这样我们就可以避免错误，并保存无法排序的值。首先，我们用破折号分割值，-然后通过lapply().

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

我们正在尝试将序列运算符:应用于 中的每个元素strsplit()，如果x[1]:x[2]无法获取，则仅返回这些元素的值，x[1]:x[2]否则继续执行序列。label然后我们只是根据结果长度复制列的值xx以获得新label列。

更新：这是我为回应您的编辑而提出的。将xx上面替换为

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，这很复杂。现在，如果我们将以下数据框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")

并在上面运行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
于 2015-10-14T01:27:25.047 回答
3

为此类代码创建排序规则：

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)

}

以@jeremycg 为例： 

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

这使

    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 的答案复制的数据：

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")
于 2015-10-14T18:33:39.260 回答
3

如果您有足够的耐心，您可能会将字符串解析为单独的部分，而不是使用 eval/parse 技巧，可惜我不是，所以：

fancy.seq = function(x) eval(parse(text=sub(', \\)', ')', sub('\\(, ', '(',
               sub('.*?([0-9]+)(.*)-(.*?)([1-9][0-9]*).*',
                   'paste0("\\3",
                           formatC(\\1:\\4, width=log10(\\4)+1, format="d", flag="0"),
                           "\\2")',
                   x)))))
# using example from jeremycg's answer
dt[, .(fancy.seq(code), label), by = 1:nrow(dt)]
#    nrow    V1             label
# 1:    1 61000   excision of CNS
# 2:    1 61001   excision of CNS
# 3:    1 61002   excision of CNS
# 4:    2 0169T ventricular shunt
# 5:    2 0170T ventricular shunt
# 6:    2 0171T ventricular shunt
# 7:    3 01000   excision of CNS
# 8:    3 01001   excision of CNS
# 9:    3 01002   excision of CNS
#10:    4 A2543    some procedure
#11:    4 A2544    some procedure
#12:    4 A2545    some procedure
#13:    5 A0543    some procedure
#14:    5 A0544    some procedure
#15:    5 A0545    some procedure

如果不清楚上述内容在做什么 - 只需sub在“代码”字符串之一上一一运行命令。

于 2015-10-14T22:10:20.167 回答
1

一种不太优雅的方法：

# the data
hcup <- data.frame(code=c("61000-61003", "0169T-0169T"),
                   label=c("excision of CNS", "ventricular shunt"), stringsAsFactors = F)
hcup
>         code             label
>1 61000-61003   excision of CNS
>2 0169T-0169T ventricular shunt

# reshaping
# split the code ranges into separate columns
seq.ends <- cbind(do.call(rbind.data.frame, strsplit(hcup$code, "-")), hcup$label)
# create a list with a data.frame for each original line
new.list <- apply(seq.ends, 1, FUN=function(x){data.frame(code=if(grepl("\\d{5}", x[1])){
                     z<-x[1]:x[2]}else{z<-x[1]}, label=rep(x[3], length(z)),
                     stringsAsFactors = F)})
# collapse the list into a df
new.df <- do.call(rbind, lapply(new.list, data.frame, stringsAsFactors=F))

new.df
>     code             label
>1.1 61000   excision of CNS
>1.2 61001   excision of CNS
>1.3 61002   excision of CNS
>1.4 61003   excision of CNS
>2   0169T ventricular shunt
于 2015-10-14T00:57:39.297 回答