CODEWITHSUNDEEP
rduplicates
santoku
santoku

Reputation: 3427

r how to alter several columns based on condition (duplication in another col)

i have a data frame like this

df <- data.frame(groupx=c("k1","k1","k2","k4","k3","k2"),x1=rep(1,6),x2=rep(2,6),
x3=rep(3,6),y1=rep(4,6),x12=rep(5,6))

and for each duplicate row in groups, i wanna modify the several related columns by prefixing the number with 'a'

i'm currently doing it like this and am quite sure it's not the most efficient method:

df[duplicated(df$groupx),"x1"]=paste0("a",df[duplicated(df$groupx),"x1"])
df[duplicated(df$groupx),"x2"]=paste0("a",df[duplicated(df$groupx),"x2"])
df[duplicated(df$groupx),"x3"]=paste0("a",df[duplicated(df$groupx),"x3"])

The desired output is to have "a" in front of corresponding rows of col x1,x2,x3 but not other columns.

Any recommendations? Thanks

Edit: sorry for the misunderstanding. Groupx row are not related with col names, the earlier example was a coincidence

Upvotes: 0

Views: 56

Answers (2)

akrun
akrun

Reputation: 887223

 xCols <- intersect(df$groupx, colnames(df))
 df[, xCols] <- lapply(df[,xCols], function(x) {indx <- duplicated(df$groupx)
                                     x[indx] <-paste0("a", x[indx]); x })


  df
  #  groupx x1 x2 x3 y1 x12
  #1     x1  1  2  3  4   5
  #2     x1 a1 a2 a3  4   5
  #3     x2  1  2  3  4   5
  #4     x4  1  2  3  4   5
  #5     x3  1  2  3  4   5
  #6     x2 a1 a2 a3  4   5

Or

  m1 <- as.matrix(df[,xCols])
  indx <- duplicated(df[,1])[row(df[,xCols])]
  m1[indx] <- paste0("a", m1[indx])
  df[,xCols] <- m1

Upvotes: 2

David Arenburg
David Arenburg

Reputation: 92292

I'm assuming that you want to update only the columns that appear in df$groupx, so here's a possible solution

indx <- grep(paste0("^", paste(unique(df$groupx), collapse = "$|^"), "$"), names(df))
df[duplicated(df$groupx), indx] <- paste0("a", as.matrix(df[duplicated(df$groupx), indx]))
df
#   groupx  x1  x2  x3 y1 x12
# 1     x1   1   2   3  4   5
# 2     x1  a1  a2  a3  4   5
# 3     x2   1   2   3  4   5
# 4     x4   1   2   3  4   5
# 5     x3   1   2   3  4   5
# 6     x2  a1  a2  a3  4   5

Upvotes: 2

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