CODEWITHSUNDEEP
rapply
u31889
u31889

Reputation: 351

Create new data frame column

My data frame looks like that:

df <- data.frame(INFO=c("A;B;C", "B;A;C"), METRICS=c("1;2;3", "4;5;6"))
df
   INFO METRICS
1 A;B;C   1;2;3
2 B;A;C   4;5;6

I am trying to catch the value of 'A' for each line using apply() and store the resulting vector in a new column of my original data frame:

df$M1 <- apply(
   df,
   1,
   function(x){
      info <- unlist(strsplit(x[1], ";"))
      metric <- unlist(strsplit(x[2], ";"))
      for(i in 1:3){
         ifelse(
            info[i]=="A", 
            metric[i],
            "."
         )
      }
   }
)

In order to get the following result:

df
   INFO METRICS M1
1 A;B;C   1;2;3  1
2 B;A;C   4;5;6  5

But the new column is not created.

Upvotes: 2

Views: 66

Answers (4)

tmfmnk
tmfmnk

Reputation: 39858

You can also try grepRaw():

ind <- sapply(df$INFO, grepRaw, pattern = "A", fixed = TRUE)
substring(df$METRICS, ind, ind)

[1] "1" "5"

Upvotes: 1

Sotos
Sotos

Reputation: 51582

A vectorized way to do it is to unlist after spliting and make it a named vector, i.e.

i1 <- setNames(unlist(strsplit(as.character(df$METRICS), ';')), 
               unlist(strsplit(as.character(df$INFO), ';')))

then simply,

i1[names(i1) == 'A']
#  A   A 
#"1" "5"

Or add it to your data frame,

 df$M1 <- i1[names(i1) == 'A']

#   INFO METRICS M1
#1 A;B;C   1;2;3  1
#2 B;A;C   4;5;6  5

Upvotes: 1

ThomasIsCoding
ThomasIsCoding

Reputation: 101044

Maybe you can try the apply like below

df$M1<-apply(df, 1, function(x) {
  unlist(strsplit(x[2],split = ";"))[unlist(strsplit(x[1],";"))=="A"]
})

such that

> df
   INFO METRICS M1
1 A;B;C   1;2;3  1
2 B;A;C   4;5;6  5

Upvotes: 1

Ronak Shah
Ronak Shah

Reputation: 388807

We can split the string on ";" and use mapply to get the corresponding value of METRICS where INFO == "A".

df$M1 <- mapply(function(x, y) y[x == "A"], strsplit(df$INFO, ";"),
                                            strsplit(df$METRICS, ";"))
df

#   INFO METRICS M1
#1 A;B;C   1;2;3  1
#2 B;A;C   4;5;6  5

data

Make sure the data is read as characters and not as factors. 

df <- data.frame(INFO=c("A;B;C", "B;A;C"), METRICS=c("1;2;3", "4;5;6"),
                  stringsAsFactors = FALSE)

Upvotes: 1

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