提取所有内容，直到第三阶段

Question

I have this Identifier column

structure(list(Identifier = c("NC.1.OA", "NC.1.OA.0", "NC.1.OA.1", 
"NC.1.OA.1.a", "NC.1.OA.1.b", "NC.1.OA.1.c", "NC.1.OA.2", "NC.1.OA.2.0", 
"NC.1.OA.3", "NC.1.OA.4", "NC.1.OA.4.0", "NC.1.OA.9", "NC.1.OA.6", 
"NC.1.OA.6.a", "NC.1.OA.6.b", "NC.1.OA.6.c", "NC.1.OA.6.d", "NC.1.OA.6.e", 
"NC.1.OA.6.f", "NC.1.OA.6.f.0", "NC.1.OA.7", "NC.1.OA.8")), row.names = c(NA, 
-22L), class = c("tbl_df", "tbl", "data.frame"))

I'd like to extract out the NC.1.OA from this column. Generally, that would be extracting everything from beginning to the third period, but the first row would violate that since there are only two periods.

I've tried gsub(".*\\\\.(.*)\\\\..*", "\\\\1", Identifier) with no cigar.

Answer 1

We can use str_extract

library(tidyverse)
df %>% 
  mutate(new = str_extract(Identifier, "NC\\.1\\.OA"))

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Also, if we are using sub ( gsub is not needed as we are not substituting globally), fix with a position identifier to notify the start ( ^ ) of the string. In the pattern below, we match one or more characters that are not a . ( [^.]+ ) followed by a . ( [.] - dot is a metacharacter, so we escape or place it in square brackets to evaluate literally), followed by one more digits ( \\\\d+ ), then a . and characters that not a dot (as earlier), capture as a group (wrap with brackets) and in the replacement use the backreference ( \\\\1 ) of the captured group

sub("^([^.]+[.]\\d+[.][^.]+).*", "\\1", df$Identifier)
#[1] "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA"
#[12] "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA" "NC.1.OA"