Question

我有一堆像这样的字符串：

x  <-  c("4/757.1%", "0/10%", "6/1060%", "0/0-%", "11/2055%")

它们是分数和所述分数的百分比价值，某种程度上某些地方被合并在一起。因此，示例中第一个数字的含义是7个中的4个是57.1％。我可以很容易地得到/之前的第一个数字（比如stringr::word(x, 1, sep = "/")），但第二个数字可以是一个或两个字符长，所以我很难想办法。我不需要％值，因为一旦我得到数字就很容易重新计算。

有人能看到一种方法吗？

Answer 1

正如您所指出的，一旦您获得分数，百分比就可以重新计算出来。你能利用这个事实来找出分裂的位置吗？

GuessSplit <- function(string) {

  tolerance <- 0.001 #How close should the fraction be?
  numerator <- as.numeric(word(string, 1, sep = "/"))
  second.half <-word(string, 2, sep = "/")
  second.half <- strsplit(second.half, '')[[1]]

  # assuming they all end in percent signs
  possibilities <- length(second.half) - 1

  for (position in 1:possibilities) {

    denom.guess <- as.numeric(paste0(second.half[1:position], collapse=''))
    percent.guess <- as.numeric(paste0(second.half[(position+1):possibilities], collapse='')) / 100

    value <- numerator / denom.guess

    if (abs(value - percent.guess) < tolerance) {

      return(list(numerator=numerator, denominator=denom.guess))

    }
  }
}

这需要一点爱来处理怪异的情况，如果它找不到可能性的答案，可能会更优雅。我也不确定哪种返回类型最好。也许你只需要分母，因为分子很容易获得，但我认为两者的列表最为一般。我希望这是一个合理的开始？

Answer 2

一种看起来像你想做的那种丑陋的解决方案：

x  <-  c("4/757.1%", "0/10%", "6/1060%", "0/0-%", "11/2055%")

split_perc <- function(x,signif_digits=1){
  x = gsub("%","",x)
  if(grepl("-",x)) return(list(NA,NA))
  index1 = gregexpr("/",x)[[1]][1]+1
  index2 = gregexpr("\\.",x)[[1]][1]-2
  if(index2==-3){index2=nchar(x)-1}

  found=FALSE
  indices = seq(index1,index2)
  k=1
  while(!found & k<=length(indices))
  {
    str1 =substr(x,1,indices[k])
    num1=as.numeric(strsplit(str1,"/")[[1]][1])
    num2 = as.numeric(strsplit(str1,"/")[[1]][2])
    value1 = round(num1/num2*100,signif_digits)
    value2 = round(as.numeric(substr(x,indices[k]+1,nchar(x))),signif_digits)
    if(value1==value2)
    {found=TRUE}
    else
    {k=k+1}
  }
  if(found)
    return(list(num1,num2))
  else
    return(list(NA,NA))
}

do.call(rbind,lapply(x,split_perc))

输出：

     [,1] [,2]
[1,] 4    7   
[2,] 0    1   
[3,] 6    10  
[4,] NA   NA  
[5,] 11   20

还有一些例子：

y = c("11/2055.003%","11/2055.2%","40/7057.1%")
do.call(rbind,lapply(y,split_perc))

     [,1] [,2]
[1,] 11   20   # default significant digits is 1, so match found.
[2,] NA   NA   # no match found since 55.1!=55.2
[3,] 40   70

Answer 3

来自tidyverse和stringr的解决方案。我们可以定义一个函数来分割第二个数字的所有可能位置，并计算百分比以查看哪个有意义。 df2是显示最佳分割位置的数据框，您想要的数字位于V3列。

library(tidyverse)
library(stringr)

x <- c("4/757.1%", "0/10%", "6/1060%", "0/0-%", "11/2055%")

dt <- str_split_fixed(x, pattern = "/", n = 2) %>%
  as_data_frame() %>%
  mutate(ID = 1:n()) %>%
  select(ID, V1, V2)

# Design a function to spit the second column based on position
split_df <- function(position, dt){
  dt_temp <- dt %>%
    mutate(V3 = str_sub(V2, 1, position)) %>%
    mutate(V4 =  str_sub(V2, position + 1)) %>%
    mutate(Pos = position)

  return(dt_temp)
}

# Process the data
dt2 <- map_df(1:3, split_df, dt = dt) %>%
  # Remove % in V4
  mutate(V4 = str_replace(V4, "%", "")) %>%
  # Convert V1, V3 and V4 to numeric
  mutate_at(vars(V1, V3, V4), funs(as.numeric)) %>%
  # Calculate possible percentage
  mutate(V5 = V1/V3 * 100) %>%
  # Calculate the difference between V4 and V5
  mutate(V6 = abs(V4 - V5)) %>%
  # Select the smallest difference based on V6 for each group
  group_by(ID) %>%
  arrange(ID, V6) %>%
  slice(1)

# The best split is now in V3
dt2$V3
[1]  7  1 10  0 20

我可以合理地拆分这些数字字符串吗？

3 个答案: