Re: [R] [FORGED] Re: rmgarch: source package installation problem
Dear All
I could solve the problem: I removed my .Rprofile and I could install
the package via the usual
install.packages("rmgarch")
My .Rprofile contains few settings, therefore, it was not too difficult
to understand what was conflicting with the installation. The issue in
.Rprofile was the following
q <- function (save="no", ...) {
quit(save=save, ...)
}
Regards
Pietro Coretto
On 02/01/2020 23.54, Rolf Turner wrote:
On 2/01/20 9:51 pm, Pietro Coretto wrote:
No problem Rolf. Thanks for you interest. But the problem is still
unsolved!
I experimented and found that I too could not install rmgarch. However
the string of error messages that I got was quite different from yours.
I did some scrounging around and after a bit of trial and error found
that I needed to do:
sudo apt-get install libgmp3-dev
sudo apt-get install libmpfr-dev
Then the R command
install.packages("rmgarch",lib="~/Rlib")
worked. (Note that "~/Rlib" is where I keep my "contributed" packages.)
This is under Ubuntu 18.04. I don't know if this will work for you
since, as I said, the error messages that I initially got were different
from those that you got.
*Don't* ask me about what to do under Mac OSX or (God save us!) under
Windoze!!! :-)
cheers,
Rolf
P.S. I have taken the liberty of CC-ing this to the r-help list in case
it is of interest to others or in case others may have useful
contributions to add.
R.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] rmgarch: source package installation problem
On 30/12/2019 12.11, Duncan Murdoch wrote:
On 30/12/2019 5:59 a.m., Pietro Coretto wrote:
[...]
You didn't show us the command you used to install it.
Duncan Murdoch
Sorry for this,
I used the following:
install.packages("rmgarch")
from the linux command line, using both a regular user account and root.
Same issue anyway
Thanks
Pietro
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] rmgarch: source package installation problem
Dear Users I tried to install the package "rmgarch". While installing binaries for Windows and MacOS didn't cause any issue. The source package installation doesn't work. I tried it on three different OS: Linux Ubuntu 18.04 (64bit), Windows 10 Pro (64bit) and MacOS Catalina. In all three cases I have worked with a fresh R 3.6.2 install. I contacted the maintainer of the package who said that it can't reproduce the same issue and that CRAN checks are fine. Below I report the console output when installing "rmgarch" on my Ubuntu 18.04 LTS (64bit). I skip the messages regarding installation of the required packages (they all installed without issues): https://pastebin.com/J6XHaP28 I have installed many other packages from source (unsing C, C++, RCpp, fortran code, etc) without any issue. Here the output of sessionInfo() at the end. https://pastebin.com/z7ZwU9iR I hope somebody can tell me what may cause this. Regards Pietro __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] No transactions/positions to chart in Quantstrat
Hello.
I get the error message that there are no transactions/positions to chart
despite the signals and rules that I inserted.
Can someone please help?
rm(list = ls(.blotter), envir = .blotter)
initdate <- "2010-01-01"
from <- "2012-01-01" #start of backtest
to <- "2017-31-12" #end of backtest
Sys.setenv(TZ= "EST") #Set up environment for timestamps
currency("USD") #Set up environment for currency to be used
symbols <- c("RUT") #symbols used in our backtest
getSymbols("^RUT",src="yahoo", from="2012-01-01", to="2017-12-31",
periodicity="daily")
stock(symbols, currency = "USD", multiplier = 1) #tells quanstrat what
instruments present and what currency to use
n <- 30
wma <- WMA(Cl(RUT), n=4, wts=c(1:4))
wmamaxt <- rollmaxr(wma, n, fill = NA)
wmamint <- - rollmaxr(- wma, n, fill = NA)
CNOwma <- function (RUT) {(wma - wmamint) / (wmamaxt - wmamint)}
tradesize <-1 #default trade size
initeq <- 10 #default initial equity in our portfolio
strategy.st <- portfolio.st <- account.st <- "firststrat" #naming strategy,
portfolio and account
#removes old portfolio and strategy from environment
rm.strat(portfolio.st)
rm.strat(strategy.st)
#initialize portfolio, account, orders and strategy objects
initPortf(portfolio.st, symbols = symbols, initDate = initdate, currency =
"USD")
initAcct(account.st, portfolios = portfolio.st, initDate = initdate, currency =
"USD", initEq = initeq)
initOrders(portfolio.st, initDate = initdate)
strategy(strategy.st, store=TRUE)
add.indicator(strategy = strategy.st,
name = 'CNOwma',
arguments = list(x = quote(Cl(mktdata)), n=4),
label = 'CNOwma4')
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "gt", cross = TRUE),
label = "longthreshold")
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "lt", cross = TRUE),
label = "shortthreshold")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"), osFUN = IKTrading::osMaxDollar,
tradeSize = tradesize, maxSize = tradesize, type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"),
type = "exit")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),osFUN = IKTrading::osMaxDollar,
tradeSize = tradesize, maxSize = tradesize, type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),
type = "exit")
out <- applyStrategy(strategy = strategy.st, portfolios = portfolio.st)
updatePortf(portfolio.st)
daterange <- time(getPortfolio(portfolio.st)$summary)[-1]
updateAcct(account.st, daterange)
updateEndEq(account.st)
for(symbol in symbols){
chart.Posn(Portfolio = portfolio.st, Symbol = symbol,
TA= c("add_SMA(n=50, col='blue')", "add_SMA(n=200, col='red')"))
}
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] Quantstrat custom indicator colnames error
I apologize if the data I will insert will not be enough.
