英文:
Extract leading numbers from string, but length varies R
问题
有一列包含字母和数字的字符字符串。该字符串始终以一个或两个数字开头，然后是多个字符。我尝试根据第一个字符的位置将字符串分开。have <-tribble(~string,'12main','6six','42go','5to9')want <-tribble(~prefix, ~rest,'12', 'main','6', 'six','42', 'go','5', 'to9')我确定有一个使用 `separate` 函数的正则表达式解决方案，但我在使其工作方面遇到了困难。want <-have %>%separate(string,into = c('prefix', 'rest'),sep = "(?=[0-9])(?<=[a-zA-Z])")
英文:
I have a column which contains a character string containing letters and numbers. The string always starts with one or two numbers, followed by multiple characters. I am trying to separate the string based on where that first character is.
have <-tribble(~string,'12main','6six','42go','5to9')want <-tribble(~prefix, ~rest,'12', 'main','6', 'six','42', 'go','5', 'to9')
I'm sure there is a regex with separate solution but having trouble getting it working.
want <-have %>%separate(string,into = c('prefix', 'rest'),sep = "(?=[0-9])(?<=[a-zA-Z])")
答案1
得分: 2
你几乎就要成功了，我们可以通过一个向前查找（用于数字）和一个向后查找（用于非数字）来实现它：
have %>%separate(string, sep = "(?<=[0-9])(?=[^0-9])", into = c("prefix", "rest"))# # A tibble: 4 × 2# prefix rest# <chr> <chr># 1 12 main# 2 6 six# 3 42 go# 4 5 to9
我认为你把查找方向搞反了：?<= 是用于前面的字符串（应该与 [0-9] 一起使用），而 ?= 是用于后面的字符串（应该与 [^0-9] 或 [A-Za-z] 一起使用）。
个人觉得这有点有趣：我们基于零长度的模式拆分字符串：前面是数字，后面是非数字，所以拆分实际上是零长度的。
顺便说一下，如果字符串中有两个这样的位置，比如 5to9to5，会出现警告：
have <- structure(list(string = c("12main", "6six", "42go", "5to9", "5to9to5")), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L))have# # A tibble: 5 × 1# string# <chr># 1 12main# 2 6six# 3 42go# 4 5to9# 5 5to9to5have %>%separate(string, sep = "(?<=[0-9])(?=[^0-9])", into = c("prefix", "rest"))# Warning: Expected 2 pieces. Additional pieces discarded in 1 rows [5].# # A tibble: 5 × 2# prefix rest# <chr> <chr># 1 12 main# 2 6 six# 3 42 go# 4 5 to9# 5 5 to9
这是一个警告，说明你正在丢弃一些信息，你可以自行决定是否需要处理这种情况。
另一个选择，因为你在实际数据中有 5to9to5：
have %>%mutate(strcapture("^([0-9]+)([^0-9].*)", string, list(prefix="", rest="")))# # A tibble: 5 × 3# string prefix rest# <chr> <chr> <chr># 1 12main 12 main# 2 6six 6 six# 3 42go 42 go# 4 5to9 5 to9# 5 5to9to5 5 to9to5
现在你可以移除 string。
另一个注意事项：如果你打算将 prefix 转换为整数或数字，那么你可以通过使用 list(prefix=0L, rest="")（或只是 =0）来避免这种需要。那就是 proto= 参数，虽然它的 data 被丢弃了，但它被用于结果列的名称和目标类。
英文:
You were close, we can achieve it with one look-behind (for a number) and one look-ahead (for a non-number):
have %>%separate(string, sep = "(?<=[0-9])(?=[^0-9])", into = c("prefix", "rest"))# # A tibble: 4 × 2# prefix rest# <chr> <chr># 1 12 main# 2 6 six# 3 42 go# 4 5 to9
I think you had the look-around reversed: ?<= is for preceding string (should be used with [0-9]), and ?= is for following string (should be used with [^0-9] or [A-Za-z]).
Personally I find this a bit intriguing: we are splitting strings based on a 0-length pattern: there is nothing between where the previous is a number and the following is a non-number, so the split is effectively 0-length.
FYI, this does run into warnings if there are two such places in a string, such as 5to9to5:
have <- structure(list(string = c("12main", "6six", "42go", "5to9", "5to9to5")), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L))have# # A tibble: 5 × 1# string# <chr># 1 12main# 2 6six# 3 42go# 4 5to9# 5 5to9to5have %>%separate(string, sep = "(?<=[0-9])(?=[^0-9])", into = c("prefix", "rest"))# Warning: Expected 2 pieces. Additional pieces discarded in 1 rows [5].# # A tibble: 5 × 2# prefix rest# <chr> <chr># 1 12 main# 2 6 six# 3 42 go# 4 5 to9# 5 5 to9
It's a warning that you are discarding some information, over to you if you want/need to guard against this.
An alternative, since you have 5to9to5 in your real data:
have %>%mutate(strcapture("^([0-9]+)([^0-9].*)", string, list(prefix="", rest="")))# # A tibble: 5 × 3# string prefix rest# <chr> <chr> <chr># 1 12main 12 main# 2 6six 6 six# 3 42go 42 go# 4 5to9 5 to9# 5 5to9to5 5 to9to5
where you can now remove string if you want.
Another note: if you are intending to convert prefix into an integer or a number, then you can preclude that need by using list(prefix=0L, rest="") (or just =0) instead. That's the proto= argument, and while its data is discarded, it is used for its names and target classes for each resulting column).
答案2
得分: 1
你也可以使用 extract。
有 %>%提取(string, c('prefix', 'rest'), "(\\d+)(.*)")＃ 一个数据框: 4 × 2前缀 剩余<chr> <chr>1 12 主要2 6 六3 42 去4 5 到9
英文:
You can also use extract
have %>%extract(string, c('prefix', 'rest'), "(\\d+)(.*)")# A tibble: 4 × 2prefix rest<chr> <chr>1 12 main2 6 six3 42 go4 5 to9
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