Displaying 20 results from an estimated 100 matches similar to: "What is cast telling me?"
2009 Jul 10
2
ReShape/cast question - sum of value in table
Hi,
I've tried to capture the basics of this problem I'm having. Been
working on this for a couple of days and just cannot get past it. As a
test of this list software I've attached is a small text file zipped
up. I hope it gets through but if it doesn't I'll post the actual text
which is only 26 lines. Put it somewhere sensible and change the first
line in the code to point
2009 Jul 07
4
Test for X=1 fails, test for >0 works, data in text file is 1
Hi,
I am apparently not understanding some nuance about either the use
of subset or more likely my ability to test for a numerical match
using '='. Which is it? Thanks in advance.
I've read a data file, reshaped it and then created MyResults by
keeping only lines where the value column is greater than 0. So far so
good. The data in MyResults looks good to me by eye.
The
2009 Jul 11
1
sort data.frame by specific date column
I have a data.frame that was built from a number of smaller
data.frames with rbind. Each ssmaller data.frame bound together runs
over the same date ranges. The format of the whole thing looks like
this:
Trade PosType EnDate EnTime ExDate ExTime PL_Pos
1 1 1 1040107 915 1040107 1300 164
2 2 1 1040108 909 1040108 1300 184
3 3 1 1040115
2009 Jul 11
2
Date conversions
Hi all,
I'm having a little bit of trouble with some date conversions and
am hoping someone can help me out. Thanks in advance.
OK, I have two sources of data that provide date info in a csv file
differently. I've attached a small zipped file with two text files
that illustrate both. (Is it ok to send attachments to this list? Not
sure. It's very small.) I need to be able to
2009 Jul 12
2
for (n in SystemResults$EnTime) return EnTime[n] until reaching "(all)"
Hi,
Newbie alert on for loops...
I have a bunch of data.frames built using rbind that have repeated
values in the EnTime column. I want to read the value in the EnTime
column and use it as an input to a function, but only down to the
first occurrence of the string "(all)" where I want to break off and
do other things. From the results I'm getting from some test code I
clearly
2009 Jul 13
0
Reducing arrays for comparison with each other...
Hi,
First up, thanks again for all the help I'm getting on this list.
I'm making great headway in analyzing my experimental data on an
experiment by experiment basis. No way I could have done this in the
time I've done it without your help.
This email is partially a question about R but is also soliciting
overall guidance if anyone wants to give it. I somehow managed to get
an
2009 Jul 09
2
How to Populate List
Hi,
I'm new to R and would like to know, how one can populate the list with array data.
I'm reading a tab separated table in R. The data in the table looks something like this.
#Table Data
Comp A B C
Extracellular 103 268 535759
Nucleus 45603 47783 442744
#R code
myData <- read.table("table.data",
header=T,
2009 Aug 28
2
new data.frame summed by date
Hi,
I wonder if someone can suggest how to create a new data.frame Y
from X where X$PL_Pos is summed by each unique X$MyDate. Y should end
up with two (or more) columns Y$MyDate and Y$PL_Sum with its value
being the cumsum of all the values in X for that date. - a 'daily
cumsum'.
Thanks,
Mark
TStoDate = function (TSDate) {
X = strptime(TSDate + 19e6L, "%Y%m%d")
2005 May 19
2
ARIMA estimation
Good morning,
(sorry for my english)
i have some problems to put off by extimation ARIMA coefficients
the ones not significatives.
Exist a method to extimate only that significatives?
i use the command: arima().
thanks to all
Stefano
[[alternative HTML version deleted]]
2014 Sep 03
2
timings for examples in R CMD check
I'm having a very hard time making R CMD check produce a clean check on
examples because of the timings inserted into examples by R CMD check.
I am getting a difference on every example output caused by timing
information being inserted by 'R CMD check'.
The current 'Writing R Extensions' manual[1] states on p. 14:
If directory tests has a subdirectory Examples
2023 Jul 02
1
Strange error in R CMD check --timings
This SO post: https://stackoverflow.com/q/76583828 describes a strange R
CMD check error. Depending on the contents of a comment in one of the
examples sections of a help page, an error like this could be triggered:
> base::assign(".dptime", (proc.time() - get(".ptime", pos =
"CheckExEnv")), pos = "CheckExEnv")
>
2010 Jun 09
1
equivalent of stata command in R
Dear all,
I need to use R for one estimation, and i have readily available stata command, but i need also the R version of the same command.
the estimation in stata is as following:
1. Compute mean values of relevant variables
. sum inno lnE lnM
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
2001 Sep 08
1
t.test (PR#1086)
Full_Name: Menelaos Stavrinides
Version: 1.3. 1
OS: Windows 98
Submission from: (NULL) (193.129.76.90)
When model simplification is used in glm (binomial errors) and anova is used two
compare two competitive models one can use either an "F" or a "Chi" test.
R always performs an F test (Although when test="Chi" the test is labeled as
Chi, there isn't any
2023 Jul 02
1
Strange error in R CMD check --timings
Am 02.07.23 um 18:01 schrieb Duncan Murdoch:
> This SO post: https://stackoverflow.com/q/76583828 describes a strange R
> CMD check error. Depending on the contents of a comment in one of the
> examples sections of a help page, an error like this could be triggered:
>
> > base::assign(".dptime", (proc.time() - get(".ptime", pos =
>
2009 Apr 08
3
MLE for bimodal distribution
Hello everyone,
I'm trying to use mle from package stats4 to fit a bi/multi-modal
distribution to some data, but I have some problems with it.
Here's what I'm doing (for a bimodal distribution):
# Build some fake binormally distributed data, the procedure fails also with
real data, so the problem isn't here
data = c(rnorm(1000, 3, 0.5), rnorm(500, 5, 0.3))
# Just to check
2010 Jun 09
1
equivalent of stata command in R
From: saint-filth@hotmail.com
To: saint-filth@hotmail.com
Subject: RE:
Date: Wed, 9 Jun 2010 09:53:20 +0000
OK! sorry thats my fault,
here the translations of the stata commands
1st step is to get the mean values of the variables, well that doesnt need explanation i guess,
2nd step is to estimate the model on panel data estimation method
which is:
2009 Jul 21
4
list of lm() results
How can I get the results of lm() into a list so I can loop through the results?
e.g.
myResults[1] <- lm(...)
myResults[2] <- lm(...)
myResults[3] <- lm(...)
...
myResults[15] <- lm(...)
myResults[16] <- lm(...)
so far every attempt I've tried doesn't work throwing a "number of
items to replace is not a multiple of replacement length" error or
simply not
2012 Jul 13
1
functions of vectors : loop or vectorization
I have a read a lot about the benefits of vectorization in R. I have a
program that takes "almost forever" to run. A good way to see if I have
learned something ... My problem can be summarized like this : I have a
nonlinear function of several variables that I want to optimize over one
letting the other describe a family of curves. In short, I wan't to optimize
f(x,a,b) for several
2003 Sep 03
2
problem with HoltWinters
Dear helpers
I'm having a problem with function HoltWinters from package ts. I have a time series that I want to fit an Holt-Winters model and make predictions for the next values. I've already built an object of class ts to serve as input to HoltWinters. But then I get an error; I've used HoltWinters a lot of times and this never hapened
> data.HW<-HoltWinters(data.ts)
Error
2009 Mar 22
2
Following progress in a lapply() function
Dear all,
I am processing a very long and complicated list using lapply through a custom function and I would like to generate some sort of progress report. For instance, print a dot on the screen every time 1000 item have been process. Or even better, reporting the percent of the list that have been process every 10%. However, I can't seem to figure out a way to achieve that.
For instance,