In R, Check If String Appears in Row of Dataframe (In Any Column)

in R, check if string appears in row of dataframe (in any column)

Here is an option. Use apply and match (%in%).

apply(temp, 1, function(x) any(x %in% "Matt")) 
[1]  TRUE  TRUE FALSE  TRUE FALSE

return TRUE when character string present in any column of data.frame

You can do:

library(tidyverse)
df %>%
  rowwise() %>%
  mutate(col4 = any(str_detect(c_across(c(col1, col2, col3)), 'chicks')),
         col5 = any(str_detect(c_across(c(col1, col2, col3)), 'chicks|pair'))) %>%
  ungroup()

# A tibble: 5 x 5
  col1  col2  col3   col4  col5 
  <chr> <chr> <chr>  <lgl> <lgl>
1 no    no    chicks TRUE  TRUE 
2 no    no    no     FALSE FALSE
3 no    pair  pair   FALSE TRUE 
4 no    no    no     FALSE FALSE
5 pair  no    no     FALSE TRUE 

If Column Contains String then enter value for that row

We can use grepl to return a logical index by matching the 'D' in the 'A' column, and then with ifelse, change the logical vector to 'yes' and 'no'

df$C <- ifelse(grepl("D", df$A), "yes", "no")

R function that detects if a dataframe column contains string values from another dataframe column and adds a column that contains the detected str

First you could a column for each of the possible categories to the dataframe with the names, as placeholders (just filled with NA). Then for each of those columns, check whether the column name (so the category) appears in the name. Turn it into a long dataframe, and then remove the FALSE rows -- those that didn't detect the category in the name.

library(tidyverse)

df1 <- tribble(
  ~name,
  "Apple page",
  "Mango page",
  "Lychee juice",
  "Cranberry club"
)
df2 <- tribble(
  ~fruit,
  "Apple",
  "Grapes",
  "Strawberry",
  "Mango",
  "lychee",
  "cranberry"
)

fruits <- df2$fruit %>%
  str_to_lower() %>% 
  set_names(rep(NA_character_, length(.)), .)

df1 %>% 
  add_column(!!!fruits) %>% 
  mutate(across(-name, ~str_detect(str_to_lower(name), cur_column()))) %>% 
  pivot_longer(-name, names_to = "category") %>% 
  filter(value) %>% 
  select(-value)

#> # A tibble: 4 × 2
#>   name           category 
#>   <chr>          <chr>    
#> 1 Apple page     apple    
#> 2 Mango page     mango    
#> 3 Lychee juice   lychee   
#> 4 Cranberry club cranberry

R - If column contains a string from vector, append flag into another column

Update:
If a list is preferred: Using str_extract_all:

df %>%  
  transmute(across(-id, ~case_when(str_detect(., pattern) ~ str_extract_all(., pattern)), .names = "new_col{col}")) 

gives:

  new_colonetext new_colcop new_coltext3
  <list>         <list>     <list>      
1 <chr [1]>      <NULL>     <chr [2]>   
2 <chr [2]>      <chr [2]>  <NULL>      
3 <chr [2]>      <chr [4]>  <chr [5]>  

Here is how you could achieve the result:

1. create a pattern of the vector
2. use mutate across to check the needed columns
3. if the desired string is detected then extract to a new column !

myvec <- c("cat", "dog", "bird")

pattern <- paste(myvec, collapse="|")

library(dplyr)
library(tidyr)
df %>% 
  mutate(across(-id, ~case_when(str_detect(., pattern) ~ str_extract_all(., pattern)), .names = "new_col{col}")) %>% 
  unite(topic, starts_with('new'), na.rm = TRUE, sep = ',')

    id onetext                cop                                                                        text3                                                                              topic                                     
  <dbl> <chr>                  <chr>                                                                      <chr>                                                                              <chr>                                     
1     1 cat furry pink british Little Grey Cat is the nickname given to a kitten of the British Shorthai~ On October 4th the first single topic blog devoted to the little grey cat was lau~ "cat,NULL,c(\"cat\", \"cat\")"            
2     2 dog cat fight          Dogs have soft fur and tails so do cats Do cats like to chase their tails  there are many fights going on and this is just an example text                    "c(\"dog\", \"cat\"),c(\"cat\", \"cat\"),~
3     3 bird cat issues        A cat and bird can coexist in a home but you will have to take certain me~ Some cats will not care about a pet bird at all while others will make it its lif~ "c(\"bird\", \"cat\"),c(\"cat\", \"bird\"~                                                                                    

If any string values in a character vector are in a column of a data frame, return the string that matches in a new column

library(tidyverse)

df <-
  tibble(
    Category = c("Sales","Marketing"),
    Comment = c("i have to use my email everyday and they dont work the poor communication is not acceptable",
                "i think the tools are not adequate for the tasks we want to achieve"   
                )
  )

keywords <- c('poor communication', 'email', 'tools', 'hardware', 'software')

df %>% 
  #Applyng for each row
  rowwise() %>% 
  mutate(
    Topic =
      #Extract keyword from the string
      str_extract(Comment,keywords) %>%
      #Remoing NA's 
      na.omit() %>% 
      #Paste keywords
      paste0(collapse = ", ")
  )

check which columns in list have exact string value and extract the column and row

I am not sure if you want things like below

a <- transform(
  as.data.frame(
    which(matrix(grepl("^11$", as.matrix(df)), nrow = nrow(df)),
    arr.ind = TRUE
  )),
  col = names(df)[col]
)

which gives

> a
  row col
1   1   B
2   1   C
3   2   C

Finding rows containing a value (or values) in any column

How about

apply(df, 1, function(r) any(r %in% c("M017", "M018")))

The ith element will be TRUE if the ith row contains one of the values, and FALSE otherwise. Or, if you want just the row numbers, enclose the above statement in which(...).

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