Re: [R] median of grouped data
Check the package doBy as well (pretty handy for analysis of grouped data). library(doBy) summaryBy(PM~ID,data=d,FUN=median) HTH, -Girish -- View this message in context: http://n4.nabble.com/median-of-grouped-data-tp1311971p1312386.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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.
Re: [R] median of grouped data
Like this?
d = read.table(textConnection("ID PM
adt_01 25.255
adt_01 225.36
adt_01 14.2325
adt_02 15
adt_02 12.3
adt_03 15.2
adt_03 148.3
adt_03 25.5
adt_03 14.25"),head=T)
d
d$PM <- as.numeric(d$PM)
ddply(d,.(ID),numcolwise(median))
Felipe D. Carrillo
Supervisory Fishery Biologist Department of the Interior US Fish & Wildlife
Service California, USA
--- On Wed, 1/27/10, "María del Pilar Márquez" wrote:
> From: "María del Pilar Márquez"
> Subject: [R] median of grouped data
> To: r-help@r-project.org
> Date: Wednesday, January 27, 2010, 10:05 AM
> Good morning,
> I would like to know if is possible somebody can help me
> with this please:
> I would like to calculate the median of grouped data, this
> is an example:
>
> ID PM
> adt_01 25.255
> adt_01 225.36
> adt_01 14.2325
> adt_02 15
> adt_02 12.3
> adt_03 15.2
> adt_03 148.3
> adt_03 25.5
> adt_03 14.25
>
> I need the median of adt_01, 02 and 03 , I know that I have
> to split first my data and then ask for the median but the
> code I have is not working.
>
> Thank you very much if you can help me with the code
> maria
>
> __
> R-help@r-project.org
> 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@r-project.org 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.
Re: [R] median of grouped data
Tena koe Maria ?tapply Specifically: tapply(yourData$PM, yourData$ID, median) HTH Peter Alspach > -Original Message- > From: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] On Behalf Of "María del > Pilar Márquez" > Sent: Thursday, 28 January 2010 7:06 a.m. > To: r-help@r-project.org > Subject: [R] median of grouped data > > Good morning, > I would like to know if is possible somebody can help me with > this please: > I would like to calculate the median of grouped data, this is > an example: > > IDPM > adt_01 25.255 > adt_01 225.36 > adt_01 14.2325 > adt_02 15 > adt_02 12.3 > adt_03 15.2 > adt_03 148.3 > adt_03 25.5 > adt_03 14.25 > > I need the median of adt_01, 02 and 03 , I know that I have > to split first my data and then ask for the median but the > code I have is not working. > > Thank you very much if you can help me with the code maria > > __ > R-help@r-project.org 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@r-project.org 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] median of grouped data
Good morning, I would like to know if is possible somebody can help me with this please: I would like to calculate the median of grouped data, this is an example: IDPM adt_01 25.255 adt_01 225.36 adt_01 14.2325 adt_02 15 adt_02 12.3 adt_03 15.2 adt_03 148.3 adt_03 25.5 adt_03 14.25 I need the median of adt_01, 02 and 03 , I know that I have to split first my data and then ask for the median but the code I have is not working. Thank you very much if you can help me with the code maria __ R-help@r-project.org 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.
Re: [R] median of grouped data
On 28 июн, 03:19, "Bricklemyer, Ross S" <[EMAIL PROTECTED]> wrote: > I am having difficulty calculating the median of grouped data. I have 8 to > 10 repeated measures per sample and I have successfully used the following > code to calculate the average for each sample. > > libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, > FUN=ave,libs.norm.preds.median[,1]) > You might want to create a wrapper around "ave" like this: myave <- function(..., ave.FUN = mean) ave(..., FUN = ave.FUN) and pass needed ave.FUN parameter to "apply". Andrey __ R-help@r-project.org 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.