So, I am trying to run a strategy through the package Quantstrat.
install.packages("quantstrat")
My problem is that I get the following error
Error incolnames<-(tmp, value = seq(ncol(tmp_val))) :
attempt to set 'colnames' on an object with less than two dimensions
when I try to run the following command:
> out <- applyStrategy(strategy=strategy.st,portfolios=portfolio.st)
I do not have this problem if I use, as indicator, one or more indicators,
which are already defined by the package TTR.
I have this error only when I try to use a custom indicator. Here is the code
for the custom indicator that I use:
wma <- WMA(Cl(mktdata), 4, wts=c(1:4))
wmamaxt <- rollmaxr(wma, 30, fill = NA)
wmamint <- - rollmaxr(- wma, 30, fill = NA)
CNOwma <- function (mktdata=quote(mktdata),x) {(wma - wmamint) / (wmamaxt -
wmamint)}
Please refer to the following code:
library(devtools)
library(quantmod)
library(quantstrat)
library(TTR)
library(png)
library(IKTrading)
wma <- WMA(Cl(mktdata), 4, wts=c(1:4))
wmamaxt <- rollmaxr(wma, 30, fill = NA)
wmamint <- - rollmaxr(- wma, 30, fill = NA)
CNOwma <- function (mktdata=quote(mktdata),x) {(wma - wmamint) / (wmamaxt -
wmamint)}
initdate <- "2010-01-01"
from <- "2012-01-01" #start of backtest
to <- "2017-31-12" #end of backtest
Sys.setenv(TZ= "EST") #Set up environment for timestamps
currency("USD") #Set up environment for currency to be used
symbols <- c("RUT", "IXIC") #symbols used in our backtest
getSymbols(Symbols = symbols, src = "google", from=from, to=to, adjust = TRUE)
#receive data from google finance, adjusted for splits/dividends
stock(symbols, currency = "USD", multiplier = 1) #tells quanstrat what
instruments present and what currency to use
tradesize <-1 #default trade size
initeq <- 10 #default initial equity in our portfolio
strategy.st <- portfolio.st <- account.st <- "firststrat" #naming strategy,
portfolio and account
#removes old portfolio and strategy from environment
rm.strat(portfolio.st)
rm.strat(strategy.st)
#initialize portfolio, account, orders and strategy objects
initPortf(portfolio.st, symbols = symbols, initDate = initdate, currency =
"USD")
initAcct(account.st, portfolios = portfolio.st, initDate = initdate, currency =
"USD", initEq = initeq)
initOrders(portfolio.st, initDate = initdate)
strategy(strategy.st, store=TRUE)
add.indicator(strategy = strategy.st,
name = 'CNOwma',
arguments = list(x = quote(Cl(mktdata)), n=4),
label = 'CNOwma4')
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "gt", cross = TRUE),
label = "longthreshold")
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "lt", cross = TRUE),
label = "shortthreshold")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"),
type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"),
type = "exit")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),
type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),
type = "exit")
out <- applyStrategy(strategy = strategy.st, portfolios = portfolio.st)
When I run the traceback() of the error, this is what I get:
> traceback()
4: stop("attempt to set 'colnames' on an object with less than two dimensions")
3: `colnames<-`(`*tmp*`, value = seq(ncol(tmp_val)))
2: applyIndicators(strategy = strategy, mktdata = mktdata, parameters =
parameters,
...)
1: applyStrategy(strategy = strategy.st, portfolios = portfolio.st
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] Error custom indicator Quantstrat colnames
I will try to be as clear as possible as I have been rebuked by some users. I
deleted the last questions and I will try to be sufficiently explicative in
this one. I apologize if the data I will insert will not be enough.
