Creating a new dataframe based on another data frame (Using a loop or otherwise)

Question

I have a data frame like this:

Site    speciescode abundance   times
LM1           MkI        9      3
LM2           KiU        8      4

I want to repeat the rows depending on the times values. I was thinking of using a loop to make the new dataframe as follows:

Site   speciescode   abundance
LM1    MkI           9
LM1    MkI           9
LM1    MkI           9
LM2    KiU           8
LM2    KiU           8
LM2    KiU           8
LM2    KiU           8

Help.

Edit your question. I hope you meant to ask something like this. I thought this may be helpful to more people — Stat-R
– Stat-R, Commented Aug 24, 2012 at 1:42

dickoa · Accepted Answer · 2012-08-24 07:14:42Z

4

You can avoid loop if you want (and it's more R'ish)

d <- data.frame(Site=c("LM1","LM2"),speciescode=c("MkI","KiU"),abundance=c(3,4),time=c(3, 4))

dnew <- d[rep(1:nrow(d), d$time), ]
dnew[ ,1:3]
##     Site speciescode abundance
## 1    LM1         MkI         3
## 1.1  LM1         MkI         3
## 1.2  LM1         MkI         3
## 2    LM2         KiU         4
## 2.1  LM2         KiU         4
## 2.2  LM2         KiU         4
## 2.3  LM2         KiU         4

You can also change the abundance value using @BlueMagister method.

edited Aug 24, 2012 at 7:14

answered Aug 23, 2012 at 12:35

dickoa

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1 Comment

Blue Magister Over a year ago

Oh, nifty. That's probably more efficient speed-wise.

Stat-R · Accepted Answer · 2012-08-24 02:17:45Z

2

df <- data.frame(Site=c("LM1","LM2"),speciescode=c("MkI","KiU"),abundance=c(9,8),times=c(3,4))
df <- df[rep(1:nrow(df), df$abundance),]
df$abundance <- rep(c(9,8), c(3,4))

#    Site speciescode abundance
#1    LM1         MkI         9
#1.1  LM1         MkI         9
#1.2  LM1         MkI         9
#2    LM2         KiU         8
#2.1  LM2         KiU         8
#2.2  LM2         KiU         8
#2.3  LM2         KiU         8

edited Aug 24, 2012 at 2:17

Stat-R

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answered Aug 23, 2012 at 12:42

Andy

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Comments

Stat-R · Accepted Answer · 2012-08-24 02:17:54Z

1

Try this (edit the row numbers as necessary):

d <- data.frame(Site=c("LM1","LM2"),speciescode=c("MkI","KiU"),abundance=c(3,4),times=c(3,4))
for(i in seq(from=3,to=7,by=2)) {
  d[i,] <- d[1,]
  d[i+1,] <- d[2,]
}
d$abundance[d$times == 3] <- 9
d$abundance[d$times == 4] <- 8

edited Aug 24, 2012 at 2:17

Stat-R

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answered Aug 23, 2012 at 12:30

Blue Magister

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