Re: [R] median of grouped data
jim holtman wrote: Is this what you want: # split the dataframe by the grouping (z was your sample data) z.s <- split(z, z[[1]]) # calculate the median (ans <- lapply(z.s, function(.grp) apply(.grp[,7:9], 2, median))) $HOR006_3 TC.15_comps IC.16_comps SOC.17_comps 10.549669 4.224790 7.012470 $HOR006_4 TC.15_comps IC.16_comps SOC.17_comps 14.428948 7.557801 6.568626 $HOR006_5 TC.15_comps IC.16_comps SOC.17_comps 22.37852313.666364 7.290354 do.call(rbind, ans) TC.15_comps IC.16_comps SOC.17_comps HOR006_310.549674.224790 7.012470 HOR006_414.428957.557801 6.568626 HOR006_522.37852 13.666364 7.290354 Isn't this essentially the same as > aggregate(airquality[1:4],airquality["Month"], median, na.rm=T) Month Ozone Solar.R Wind Temp 1 518 194.0 11.5 66 2 623 188.5 9.7 78 3 760 253.0 8.6 84 4 852 197.5 8.6 82 5 923 192.0 10.3 76 ? On Fri, Jun 27, 2008 at 7:19 PM, Bricklemyer, Ross S <[EMAIL PROTECTED]> wrote: I am having difficulty calculating the median of grouped data. I have 8 to 10 repeated measures per sample and I have successfully used the following code to calculate the average for each sample. libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, FUN=ave,libs.norm.preds.median[,1]) I then use the unique function to collapse the data into one line per sample. I would also like to calculate the median, standard error, and coefficient of variation as well. I have not been able to get median to work properly. I have tried this and variants: libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, FUN=median,libs.norm.preds.median[,1]) I receive the following error: Warning messages: 1: In if (na.rm) x <- x[!is.na(x)] else if (any(is.na(x))) return(x[FALSE][NA]) :the condition has length > 1 and only the first element will be used Here is a subset of my data (tab delimited): samp.id corefield TC IC SOC TC.15 comps IC.16 comps SOC.17 compsTC.15 comps IC.16 comps SOC.17 comps HOR006_3HOR006 HOR 7.157 0 7.157 8.008273281 0.786161341 6.402343153 8.008273281 0.786161341 6.402343153 HOR006_3HOR006 HOR 7.157 0 7.157 6.258510623 -1.1175672686.987405984 6.258510623 0 6.987405984 HOR006_3HOR006 HOR 7.157 0 7.157 14.21306811 7.968072165 6.818917226 14.21306811 7.968072165 6.818917226 HOR006_3HOR006 HOR 7.157 0 7.157 17.73301788 9.017994045 9.035508792 17.73301788 9.017994045 9.035508792 HOR006_3HOR006 HOR 7.157 0 7.157 12.54204929 6.285521186 6.052762372 12.54204929 6.285521186 6.052762372 HOR006_3HOR006 HOR 7.157 0 7.157 10.07603128 3.485872902 6.93459 10.07603128 3.485872902 6.93459 HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 HOR006_3HOR006 HOR 7.157 0 7.157 9.249550001 1.92641169 7.675586354 9.249550001 1.92641169 7.675586354 HOR006_3HOR006 HOR 7.157 0 7.157 7.414208739 -0.0205335687.057048733 7.414208739 0 7.057048733 HOR006_4HOR006 HOR 11.73 0 11.73 14.42894814 8.998403641 5.752994239 14.42894814 8.998403641 5.752994239 HOR006_4HOR006 HOR 11.73 0 11.73 13.65284466 6.757373476 6.388413921 13.65284466 6.757373476 6.388413921 HOR006_4HOR006 HOR 11.73 0 11.73 10.72185703 5.053095924 6.016783029 10.72185703 5.053095924 6.016783029 HOR006_4HOR006 HOR 11.73 0 11.73 14.68382689 7.557801473 6.667911142 14.68382689 7.557801473 6.667911142 HOR006_4HOR006 HOR 11.73 0 11.73 2.287381003 -3.0741746566.654986023 2.287381003 0 6.654986023 HOR006_4HOR006 HOR 11.73 0 11.73 14.57145428 8.812845515 6.625453309 14.57145428 8.812845515 6.625453309 HOR006_4HOR006 HOR 11.73 0 11.73 21.12964238 13.27394496 6.568626499 21.12964238 13.27394496 6.568626499 HOR006_4HOR006 HOR 11.73 0 11.73 19.46136803 8.03100103 6.910126723 19.46136803 8.03100103 6.910126723 HOR006_4HOR006 HOR 11.73 0 11.73 13.16591198 4.738398449 6.051036242 13.16591198 4.738398449
Re: [R] median of grouped data