So, I am trying to run a strategy through the package Quantstrat.
install.packages("quantstrat")
My problem is that I get the following error
Error incolnames<-(tmp, value = seq(ncol(tmp_val))) :
attempt to set 'colnames' on an object with less than two dimensions
when I try to run the following command:
> out <- applyStrategy(strategy=strategy.st,portfolios=portfolio.st)
I do not have this problem if I use, as indicator, one or more indicators,
which are already defined by the package TTR.
I have this error only when I try to use a custom indicator. Here is the code
for the custom indicator that I use:
wma <- WMA(Cl(mktdata), 4, wts=c(1:4))
wmamaxt <- rollmaxr(wma, 30, fill = NA)
wmamint <- - rollmaxr(- wma, 30, fill = NA)
CNOwma <- function (mktdata=quote(mktdata),x) {(wma - wmamint) / (wmamaxt -
wmamint)}
Please refer to the following code:
library(devtools)
library(quantmod)
library(quantstrat)
library(TTR)
library(png)
library(IKTrading)
wma <- WMA(Cl(mktdata), 4, wts=c(1:4))
wmamaxt <- rollmaxr(wma, 30, fill = NA)
wmamint <- - rollmaxr(- wma, 30, fill = NA)
CNOwma <- function (mktdata=quote(mktdata),x) {(wma - wmamint) / (wmamaxt -
wmamint)}
initdate <- "2010-01-01"
from <- "2012-01-01" #start of backtest
to <- "2017-31-12" #end of backtest
Sys.setenv(TZ= "EST") #Set up environment for timestamps
currency("USD") #Set up environment for currency to be used
symbols <- c("RUT", "IXIC") #symbols used in our backtest
getSymbols(Symbols = symbols, src = "google", from=from, to=to, adjust = TRUE)
#receive data from google finance, adjusted for splits/dividends
stock(symbols, currency = "USD", multiplier = 1) #tells quanstrat what
instruments present and what currency to use
tradesize <-1 #default trade size
initeq <- 10 #default initial equity in our portfolio
strategy.st <- portfolio.st <- account.st <- "firststrat" #naming strategy,
portfolio and account
#removes old portfolio and strategy from environment
rm.strat(portfolio.st)
rm.strat(strategy.st)
#initialize portfolio, account, orders and strategy objects
initPortf(portfolio.st, symbols = symbols, initDate = initdate, currency =
"USD")
initAcct(account.st, portfolios = portfolio.st, initDate = initdate, currency =
"USD", initEq = initeq)
initOrders(portfolio.st, initDate = initdate)
strategy(strategy.st, store=TRUE)
add.indicator(strategy = strategy.st,
name = 'CNOwma',
arguments = list(x = quote(Cl(mktdata)), n=4),
label = 'CNOwma4')
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "gt", cross = TRUE),
label = "longthreshold")
add.signal(strategy.st, name = "sigThreshold",
arguments = list(column = "CNOwma4", threshold = 0.6,
relationship = "lt", cross = TRUE),
label = "shortthreshold")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"),
type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "long", replace = FALSE,
prefer = "Open"),
type = "exit")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "shortthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),
type = "enter")
add.rule(strategy.st, name = "ruleSignal",
arguments = list(sigcol = "longthreshold", sigval = TRUE,
orderqty = "all", ordertype = "market",
orderside = "short", replace = FALSE,
prefer = "Open"),
type = "exit")
out <- applyStrategy(strategy = strategy.st, portfolios = portfolio.st)
When I run the traceback() of the error, this is what I get:
> traceback()
4: stop("attempt to set 'colnames' on an object with less than two dimensions")
3: `colnames<-`(`*tmp*`, value = seq(ncol(tmp_val)))
2: applyIndicators(strategy = strategy, mktdata = mktdata, parameters =
parameters,
...)
