similar to: question on aggregate

I have > df quantity branch client date name 1 10 1 1 2010-01-01 one 2 20 2 1 2010-01-01 one 3 30 3 2 2010-01-01 two 4 15 4 1 2010-01-01 one 5 10 5 2 2010-01-01 two 6 20 6 3 2010-01-01 three 7 1000 1 1 2011-01-01 one 8 2000 2 1 2011-01-01

I have a (very big - 1.5 rows) dataframe with a (POSIXt" "POSIXlt") column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. > n.h1 = sqldf("select distinct h, count(*) from x group by h") Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In

I have a data frame that reads client ID date transcations 323232 11/1/2010 22 323232 11/2/2010 0 323232 11/3/2010 missing 121212 11/10/2010 32 121212 11/11/2010 15 ................................. I want to order the rows by client ID and date and using a black-box forecasting method create the data fcst(client,date of forecast, date for which forecast applies). Assume that I

Hi, I am trying to aggregate max a Date type column but have weird result, how do I fix this? > a <- rbind( + data.frame(name='Tom', payday=as.Date('1999-01-01')), + data.frame(name='Tom', payday=as.Date('2000-01-01')), + data.frame(name='Pete', payday=as.Date('1998-01-01')), + data.frame(name='Pete',

Hi, I'm removing non-unique time indices in a zoo time series by means of aggregate. The time series is bivariate, and the row to be kept only depends on the maximum of one of the two columns. Here's an example: x <- zoo(rbind( c(1,1), c(1.1, 0.9), c(1.1, 1.1), c(1,1) ), order.by=c(1,1,2,2)) The eventual aggregated result should be 1 1.1 0.9 2 1.1 1.1 that is, in

I'm having a problem with aggregate.formula when I call it in a function and the function is converted from a string in the funtion I think my problem may also only occur when the left hand side of the formula is cbind(...) Here is example code that generates a dataset and then the error. The first function "agg2" fails > agg2(FALSE) do agg 2 Error in m[[2L]][[2L]] : object

Hello everybody, Data is myd <- data.frame(id1=rep(c("a","b","c"),each=3),id2=rep(1:3,3),val=rnorm(9)) I want to get a cumulative sum over each of id1. trying aggregate does not work myd$pcum <- aggregate(myd[,c("val")],list(orig=myd$id1),cumsum) Please suggest a solution. In real the dataframe is huge so looping with for and subsetting is not a

I have the following time series: > class(CCasadesz2) [1] "zoo" > setmanes <- cut(time(CCasadesz2),breaks="weeks") > CCasadeswz <- aggregate(CCasadesz2,sum,by=setmanes) > class(CCasadeswz) [1] "zoo" > summary(CCasadeswz) Index CCasadeswz 2009-01-12 00:00:00: 1 Min. : 4.0 2009-01-19 00:00:00: 1 1st Qu.:

stack() seems to drop empty levels. Perhaps there could be a drop=FALSE argument if one wanted all the original levels. In the example below, we may wish to retain level "b" in s$ind even though component LL$b has length 0. > LL <- list(a = 1:3, b = list()) > s <- stack(LL) > str(s) 'data.frame': 3 obs. of 2 variables: $ values: int 1 2 3 $ ind : Factor

All: I have a SQlite database where I have stored some verification data by date & time (cycle Z/UTC), lead_time as well as type, duration, etc. I would like to analyze & plot the data as monthly averages. I have looked at a bunch of examples which use some combination of zoo and aggregate, but I have not been able to successfully apply bits and pieces from the examples I have found. Any

Dear R community, I have the following two zoo objects: MONTHLY CPI > plot(z) > par("usr") [1] 1977.76333 2011.15333 70.39856 227.03744 > z=zooreg(cpius$Value,as.yearmon("1979-11"),frequency=12) > str(z) ?zooreg? series from Nov 1979 to Oct 2010 Data: num [1:372] 76.2 77 77.8 78.5 79.5 80.3 81.1 82 82 82.6 ... Index: Class 'yearmon' num [1:372]

One would normally want the original order that so that one can stack a list, operate on the result and then unstack it back with the unstacked result having the same ordering as the original. LL <- list(z = 1:3, a = list()) # since we can't do s <- stack(LL,. drop = FALSE) do this instead: s <- transform(stack(LL), ind = factor(as.character(ind), levels = names(LL))) unstack(s)

Hi, I need to perform calculations on subsets of a data frame: DF = data.frame(read.table(textConnection(" A B C D E F 1 a 1995 0 4 1 2 a 1997 1 1 3 3 b 1995 3 7 0 4 b 1996 1 2 3 5 b 1997 1 2 3 6 b 1998 6 0 0 7 b 1999 3 7 0 8 c 1997 1 2 3 9 c 1998 1 2 3 10 c 1999 6 0 0 11 d 1999 3 7 0 12 e 1995 1 2 3 13 e 1998 1 2 3 14 e 1999 6

Hi Thanks Gabor for your suggestion. I am posting the code that worked for me. dataframe1 = data.frame(cbind(Src = c(1,1,1,2,3), Target1 = c('aaa','bbb','ccc','aaa','ddd'))); #must be data frame dataframe2 = data.frame(cbind(Src = c(2,3,4,4,4), Target2 = c('aaaa','dddd','bbbb','eeee','ffff'))); dataframe3 =

Here's my code: http://pastebin.com/0yRxEVtm The important parts are uncommented and should be easy to find using the link above. For the following line of code, I plan on looking for a way to offset it up 7 rows so that the 15 minute timestamp would be considered the "median" of the subset being averaged to find the mean: avgCool = aggregate(intCool, trunc(time(intCool),

Another newbie question I got the 1 minute spine interpolation and 15 mean aggregation working with many thanks to Gabor Grothendieck using Zoo functions. I got a tip from Hasan Diwan to look at xts but it seemed I would make better progress using code from Gabor. Now I'm having trouble plotting this zoo object. I'm thinking I want a function to "split" the zoo object back to

Dear all, I have following 2 zoo objects. However when I try to merge those 2 objects into one, nothing is coming as intended. Please see below the objects as well as the merged object: > dat11 V2 V3 V4 V5 2010-10-15 13:43:54 73.8 73.8 73.8 73.8 2010-10-15 13:44:15 73.8 73.8 73.8 73.8 2010-10-15 13:45:51 73.8 73.8 73.8 73.8 2010-10-15 13:46:21 73.8 73.8 73.8 73.8

I am trying to aggregate data in column 2 to identifiers in col 1 eg.. take this> identifier quantity 1 10 1 20 2 30 1 15 2 10 3 20 and make this> identifier quantity 1 45 2 40 3 20 Thanks in

>From what I read this should work. So please help my misunderstanding: > x <- data.frame(Name=c("A","A","C"), Category=c("a","a","b"), Quantity=c(1,2,3)) > x Name Category Quantity 1 A a 1 2 A a 2 3 C b 3 > aggregate(x,

Quick newb question about R relating to the line of code below: rawCool = read.zoo("cooling.txt", FUN = as.chron, format = "%m/%d/%Y %H:%M", sep = "\t", aggregate = function(x) tail(x, 1)) I'm wondering what the specifics are for the argument where it has "aggregate = function(x) tail(x, 1)". I understand that it removes the last row of