Is this what you want: > # split the dataframe by the grouping (z was your sample data) > z.s <- split(z, z[[1]]) > # calculate the median > (ans <- lapply(z.s, function(.grp) apply(.grp[,7:9], 2, median))) $HOR006_3 TC.15_comps IC.16_comps SOC.17_comps 10.549669 4.224790 7.012470 $HOR006_4 TC.15_comps IC.16_comps SOC.17_comps 14.428948 7.557801 6.568626 $HOR006_5 TC.15_comps IC.16_comps SOC.17_comps 22.37852313.666364 7.290354 > do.call(rbind, ans) TC.15_comps IC.16_comps SOC.17_comps HOR006_310.549674.224790 7.012470 HOR006_414.428957.557801 6.568626 HOR006_522.37852 13.666364 7.290354 On Fri, Jun 27, 2008 at 7:19 PM, Bricklemyer, Ross S <[EMAIL PROTECTED]> wrote: > I am having difficulty calculating the median of grouped data. I have 8 to > 10 repeated measures per sample and I have successfully used the following > code to calculate the average for each sample. > > libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, > FUN=ave,libs.norm.preds.median[,1]) > > I then use the unique function to collapse the data into one line per sample. > > I would also like to calculate the median, standard error, and coefficient of > variation as well. I have not been able to get median to work properly. I > have tried this and variants: > > libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, > FUN=median,libs.norm.preds.median[,1]) > > > I receive the following error: >Warning messages: > 1: In if (na.rm) x <- x[!is.na(x)] else if (any(is.na(x))) > return(x[FALSE][NA]) :the condition has length > 1 and only the first > element will be used > > Here is a subset of my data (tab delimited): > > samp.id corefield TC IC SOC TC.15 comps IC.16 comps > SOC.17 compsTC.15 comps IC.16 comps SOC.17 comps > HOR006_3HOR006 HOR 7.157 0 7.157 8.008273281 > 0.786161341 6.402343153 8.008273281 0.786161341 6.402343153 > HOR006_3HOR006 HOR 7.157 0 7.157 6.258510623 > -1.1175672686.987405984 6.258510623 0 6.987405984 > HOR006_3HOR006 HOR 7.157 0 7.157 14.21306811 > 7.968072165 6.818917226 14.21306811 7.968072165 6.818917226 > HOR006_3HOR006 HOR 7.157 0 7.157 17.73301788 > 9.017994045 9.035508792 17.73301788 9.017994045 9.035508792 > HOR006_3HOR006 HOR 7.157 0 7.157 12.54204929 > 6.285521186 6.052762372 12.54204929 6.285521186 6.052762372 > HOR006_3HOR006 HOR 7.157 0 7.157 10.07603128 > 3.485872902 6.93459 10.07603128 3.485872902 6.93459 > HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 > 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 > HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 > 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 > HOR006_3HOR006 HOR 7.157 0 7.157 9.249550001 > 1.92641169 7.675586354 9.249550001 1.92641169 7.675586354 > HOR006_3HOR006 HOR 7.157 0 7.157 7.414208739 > -0.0205335687.057048733 7.414208739 0 7.057048733 > HOR006_4HOR006 HOR 11.73 0 11.73 14.42894814 > 8.998403641 5.752994239 14.42894814 8.998403641 5.752994239 > HOR006_4HOR006 HOR 11.73 0 11.73 13.65284466 > 6.757373476 6.388413921 13.65284466 6.757373476 6.388413921 > HOR006_4HOR006 HOR 11.73 0 11.73 10.72185703 > 5.053095924 6.016783029 10.72185703 5.053095924 6.016783029 > HOR006_4HOR006 HOR 11.73 0 11.73 14.68382689 > 7.557801473 6.667911142 14.68382689 7.557801473 6.667911142 > HOR006_4HOR006 HOR 11.73 0 11.73 2.287381003 > -3.0741746566.654986023 2.287381003 0 6.654986023 > HOR006_4HOR006 HOR 11.73 0 11.73 14.57145428 > 8.812845515 6.625453309 14.57145428 8.812845515 6.625453309 > HOR006_4HOR006 HOR 11.73 0 11.73 21.12964238 > 13.27394496 6.568626499 21.12964238 13.27394496 6.568626499 > HOR006_4HOR006 HOR 11.73 0 11.73 19.46136803 > 8.03100103 6.910126723 19.46136803 8.03100103 6.910126723 > HOR006_4HOR006 HOR 11.73 0 11.73 13.16591198 > 4.738398449 6.051036242 13.16591198 4.738398449 6.051036242 > HOR006_5HOR006 HOR 20.339 14.383 5.956 24.17001811 > 15.44634892 8.095868636 24.17001811 15.44634892 8.095868636 > HOR006_5HOR006 HOR 20.339 14.383
