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Displaying 20 results from an estimated 1000 matches similar to: "Can I call MySql statements directly??"

2017 Nov 09
1
weighted average grouped by variables
Hello, Using base R only, the following seems to do what you want. with(mydf, ave(speed, date_time, type, FUN = weighted.mean, w = n_vehicles)) Hope this helps, Rui Barradas Em 09-11-2017 13:16, Massimo Bressan escreveu: > Hello > > an update about my question: I worked out the following solution (with the package "dplyr") > > library(dplyr) > > mydf%>% >
2017 Nov 09
0
weighted average grouped by variables
Hello an update about my question: I worked out the following solution (with the package "dplyr") library(dplyr) mydf%>% mutate(speed_vehicles=n_vehicles*mydf$speed) %>% group_by(date_time,type) %>% summarise( sum_n_times_speed=sum(speed_vehicles), n_vehicles=sum(n_vehicles), vel=sum(speed_vehicles)/sum(n_vehicles) ) In fact I was hoping to manage everything in a
2017 Nov 09
4
weighted average grouped by variables
hi all I have this dataframe (created as a reproducible example) mydf<-structure(list(date_time = structure(c(1508238000, 1508238000, 1508238000, 1508238000, 1508238000, 1508238000, 1508238000), class = c("POSIXct", "POSIXt"), tzone = ""), direction = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("A", "B"), class =
2006 Feb 08
1
Possible AGI Bug in Asterisk?
Dear All, I seem to have stumbled across an AGI problem; I have written an AGI Script (bottom of this email); The script does the following; Makes a CDR entry when called Records the call Updates the CDR Finds a corresponding DNIS from the SMDR table (captured via a serial port logger) Matches up the record and updates the CDR. The script works perfectly in my test lab and has been doing so
2009 Oct 06
1
ggplot2 applying a function based on facet
Look at the bottom of the message for my question #here is a little function that I wrote USGS <- function(input="discharge", days=7){ library(chron) library(gsubfn) #021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS) #02102908 is the Flat Creek Gauge (ftbrfcms) #02133500 is the Drowning Creek (ftbrbmcm) #02341800 is the Upatoi Creek Near Columbus (ftbn) #02342500 is
2010 Oct 27
1
Fill in missing times in a timeseries with NA
Hi, I have a irregularly spaced time series dataset, which reads in from a .csv. I need to convert this to a regularly spaced time series by filling in missing rows of data with NAs. So my data, called NtuMot, looks like this (I've removed some of the additional rows for simplicity).... ELEID date_time height slope 1 2009-06-24 00:00:00
2010 Mar 17
2
How can I return rows from a data frame with maximum value by factor?
Hi, I'm new to R and new to this forum. I'm struggling with trying to extract certain rows of data from my data.frame. The data.frame has eleven columns. Among those columns are "FISH_ID" and "DATE_TIME". FISH_ID is a factor. For each of my 21 unique FISH_IDs (levels) I have a few to a few thousand rows, each row with a unique DATE_TIME value. I would like to obtain,
2017 Nov 09
2
weighted average grouped by variables
Hi Thanks for working example. you could use split/ lapply approach, however it is probably not much better than dplyr method. sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) gives you averages aggregate(mydf$n_vehicles, list(mydf$type), sum)$x gives you sums Cheers Petr > -----Original Message----- > From: R-help
2017 Nov 11
0
weighted average grouped by variables
> On 9 Nov 2017, at 14:58, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > Thanks for working example. > > you could use split/ lapply approach, however it is probably not much better than dplyr method. > > sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) > gives you averages > The
2010 May 18
2
Function that is giving me a headache- any help appreciated (automatic read )
note: whole function is below- I am sure I am doing something silly. when I use it like USGS(input="precipitation") it is choking on the precip.1 <- subset(DF, precipitation!="NA") b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum) DF.precip <- precip.1 DF.precip$precipitation <- b$.data part, but runs fine outside of the function: days=7
2017 Nov 09
1
weighted average grouped by variables
Dear Massimo, It seems straightforward to use weighted.mean() in a dplyr context library(dplyr) mydf %>% group_by(date_time, type) %>% summarise(vel = weighted.mean(speed, n_vehicles)) Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team
2012 Nov 12
5
Matrix to data frame conversion
I have a matrix which I wanted to convert to a data frame. As I could not succeed and resorted to export to csv and reimport it again. Why did I fail in the attempt and how can I achieve what I wanted without this roundabouts? The original matrix: > str(comb_model0) num [1:90, 1:4] 3.5938 0.0274 0.0342 0.0135 0.0207 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:90]
2010 Feb 26
7
How to add a variable to a dataframe whose values are conditional upon the values of an existing variable
Hi everyone, I am at my wits end with what I believe would be considered simple by a more experienced R user. I want to know how to add a variable to a dataframe whose values?are conditional?on the values of an existing?variable.?I can't seem to make an ifelse statement work?for my situation.?The existing variable?in my dataframe is?a character variable named DOW which contains abbreviated
2009 Oct 06
2
ggplot cumsum refined question (?)
