CODEWITHSUNDEEP
rdataframe
Adrian Mak
Adrian Mak

Reputation: 157

Create new columns for df in for-loop with name of existing columns + string

My dataframe is structured like the following:

ID   A_L   A_R   B_L   B_R
1     7     5     6     3
2     3     2     3     1
3     6     3     4     5

The goal is to create a new column for each existing column (besides the first column ID) dividing the value of the existing column through its L/R counterpart. So A_L_ratio = A_L/A_R and A_R_ratio = A_R/A_L etc.

I've tried to create a for-loop, using if/elseto differentiate between odd and even indices.

for (col in 2:length(df)) {
  if( (col%%2) == 0){
    a <- df[,col] / df[,col+1]}
    else{
    a <- df[,col] / df[,col-1]}
df[colnames(df[col])"_ratio"] <- a
}

But I seem to fail at R's syntax when it comes to naming the columns. Name should be the name of the column that is called in each loop df[,col] + the string _ratio. At the end I want to append that columne to df. Could someone tell me the right syntax to do this? Thanks a lot!

Upvotes: 0

Views: 306

Answers (2)

Thiago Jacomasso
Thiago Jacomasso

Reputation: 51

You need to paste the colnames to the string "_ratio". Something like this, maybe:

# Create the data.frame
df <- data.frame(
  ID = 1:3,
  A_L = c(7, 3, 6),
  A_R = c(5, 2, 3),
  B_L = c(6, 3, 4),
  B_R = c(3, 1, 5)
)

# create the cols with "_ratio" character appended
for (col in 2:length(df)) {
  if( (col%%2) == 0){
    a <- df[,col] / df[,col+1]
    } else {(a <- df[,col] / df[,col-1])}
df[paste(colnames(df[col]), "_ratio", sep = "")] <- a
}

There are easier and more efficient ways to do this using the dplyr package, though.

Upvotes: 1

Sirius
Sirius

Reputation: 5429

Don't konw how important this is, but I got the imrpession you have many more columns than what is shown? If so better take it nice and slow so you don't get errors.

If safety checks are not needed, then disregard this.


df <- read.table( text="
ID   A_L   A_R   B_L   B_R
1     7     5     6     3
2     3     2     3     1
3     6     3     4     5
", header=TRUE )

var.names.L <- grep( "_L$", colnames(df) , value=TRUE )
var.names.R <- sub( "_L", "_R", var.names.L )

i.L.name.ok <- var.names.R %in% colnames(df)

ok.L.names <- var.names.L[i.has.R.name]
ok.R.names <- var.names.R[i.has.R.name]

new.columns.1 <- df[, ok.L.names ] / df[, ok.R.names ]
colnames(new.columns.1) <- paste0( colnames(new.columns.1), "_ratio" )
new.columns.2 <- df[, ok.R.names ] / df[, ok.L.names ]
colnames(new.columns.2) <- paste0( colnames(new.columns.2), "_ratio" )


cbind.data.frame(
    df,
    new.columns.1,
    new.columns.2
)

The above code nice and neatly checks that for every _L column there is a coresponding _R column, and then it performs the divition with only those columns.

Output:


  ID A_L A_R B_L B_R A_L_ratio B_L_ratio A_R_ratio B_R_ratio
1  1   7   5   6   3       1.4       2.0 0.7142857 0.5000000
2  2   3   2   3   1       1.5       3.0 0.6666667 0.3333333
3  3   6   3   4   5       2.0       0.8 0.5000000 1.2500000

Upvotes: 1

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