[R] median of grouped data
I am having difficulty calculating the median of grouped data. I have 8 to 10 repeated measures per sample and I have successfully used the following code to calculate the average for each sample. libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, FUN=ave,libs.norm.preds.median[,1]) I then use the unique function to collapse the data into one line per sample. I would also like to calculate the median, standard error, and coefficient of variation as well. I have not been able to get median to work properly. I have tried this and variants: libs.norm.preds.median[,7:9]<-apply(libs.norm.preds.median[,7:9],MARGIN=2, FUN=median,libs.norm.preds.median[,1]) I receive the following error: Warning messages: 1: In if (na.rm) x <- x[!is.na(x)] else if (any(is.na(x))) return(x[FALSE][NA]) :the condition has length > 1 and only the first element will be used Here is a subset of my data (tab delimited): samp.id corefield TC IC SOC TC.15 comps IC.16 comps SOC.17 compsTC.15 comps IC.16 comps SOC.17 comps HOR006_3HOR006 HOR 7.157 0 7.157 8.008273281 0.786161341 6.402343153 8.008273281 0.786161341 6.402343153 HOR006_3HOR006 HOR 7.157 0 7.157 6.258510623 -1.1175672686.987405984 6.258510623 0 6.987405984 HOR006_3HOR006 HOR 7.157 0 7.157 14.21306811 7.968072165 6.818917226 14.21306811 7.968072165 6.818917226 HOR006_3HOR006 HOR 7.157 0 7.157 17.73301788 9.017994045 9.035508792 17.73301788 9.017994045 9.035508792 HOR006_3HOR006 HOR 7.157 0 7.157 12.54204929 6.285521186 6.052762372 12.54204929 6.285521186 6.052762372 HOR006_3HOR006 HOR 7.157 0 7.157 10.07603128 3.485872902 6.93459 10.07603128 3.485872902 6.93459 HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 HOR006_3HOR006 HOR 7.157 0 7.157 11.02330763 4.963708049 7.03753441 11.02330763 4.963708049 7.03753441 HOR006_3HOR006 HOR 7.157 0 7.157 9.249550001 1.92641169 7.675586354 9.249550001 1.92641169 7.675586354 HOR006_3HOR006 HOR 7.157 0 7.157 7.414208739 -0.0205335687.057048733 7.414208739 0 7.057048733 HOR006_4HOR006 HOR 11.73 0 11.73 14.42894814 8.998403641 5.752994239 14.42894814 8.998403641 5.752994239 HOR006_4HOR006 HOR 11.73 0 11.73 13.65284466 6.757373476 6.388413921 13.65284466 6.757373476 6.388413921 HOR006_4HOR006 HOR 11.73 0 11.73 10.72185703 5.053095924 6.016783029 10.72185703 5.053095924 6.016783029 HOR006_4HOR006 HOR 11.73 0 11.73 14.68382689 7.557801473 6.667911142 14.68382689 7.557801473 6.667911142 HOR006_4HOR006 HOR 11.73 0 11.73 2.287381003 -3.0741746566.654986023 2.287381003 0 6.654986023 HOR006_4HOR006 HOR 11.73 0 11.73 14.57145428 8.812845515 6.625453309 14.57145428 8.812845515 6.625453309 HOR006_4HOR006 HOR 11.73 0 11.73 21.12964238 13.27394496 6.568626499 21.12964238 13.27394496 6.568626499 HOR006_4HOR006 HOR 11.73 0 11.73 19.46136803 8.03100103 6.910126723 19.46136803 8.03100103 6.910126723 HOR006_4HOR006 HOR 11.73 0 11.73 13.16591198 4.738398449 6.051036242 13.16591198 4.738398449 6.051036242 HOR006_5HOR006 HOR 20.339 14.383 5.956 24.17001811 15.44634892 8.095868636 24.17001811 15.44634892 8.095868636 HOR006_5HOR006 HOR 20.339 14.383 5.956 19.17125764 12.28559645 7.468646662 19.17125764 12.28559645 7.468646662 HOR006_5HOR006 HOR 20.339 14.383 5.956 20.18713584 13.12584843 6.985808635 20.18713584 13.12584843 6.985808635 HOR006_5HOR006 HOR 20.339 14.383 5.956 25.58402927 18.23958469 6.960777883 25.58402927 18.23958469 6.960777883 HOR006_5HOR006 HOR 20.339 14.383 5.956 24.04109959 16.32371239 7.12821025 24.04109959 16.32371239 7.12821025 HOR006_5HOR006 HOR 20.339 14.383 5.956 19.809507 12.28987767 7.290354063 19.809507 12.28987767 7.290354063 HOR006_5HOR006 HOR 20.339 14.383 5.956 22.37852335 13.66636406 7.814588276 22.37852335 13.66636406 7.814588276 HOR006_5HOR006 HOR 20.339 14
