R：根据另一列的文本条件更新列

Question

I would like to make a new column in my data frame by using a conditional statement that would say "If Column_y contains Column_x then 1 else 0"

For example:

Event   Name     Winner       Loser          New Column
1       James    James,Bob    John,Steve     1
1       Bob      James,Bob    John,Steve     1
1       John     James,Bob    John,Steve     0
1       Steve    James,Bob    John,Steve     0

I want to have New Column<- "If Winner contains Name then 1 else 0"

Keep in mind this is for 100,000 rows and probably 700 unique names. When I try things like

df$NewColumn<-ifelse(grepl(df$Name,df$Winner)==TRUE,1,0) 

or variations I get the "pattern has a length > 1" error.

Answer 1

I think you just want to compare the Name column against the Winner column:

df$NewColumn <- ifelse(df$Name == df$Winner, 1, 0)

Note that because df$Name == df$Winner is actually a boolean expression, you might also be able to simplify to:

df$NewColumn <- df$Name == df$Winner

Answer 2

In your example, exact string matching works. But I am assuming it does not hold true for your entire data.

Implementing the contains condition would be something like this:

library(dplyr)
library(purrr)

df = df %>% 
  dplyr::mutate(NewColumn = purrr::map2_dbl(.x=Winner,.y=Name,~ifelse(grepl(.y,.x),1,0)))

Adding an alternate solution with stringr :

df = df %>% 
  dplyr::mutate(NewColumn=ifelse(str_detect(Winner,Name),1,0))

Let me know if this works.

PS: str_detect is faster.