1: applyStrategy(strategy = strategy.st, portfolios = portfolio.st)
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] SARIMA in rpy2
Hello, Am trying to estimate a seasonal Arima by calling the R forecast package in Rpy2: fit = forecast.Arima(x = h02, order = order, seasonal = seasonal) Strangely I get the estimates of the non-seasonal part, only, even if the model is specified as (3,0,1)x(0,1,2). Can anyone tell me where the issue is? Regards, P. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] help with xts
Thanks to all for your ideas and inspirations Il 18/05/2014 18.35, Joshua Ulrich ha scritto: Using subset assignment with an array usually doesn't work well with xts/zoo objects. Your case wouldn't even work with a matrix because you have NA in your array. In this case, you can achieve the same result using multiplication: pos.neg.1 - ((cond1 cond2)*-2+1) test - test * pos.neg.1 Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.com On Sun, May 18, 2014 at 9:25 AM, Pete freeri...@gmail.com wrote: I have 3 xts objects: test, cond1, cond2 You can download here: https://dl.dropboxusercontent.com/u/102669/obj.rar My problem is very simple. test [ cond1 cond2] = NA THIS WORKS test [ cond1 cond2] = -test [ cond1 cond2] THIS DOESN'T WORKS Why? My objective is to substitute all values in test (when cond1 cond2) with the corresponding values of test but with negative sign __ 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] Help with multiple use of quantile
Hello I have a matrix aaa like this: aa1 aa2 aa3 [1,] 8371.417 27613.57 1170.466 [2,] 14317.999 42421.82 3423.934 [3,] 22026.466 59874.14 8103.084 [4,] 33884.986 84506.34 19176.764 [5,] 57954.968 129824.30 56097.310 where each column represents an empirical distribution of random variates (normally the number of rows would be, say, 10,000 or 100,000) and a matrix bbb of percentiles like this: bbb [,1] [,2] [,3] [1,] 0.70092980 0.8144194 0.6200732 [2,] 0.77968803 0.5804948 0.5463661 [3,] 0.01509415 0.9313509 0.8611973 [4,] 0.22654757 0.6183386 0.4962867 [5,] 0.36548835 0.6608696 0.3062784 What I'd like to do is to apply the quantiles in the three columns of bbb to the columns of aaa independently, so as to obtain a matrix ccc such that, for example, ccc[3,1]=quantile(aaa[,1],bbb[3,1]). THe complete matrix is: ccc =quantile(aaa[,1],bbb[1,1]), quantile(aaa[,2],bbb[1,2]), quantile(aaa[,3],bbb[1,3]) quantile(aaa[,1],bbb[2,1]), quantile(aaa[,2],bbb[2,2]), quantile(aaa[,3],bbb[2,3]) quantile(aaa[,1],bbb[3,1]), quantile(aaa[,2],bbb[3,2]), quantile(aaa[,3],bbb[3,3]) quantile(aaa[,1],bbb[4,1]), quantile(aaa[,2],bbb[4,2]), quantile(aaa[,3],bbb[4,3]) quantile(aaa[,1],bbb[5,1]), quantile(aaa[,2],bbb[5,2]), quantile(aaa[,3],bbb[5,3]) Now if I only two vectors, it would be enough for me to define ccc=quantile(aaa,bbb). However, if I do this when aaa and bbb are matrix, quantile(aaa,bbb) does something else -- it gives the percentiles bbb of the whole set of matrix elements rather than considering separately the quantiles of the different columns. I suspect I'll need to use apply in some form but have been able to come up with the correct form. As you might have imagined, this arises in the context of simulating random variables from different distribution in empirical form that are correlated through a copula. Thanks in advance for your help! Pietro _ __ For information pertaining to Willis' email confidentiality and monitoring policy, usage restrictions, or for specific company registration and regulatory status information, please visit http://www.willis.com/email_trailer.aspx We are now able to offer our clients an encrypted email capability for secure communication purposes. If you wish to take advantage of this service or learn more about it, please let me know or contact your Client Advocate for full details. ~W67897 __ [[alternative HTML version deleted]] __ 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] data.frame with NA
