Hello
thanks to my friend Diego, a computer agronomist,
I was able to figure out how to do it.
I still have only one problem: how do I calculate the data for the
first week? The traps were set up on the fields on May 5th. On that
date, a 0 could be entered for all time-series. But how to do it with
sorry, i'm a bit of a bummer at copying and saving various pieces of
code
and I'm not very experienced with R
- I am trying to build a partnership with a local IT cooperative, but
this is a job that was not foreseen and it is not covered by budget, so
I am trying to solve it myself -
I do not want
I have *not* followed this in any detail, but this line seems wrong:
arvaia_catture_order <- arvaia_catture[order(arvaia_catture$tempo)]
Perhaps it should be:
arvaia_catture_order <- arvaia_catture[order(arvaia_catture$tempo), ] ##
note the comma!
If I am mistaken, just ignore and move on.
Chee
Hello,
I'm not getting your simple sum:
arvaia_catture_order[, week := as.integer(format(tempo, "%U"))]
aggregate(catture ~ week, arvaia_catture_order, sum)
# week catture
#1 19 15
#2 20 5
#3 21 78
#4 22 120
Can you explain your result better?
Hope this helps,
R
Hello
I just registered on the list.
I am an agricultural technician and I am collaborating on a research
project on agroforestry and Brown Marmorated Stink Bug (Halyomorpha
halys, abbreviated BMSB).
Through kobotoolbox we are collecting data of catches in traps on
farms. Farms register inconsiste
ay, December 14, 2017 8:48 AM
> To: r-help@r-project.org
> Cc: Michael Haenlein
> Subject: [R] Aggregation across two variables in data.table
>
> Dear all,
>
> I have a data.frame that includes a series of demographic variables for a
> set of respondents plus a dependent va
Dear all,
I have a data.frame that includes a series of demographic variables for a
set of respondents plus a dependent variable (Theta). For example:
AgeEducation Marital Familysize
IncomeHousingTheta
1: 50 Ass
Hi,
May be this helps:
DataA <- read.table(text="ID,Var1,Var2
1,A,100
1,B,50
2,A,200
2,B,100
2,B,50",sep=",",header=TRUE,stringsAsFactors=FALSE)
dcast(DataA,ID~Var1,value.var="Var2") ## I guess this is what you mentioned
#Aggregation function missing: defaulting to length
# ID A B
#1 1 1 1
#2
Thanks:)
Regards, Farnoosh Sheikhi
On Thursday, February 13, 2014 1:29 PM, arun wrote:
Sorry, the library should be
library(reshape2)
On Thursday, February 13, 2014 4:27 PM, arun wrote:
HI Farnoosh,
You can use ?dcast()
library(plyr)
dcast(DataA,ID~Var1,value.var="Var2")
# ID
Sorry, the library should be
library(reshape2)
On Thursday, February 13, 2014 4:27 PM, arun wrote:
HI Farnoosh,
You can use ?dcast()
library(plyr)
dcast(DataA,ID~Var1,value.var="Var2")
# ID A B
#1 1 100 50
#2 2 200 100
A.K.
On Thursday, February 13, 2014 2:59 PM, farnoosh she
HI Farnoosh,
You can use ?dcast()
library(plyr)
dcast(DataA,ID~Var1,value.var="Var2")
# ID A B
#1 1 100 50
#2 2 200 100
A.K.
On Thursday, February 13, 2014 2:59 PM, farnoosh sheikhi
wrote:
Hi Arun,
I hope all is well. I need to aggregate a data like below:
DataA
ID Var1
Hello,
With the following, the first instruction will give you correlations
matrices, the second coefficients.
dat <- read.table(text = "
x y group
1 0.876751503 0.6518345 a
2 0.627067150 0.8801790 a
3 0.632465192 0.1768305 a
4 0.060359554 0.8835652 a
5 0.67586
d.cars...@kcl.ac.uk
> Sent: Fri, 8 Feb 2013 08:33:45 -0800 (PST)
> To: r-help@r-project.org
> Subject: [R] aggregation-type question
>
> I seem to have a Friday afternoon block and can't see the easiest way of
> doing this.
>
> Given a data frame like:
>
> dat &l
I seem to have a Friday afternoon block and can't see the easiest way of
doing this.
Given a data frame like:
dat <- data.frame(x = runif(100), y = runif(100), group = rep(letters[1:10],
each = 10))
> head(dat)
x y group
1 0.876751503 0.6518345 a
2 0.627067150 0.8801790
Dr Eberhard Lisse wrote:
> Hi,
>
> I am reading payment data like so
>
> 2010-01-01,100.00
> 2010-01-04,100.00
> ...
> 2011-01-01,200.00
> 2011-01-07,100.00
>
> and plot it aggregated per month like so
>
> library(zoo)
> df <- read.csv("daily.csv", colClasses=c(d="Date",s="numeric"))
> z <- zoo(d
On Thu, Jul 7, 2011 at 5:41 AM, Dr Eberhard Lisse wrote:
> Hi,
>
> I am reading payment data like so
>
> 2010-01-01,100.00
> 2010-01-04,100.00
> ...
> 2011-01-01,200.00
> 2011-01-07,100.00
>
> and plot it aggregated per month like so
>
> library(zoo)
> df <- read.csv("daily.csv", colClasses=c(d="D
Hi,
I am reading payment data like so
2010-01-01,100.00
2010-01-04,100.00
...
2011-01-01,200.00
2011-01-07,100.00
and plot it aggregated per month like so
library(zoo)
df <- read.csv("daily.csv", colClasses=c(d="Date",s="numeric"))
z <- zoo(df$s, df$d)
z.mo <- aggregate(z, as.yearmon, sum)
barp
Hi,
I am reading payment data like so
2010-01-01,100.00
2010-01-04,100.00
...
2011-01-01,200.00
2011-01-07,100.00
and plot it aggregated per month like so
library(zoo)
df <- read.csv("daily.csv", colClasses=c(d="Date",s="numeric"))
z <- zoo(df$s, df$d)
z.mo <- aggregate(z, as.yearmon, sum)
barp
Dear R-ers,
My aggregation saga continues.
Using the following sequence, I can calculate any statistic for row
groups and merge the result back to all associated rows ...
> WM = by( D60, D60[ "KeyProfA"], FUN=function(x) weighted.mean( x$IAC, x$Wt))
> D60$IAC.WM = as.numeric( WM[ D60$KeyProf
On Dec 25, 2007 4:41 AM, Pfeiffer & Koberstein Immobilien GmbH - Ralf
Pfeiffer <[EMAIL PROTECTED]> wrote:
> thank you, I downloaded and installed the package. With library(help="doBy")
> I got the information about this package.
>
> But when I use the syntax
>
> summaryBy(daten[,c(3:4)], list(A,B)
ryBy)
I tried before
require(doBy), but this doesn't work neither.
What can I do?
- Original Message -
From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
To: "Agrarimmobilien" <[EMAIL PROTECTED]>
Cc:
Sent: Monday, December 24, 2007 10:48 PM
Subject: Re:
See summaryBy in the doBy package.
On Dec 24, 2007 4:27 PM, Agrarimmobilien
<[EMAIL PROTECTED]> wrote:
> Hello,
>
> using the syntax
>
> aggregate(daten[,c(3,4)], list(A,B), mean)
>
> I'm getting the following data.frame:
>
>
>A BCD
> 1 351 6.16000
Hello,
using the syntax
aggregate(daten[,c(3,4)], list(A,B), mean)
I'm getting the following data.frame:
A BCD
1 351 6.16000 5
2 47131.24333 20
3 54126.81773 2
4 3 212.990
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