finding a row based of a word in a column. But the word in the column is either on its own or as par-CodePudding

How do I write code to find the rows with the word "Alternative" under style. BUT not just "Alternative" on its own but "Alternative" combined with other words. Here the dataframe is called metals

style
id brand_id brand_name  origin    formed   split    fans        style
5   1038    Keldian      Norway   2005     2005      54           Adult
6   2022    Dawn Of Ashes   USA   2001     2001     17          Aggrotech
7   43     Sepultura     Brazil   1984      2003    1185         Alternative
8   3388    Louna        Russia   2007     2008     5       Alternative rock
9   785    Stam1na      Finland    1992     1996     78   Alternative thrash

Its easy to find the columns under style with "Alternative" only like here:

metals[metals$style == "Alternative",]

but I want to find the rows with "Alternative" under style, regardless if it has another word besides it like rock of thrash or not, which means I want 3 rows printed with the sample I posted above

CodePudding user response：

Try this. It's also a good way to share a minimal reproducible example which enables us to be able to run and improve code.

library(tidyverse)


tribble(~id, ~style,
5, "Adult",
6, "Aggrotech",
7, "Alternative",
8, "Alternative rock",
9, "Alternative thrash"
) |> 
  filter(str_detect(style, "Alternative"))
#> # A tibble: 3 × 2
#>      id style             
#>   <dbl> <chr>             
#> 1     7 Alternative       
#> 2     8 Alternative rock  
#> 3     9 Alternative thrash

^{Created on 2022-05-11 by the reprex package (v2.0.1)}

CodePudding user response：

Use grepl to match a pattern:

metals[grepl("(?i)Alternative", metals$style),]

Note that the (?i) flag is used to make the match case-insensitive so that you would find values such as "alternative" or even " "alterNAtive" too. If that's not wanted or necessary just leave it out.