R function to search in character string-CodePudding

I have a dataframe like this

Now When I am loading this dataframe in R I want something like this

when a number in NAICS_CD col is present in top_3 , I need is_present col as 1 else 0 I need this to be done with R programming

CodePudding user response：

We could use str_detect from stringr package together with an ifelse statement:

library(dplyr)
library(stringr)

df %>% 
  mutate(is_present = ifelse(str_detect(top_3, as.character(NAICS_CD)), 1, 0))

  NAICS_CD                        top_3 is_present
1   541611 ["541611","541618","611430"]          1
2   812990 ["561720","561740","561790"]          0
3   424950 ["444120","711510","811121"]          0
4   722330 ["311991","722310","722320"]          0
5   722320 ["722320","722330","722310"]          1
6   531180 ["531110","531190","531111"]          0
7   484121 ["484121","484110","484230"]          1
8   531311 ["531110","531311","531111"]          1

data:

df <- structure(list(NAICS_CD = c(541611L, 812990L, 424950L, 722330L, 
722320L, 531180L, 484121L, 531311L), top_3 = c("[\"541611\",\"541618\",\"611430\"]", 
"[\"561720\",\"561740\",\"561790\"]", "[\"444120\",\"711510\",\"811121\"]", 
"[\"311991\",\"722310\",\"722320\"]", "[\"722320\",\"722330\",\"722310\"]", 
"[\"531110\",\"531190\",\"531111\"]", "[\"484121\",\"484110\",\"484230\"]", 
"[\"531110\",\"531311\",\"531111\"]")), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8"))

CodePudding user response：

You can use grepl to check whether any pattern (in this case any targeted number) is present or not in the targeted row, and then use any to return a single TRUE or FALSE to be suitable with conditional ifelse. Then, use ifelse to assign 1 or 0 to is_present column. For example:

top3_row1 <-'["541611", "541618","611430"]'
is_present <- ifelse(any(grepl("54161", top3_row1)), 1, 0)
is_present
[1] 1

To apply this to your data frame, you can use for loop or other ways. For example:

for(k in 1:nrow(mydf)){
mydf$is_present[k] <- ifelse(any(grepl(mydf$NAICS_CD[k], mydf$top_3)), 1, 0)
}