transmute new columns based on exact match of multiple words in string

问题

I have a data frame:

df <- data.frame(
  Otherspp = c("suck SD", "BT", "SD RS", "RSS"),
  Dominantspp = c("OM", "OM", "RSS", "CH"),
  Commonspp = c(" ", " ", " ", "OM"),
  Rarespp = c(" ", " ", "SD", "NP"),
  NP = rep("northern pikeminnow|NORTHERN PIKEMINNOW|np|NP|npm|NPM", 4),
  OM = rep("steelhead|STEELHEAD|rainbow trout|RAINBOW TROUT|st|ST|rb|RB|om|OM", 4),
  RSS = rep("redside shiner|REDSIDE SHINER|rs|RS|rss|RSS", 4),
  suck = rep("suckers|SUCKERS|sucker|SUCKER|suck|SUCK|su|SU|ss|SS", 4)
)

I need to use the columns populated with common fish codes/names (NP, OM, RSS, suck) to evaluate the expressions in the first four columns and output a 1/0 based on each of those columns, if the expression is met EXACTLY. The code I have below does not match full words (only partial) and provides incorrect data (see resulting tibble below).

df %>%
  rowwise() %>%
  transmute_at(vars(NP, OM, RSS, suck), 
               funs(case_when(
                 grepl(., Dominantspp) ~ "1",
                 grepl(., Commonspp) ~ "1",
                 grepl(., Rarespp) ~ "1",
                 grepl(., Otherspp) ~ "1",
                 TRUE ~ "0"))) %>%
  ungroup()

Result: see that in row three, both "suck" and "RSS" receive a "1".

# A tibble: 4 x 4
     NP    OM   RSS  suck
  <chr> <chr> <chr> <chr>
1     0     1     0     1
2     0     1     0     0
3     0     0     1     1
4     1     1     1     1

Desired output:

  NP OM RSS suck
1  0  1   0    1
2  0  1   0    0
3  0  0   1    0
4  1  1   1    0

回答1:

The fastest way to solve your problem using your same approach is to add word boundaries to the beginning and end of each of your regexes, with \\b:

df <- data.frame(
  Otherspp = c("suck SD", "BT", "SD RS", "RSS"),
  Dominantspp = c("OM", "OM", "RSS", "CH"),
  Commonspp = c(" ", " ", " ", "OM"),
  Rarespp = c(" ", " ", "SD", "NP"),
  NP = rep("\\b(northern pikeminnow|NORTHERN PIKEMINNOW|np|NP|npm|NPM)\\b", 4),
  OM = rep("\\b(steelhead|STEELHEAD|rainbow trout|RAINBOW TROUT|st|ST|rb|RB|om|OM\\b)", 4),
  RSS = rep("\\b(redside shiner|REDSIDE SHINER|rs|RS|rss|RSS)\\b", 4),
  suck = rep("\\b(suckers|SUCKERS|sucker|SUCKER|suck|SUCK|su|SU|ss|SS)\\b", 4),
  stringsAsFactors = FALSE
)

This makes the regular expressions only match full words, which will make your subsequent solution work.

Having said that, I don't think this is necessarily the way to approach the problem (rowwise() is rarely recommended today, and this approach won't scale well to many fish codes). I think you'd have an easier time working with this data if you standardized it to a tidy format, with one row per combination of row and code:

library(tidyr)
library(tidytext)

row_codes <- df %>%
  select(Otherspp:Rarespp) %>%
  mutate(row = row_number()) %>%
  gather(type, codes, -row) %>%
  unnest_tokens(code, codes, token = "regex", pattern = " ")

Which would result in:

   row        type code
1    1 Dominantspp   om
2    1    Otherspp suck
3    1    Otherspp   sd
4    2 Dominantspp   om
5    2    Otherspp   bt
6    3 Dominantspp  rss
7    3    Otherspp   sd
8    3    Otherspp   rs
9    3     Rarespp   sd
10   4   Commonspp   om
11   4 Dominantspp   ch
12   4    Otherspp  rss
13   4     Rarespp   np

At this point, the codes are much easier to work with (you don't need regular expressions anymore). For example, you could inner_join it to a table of the fish codes.

来源：https://stackoverflow.com/questions/47933639/transmute-new-columns-based-on-exact-match-of-multiple-words-in-string

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dplyr

tidyverse