OK, so maybe last night was a little too much at one throw, so I have reduced the data to two stations- one that has precipitation and one that does not. This is going to be in the context of a larger data set. I would like to be able to issue a ggplot command and have cum sum just act on the facets (factors) to apply this. library(chron) library(ggplot2) DF <- structure(list(date_time =
2010 Mar 05
2
How to assign week numbers to a time-series
Hello everyone, My progress has stalled on finding a way of creating a somewhat complicated variable to add to my existing dataframe and I am hoping one of you could help me out. The dataframe below contains only a fraction of the data of my complete dataframe, but all of the variables. What I want to do is add another variable named 'WEEK' to this dataframe that is assigned 1 for row 1
2007 Mar 11
2
DST changes for the US
After palying around to set this correctly (or so I think, please correct me if I got something wrong) I decided to share it with everyone. For Polycom phones in sip.cfg change the following line from: <SNTP tcpIpApp.sntp.resyncPeriod="86400" tcpIpApp.sntp.address="" tcpIpApp.sntp.gmtOffset="" tcpIpApp.sntp.daylightSavings.enable="1"
2005 Jan 25
2
Having problems running palm desktop or the uninstall program
I have a user John Dow on a local machine johnsmachine. I have a samba user jdow. When I moved the profile john dow to the domain keydomain, I can't get the install program to run. This is how it looks. I have local profile johnsmachine/John Dow and roaming profile keydomain/jdow. When logged into profile johnsmachine/John Dow, I can install and uninstall the palm software with no
2012 Sep 03
2
[LLVMdev] [NVPTX] Backend cannot handle array-of-arrays constant
Dear all, Looks like the NVPTX backend cannot handle array-of-arrays contant (please see the reporocase below). Is it supposed to work? Any ideas how to get it working? Important for our target applications. Thanks, - Dima. $ cat test.ll ; ModuleID = '__kernelgen_main_module' target datalayout =
2012 Sep 04
2
[LLVMdev] [NVPTX] Backend cannot handle array-of-arrays constant
I think our test case demonstrates that requiring the array item being initialized to be constant is incorrect. NVPTX does not crash anymore and produces correct result with the following change: --- NVPTXAsmPrinter.cpp 2012-09-03 15:14:00.000000000 +0200 +++ NVPTXAsmPrinter.cpp 2012-09-04 15:47:17.859398193 +0200 @@ -1890,17 +1890,15 @@ case Type::ArrayTyID: case Type::VectorTyID: case
2006 Jun 15
1
Array
Dear all R users, I am wondering if there is any way to define a 3 dimentional or 4 dimentional array in R. Sincerely yours, thanks in advance Send instant messages to your online friends http://in.messenger.yahoo.com Stay connected with your friends even when away from PC. Link: http://in.mobile.yahoo.com/new/messenger/ [[alternative HTML version deleted]]