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diff in a dataframe
by Vishal Belsare
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I have a dataframe say:
date price_g price_s 0.34 0.56 0.36 0.76 . . . . . . and so on. say, 1000 rows. Is it possible to add two columns to this dataframe, by computing say diff(log(price_g) and diff(log(price_s)) ? The elements in the first row of these columns cannot be computed, but can I coerce this to happen and assign a missing value there? It would be really great if I could do that, because in this case I don't have to re-index my transformed series to the dates again in a new dataframe. Thanks in anticipation. Vishal Belsare ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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Re: diff in a dataframe
by Henrique Dallazuanna
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Perhaps you can do this:
cbind(df, sapply(rbind(c(NA, NA),log(df)), diff)) On 10/01/2008, Vishal Belsare <shoot.spam@...> wrote: > I have a dataframe say: > > date price_g price_s > 0.34 0.56 > 0.36 0.76 > . . > . . > . . > > and so on. say, 1000 rows. > > Is it possible to add two columns to this dataframe, by computing say > diff(log(price_g) and diff(log(price_s)) ? > > The elements in the first row of these columns cannot be computed, but > can I coerce this to happen and assign a missing value there? It would > be really great if I could do that, because in this case I don't have > to re-index my transformed series to the dates again in a new > dataframe. > > Thanks in anticipation. > > > Vishal Belsare > > ______________________________________________ > R-help@... mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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Re: diff in a dataframe
by bartjoosen
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Maybe: cbind(df,rbind(NA,apply(log(df),2,diff))) Bart
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Re: diff in a dataframe
by Don MacQueen
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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 and sapply are overly complicated for what you requested. If you had to do this for many, many columns a looping solution would be worth it, but for just two columns, it's not. -Don At 12:46 PM +0530 1/10/08, Vishal Belsare wrote: >I have a dataframe say: > >date price_g price_s > 0.34 0.56 > 0.36 0.76 > . . > . . > . . > >and so on. say, 1000 rows. > >Is it possible to add two columns to this dataframe, by computing say >diff(log(price_g) and diff(log(price_s)) ? > >The elements in the first row of these columns cannot be computed, but >can I coerce this to happen and assign a missing value there? It would >be really great if I could do that, because in this case I don't have >to re-index my transformed series to the dates again in a new >dataframe. > >Thanks in anticipation. > > >Vishal Belsare > >______________________________________________ >R-help@... mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. -- -------------------------------------- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA 925-423-1062 ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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Re: diff in a dataframe
by Gabor Grothendieck
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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) price_g price_s 2000-01-01 NA NA 2000-01-05 0.05715841 0.3053816 See the two zoo vignettes: vignette("zoo") vignette("zoo-quickref") On Jan 10, 2008 2:16 AM, Vishal Belsare <shoot.spam@...> wrote: > I have a dataframe say: > > date price_g price_s > 0.34 0.56 > 0.36 0.76 > . . > . . > . . > > and so on. say, 1000 rows. > > Is it possible to add two columns to this dataframe, by computing say > diff(log(price_g) and diff(log(price_s)) ? > > The elements in the first row of these columns cannot be computed, but > can I coerce this to happen and assign a missing value there? It would > be really great if I could do that, because in this case I don't have > to re-index my transformed series to the dates again in a new > dataframe. > > Thanks in anticipation. > > > Vishal Belsare > > ______________________________________________ > R-help@... mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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Re: diff in a dataframe
by Vishal Belsare
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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 however. Say I have a zoo object named X and say it has 200 time series within it. Each has a unique column name. I tried to retrieve merely one column (time series) from X, trying variously: X$columnname, or X["columnname"] or X[["columnname"]] in the hope that I would be able to get that one time series, but the only way which seemed to work is X[,"columnname"] Is that the 'correct' way to retrieve a single time series from a zoo of multiple time series? I would think that it'd be cooler if we could merely do a : X$columnname sort of thing. Please enlighten. Thanks much! Cheers, Vishal On Jan 10, 2008 10:40 PM, Gabor Grothendieck <ggrothendieck@...> wrote: > 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) > price_g price_s > 2000-01-01 NA NA > 2000-01-05 0.05715841 0.3053816 > > > See the two zoo vignettes: > vignette("zoo") > vignette("zoo-quickref") > > > On Jan 10, 2008 2:16 AM, Vishal Belsare <shoot.spam@...> wrote: > > I have a dataframe say: > > > > date price_g price_s > > 0.34 0.56 > > 0.36 0.76 > > . . > > . . > > . . > > > > and so on. say, 1000 rows. > > > > Is it possible to add two columns to this dataframe, by computing say > > diff(log(price_g) and diff(log(price_s)) ? > > > > The elements in the first row of these columns cannot be computed, but > > can I coerce this to happen and assign a missing value there? It would > > be really great if I could do that, because in this case I don't have > > to re-index my transformed series to the dates again in a new > > dataframe. > > > > Thanks in anticipation. > > > > > > Vishal Belsare > > > > > ______________________________________________ > > R-help@... mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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Re: diff in a dataframe
by Gabor Grothendieck
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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 <shoot.spam@...> 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 > 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 however. > Say I have a zoo object named X and say it has 200 time series within > it. Each has a unique column name. I tried to retrieve merely one > column (time series) from X, trying variously: X$columnname, or > X["columnname"] or X[["columnname"]] in the hope that I would be able > to get that one time series, but the only way which seemed to work is > X[,"columnname"] > > Is that the 'correct' way to retrieve a single time series from a zoo > of multiple time series? I would think that it'd be cooler if we could > merely do a : X$columnname sort of thing. Please enlighten. Thanks > much! > > Cheers, > > Vishal > > > > On Jan 10, 2008 10:40 PM, Gabor Grothendieck <ggrothendieck@...> wrote: > > 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) > > price_g price_s > > 2000-01-01 NA NA > > 2000-01-05 0.05715841 0.3053816 > > > > > > See the two zoo vignettes: > > vignette("zoo") > > vignette("zoo-quickref") > > > > > > On Jan 10, 2008 2:16 AM, Vishal Belsare <shoot.spam@...> wrote: > > > I have a dataframe say: > > > > > > date price_g price_s > > > 0.34 0.56 > > > 0.36 0.76 > > > . . > > > . . > > > . . > > > > > > and so on. say, 1000 rows. > > > > > > Is it possible to add two columns to this dataframe, by computing say > > > diff(log(price_g) and diff(log(price_s)) ? > > > > > > The elements in the first row of these columns cannot be computed, but > > > can I coerce this to happen and assign a missing value there? It would > > > be really great if I could do that, because in this case I don't have > > > to re-index my transformed series to the dates again in a new > > > dataframe. > > > > > > Thanks in anticipation. > > > > > > > > > Vishal Belsare > > > > > > > > ______________________________________________ > > > R-help@... mailing list > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > > > and provide commented, minimal, self-contained, reproducible code. > > > > > > ______________________________________________ R-help@... mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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