I have a dataframe say:
date price_g price_s
0.340.56
0.360.76
. .
. .
. .
and so on. say, 1000 rows.
Is it possible to add two columns to this dataframe, by computing say
diff(log(price
Perhaps you can do this:
cbind(df, sapply(rbind(c(NA, NA),log(df)), diff))
On 10/01/2008, Vishal Belsare <[EMAIL PROTECTED]> wrote:
> I have a dataframe say:
>
> date price_g price_s
> 0.340.56
> 0.360.76
>. .
>.
Maybe:
cbind(df,rbind(NA,apply(log(df),2,diff)))
Bart
Vishal Belsare wrote:
>
> I have a dataframe say:
>
> date price_g price_s
> 0.340.56
> 0.360.76
>. .
>. .
>. .
>
> and so
So, what's the easiest way to add a column to a dataframe? Just do it.
Here is a really simple example to illustrate:
> foo <- data.frame(x=1:4,y=2:5)
> foo
x y
1 1 2
2 2 3
3 3 4
4 4 5
> foo$z <- c(NA,diff(foo$x))
> foo
x y z
1 1 2 NA
2 2 3 1
3 3 4 1
4 4 5 1
Solutions using apply
1.021651 -0.2744368
3 3.00 4.00 1.098612 1.3862944
and if you wnat to coerce the first row to a whatever
value just type
a[1,]<-NA
Cheers
A
- Messaggio originale -
Da: Vishal Belsare <[EMAIL PROTECTED]>
A: r-help@r-project.org
Inviato: Mercoledì 9 gennaio 2008, 23:16:38
Og
Represent this as a time series. Using
the zoo package:
> library(zoo)
> z <- zoo(cbind(price_g = c(0.34, 0.36), price_s = c(0.56, 0.76)),
> as.Date(c("2000-01-01", "2000-01-05")))
> diff(log(z))
price_g price_s
2000-01-05 0.05715841 0.3053816
> diff(log(z), na.pad = TRUE)
Thanks Henrique, Don and Gabor. I did come around to solving it by
using the zoo library. Very useful stuff that one for handling a bunch
of long irregular time series.
Gabor, thanks for your present and previous responses. The quickref
was indeed helpful. I do have another question regarding zoo
This is consistent with how matrices and ts series in R work: they all
use x[,j] only.
On Jan 10, 2008 1:08 PM, Vishal Belsare <[EMAIL PROTECTED]> wrote:
> Thanks Henrique, Don and Gabor. I did come around to solving it by
> using the zoo library. Very useful stuff that one for handling a bunch
>
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