CODEWITHSUNDEEP
r
crumbly
crumbly

Reputation: 79

Add multiple columns simultaneously based on a sequence

I have the following dataframe u

 u <- data.frame(a1 = c("a", "b", "c"), a2 = c(2, 1, 3))

I want to add 14 more columns simultaneously based on a simple sequence which is to create new column by adding constant 3 to the last column. the resulting data frame: 

a1  a2  a3  a4  a5  a6  a7  a8  a9  a10 a11 a12 a13 a14 a15 a16
a   2   5   8   11  14  17  20  23  26  29  32  35  38  41  44
b   1   4   7   10  13  16  19  22  25  28  31  34  37  40  43
c   3   6   9   12  15  18  21  24  27  30  33  36  39  42  45

Upvotes: 2

Views: 230

Answers (5)

joel.wilson
joel.wilson

Reputation: 8413

u[paste0("a",3:16)] = as.data.frame(lapply(1:14, function(x) u$a2+3*x))
#  a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16
#1  a  2  5  8 11 14 17 20 23  26  29  32  35  38  41  44
#2  b  1  4  7 10 13 16 19 22  25  28  31  34  37  40  43
#3  c  3  6  9 12 15 18 21 24  27  30  33  36  39  42  45

Upvotes: 4

granddejeuner
granddejeuner

Reputation: 69

  u<-data.frame(a1=c("a","b","c"),a2=c(2,1,3))
 for(i in 1:14){ 
 u[,ncol(u)+1] <-  u[,ncol( u)]+3
   names(u)[ncol(u)]<-paste0("a",ncol(u))
 }

This uses a for loop and gives you the column names you want.

Upvotes: 0

David Arenburg
David Arenburg

Reputation: 92282

I would vectorize this as follows

u[paste0("a", 3:16)] <- matrix(u$a2 + rep(3*1:14, each = 3), nrow = 3)
u
#   a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16
# 1  a  2  5  8 11 14 17 20 23  26  29  32  35  38  41  44
# 2  b  1  4  7 10 13 16 19 22  25  28  31  34  37  40  43
# 3  c  3  6  9 12 15 18 21 24  27  30  33  36  39  42  45

Upvotes: 7

niczky12
niczky12

Reputation: 5063

A simple for loop should do the trick:

for (i in 3:16){
  u[ , sprintf("a%.0d", i)] <- u[, i-1] + 3
}

Upvotes: 0

Ronak Shah
Ronak Shah

Reputation: 388797

We can use sapply to loop over every element of a2 column and generate a sequence from that number , increment it by 3 and output 15 values (column 2 is over written here)

temp = cbind(u[1], t(sapply(u$a2, function(x) seq(x, by = 3, length.out = 15))))
temp

#  a1 1 2 3  4  5  6  7  8  9 10 11 12 13 14 15
#1  a 2 5 8 11 14 17 20 23 26 29 32 35 38 41 44
#2  b 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43
#3  c 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45

Upvotes: 1

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