Thank you David and thank you Petr At 14.18 19/03/2013, David L Carlson wrote: Try this instead: Foglio1[,2:ncol(Foglio1)] - na.locf(Foglio1[,2:ncol(Foglio1)],fromLast=T) str(Foglio1) 'data.frame': 1489 obs. of 15 variables: $ Date: Date, format: 2001-08-17 2001-08-20 ... $ a : num 202 201 202 201 202 ... $ b : num 231 230 230 230 232 ... $ c : num 177 179 181 180 182 ... $ d : num 277 277 276 276 275 ... $ e : num 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 ... $ f : num 275 277 279 279 279 ... $ g : num 91.7 90.7 90.8 91.1 91 ... $ h : num 11446 11258 11280 11396 11127 ... $ i : num 388 389 393 392 393 ... $ l : num 93.2 94 92.4 93.4 93.1 ... $ m : num 128 127 126 129 130 ... $ n : num 103 103 103 103 103 ... $ o : num 133 133 133 133 133 ... $ p : num 107 107 107 107 107 ... It appears that na.locf() converts the object to a matrix at some point (but I haven't checked the source code). The first column (the Date variable) is treated as character. As a result, everything gets converted to character. This will skip the first column which does not have any missing values. -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: Pietro [mailto:freeri...@gmail.com] Sent: Tuesday, March 19, 2013 6:10 AM To: dcarl...@tamu.edu; dcarl...@tamu.edu Cc: r-h...@stat.math.ethz.ch Subject: RE: [R] data.frame with NA Yes, colClasses is the solution. Thank you very much. However i found a very strange thing. If i use: Foglio1 - read.xlsx2(mydb.xlsx, 1, colClasses=c(Date, rep(numeric,14))) i get numeric dataframe, as you said. I also get NaN (and not NA). At this point i use the function: Foglio1 = na.locf(Foglio1,fromLast=T) and it works perfectly. All NaN 's were replaced with the first numeric value, as expected. And now the enigma. After na.locf function, Foglio1 become all CHR again! It seems that na.locf convert from num to chr. Even Date is converted in chr. I'm reading the help of this function but i can't find trace about the possibility of this conversion. It seems that i can't get in anyway a numeric dataframe without NA o NaN! Ok, i admit that i'm a newbie, but i'm trying every day to gain confidence with R Can i ask you the courtesy to use na.locf function to see if also on your computer this function convert all to CHR? Thank you At 21.37 18/03/2013, David L Carlson wrote: It appears that you MUST use the colClasses= argument with read.xlsx2: Foglio1 - read.xlsx2(mydb.xlsx, 1, colClasses=c(Date, rep(numeric, 14))) However, e and n are converted to NaN not NA so you would need to convert those columns (at least, I didn't check for missing values in the other columns): Foglio1$e - ifelse(is.nan(Foglio1$e), NA, Foglio1$e) Foglio1$n - ifelse(is.nan(Foglio1$n), NA, Foglio1$n) str(Foglio1) 'data.frame': 1489 obs. of 15 variables: $ Date: Date, format: 2001-08-17 2001-08-20 ... $ a : num 202 201 202 201 202 ... $ b : num 231 230 230 230 232 ... $ c : num 177 179 181 180 182 ... $ d : num 277 277 276 276 275 ... $ e : num NA NA NA NA NA NA NA NA NA NA ... $ f : num 275 277 279 279 279 ... $ g : num 91.7 90.7 90.8 91.1 91 ... $ h : num 11446 11258 11280 11396 11127 ... $ i : num 388 389 393 392 393 ... $ l : num 93.2 94 92.4 93.4 93.1 ... $ m : num 128 127 126 129 130 ... $ n : num NA NA NA NA NA NA NA NA NA NA ... $ o : num 133 133 133 133 133 ... $ p : num 107 107 107 107 107 ... --- David -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of David L Carlson Sent: Monday, March 18, 2013 3:22 PM To: 'Pietro'; 'Berend Hasselman' Cc: r-h...@stat.math.ethz.ch Subject: Re: [R] data.frame with NA Try this Open the spreadsheet in Excel. Select all of the data click Copy. Don't close Excel. Open R and type the following command: Foglio1 - read.table(clipboard-128, header=TRUE, sep=\t) Now take a look at the structure of the data.frame str(Foglio1) 'data.frame': 1489 obs. of 15 variables: $ Date: Factor w/ 1489 levels 1/10/2002,1/10/2003,..: 1275 1291 1295 1299 1304 1309 1321 1325 1329 1337 ... $ a : num 202 201 202 201 202 ... $ b : num 231 230 230 230 232 ... $ c : num 177 179 181 180 182 ... $ d : num 277 277 276 276 275 ... $ e : num NA NA NA NA NA NA NA NA NA NA ... $ f : num 275 277 279 279 279 ... $ g : num 91.7 90.7 90.8 91.1 91 ... $ h : num 11446 11258 11280 11396 11127 ... $ i : num 388 389 393 392 393 ... $ l : num 93.2 94 92.4 93.4 93.1 ... $ m : num 128 127 126 129 130 ... $ n : num NA NA NA NA NA NA NA NA NA NA ... $ o
Re: [R] data.frame with NA
Yes, colClasses is the solution. Thank you very much. However i found a very strange thing. If i use: Foglio1 - read.xlsx2(mydb.xlsx, 1, colClasses=c(Date, rep(numeric,14))) i get numeric dataframe, as you said. I also get NaN (and not NA). At this point i use the function: Foglio1 = na.locf(Foglio1,fromLast=T) and it works perfectly. All NaN 's were replaced with the first numeric value, as expected. And now the enigma. After na.locf function, Foglio1 become all CHR again! It seems that na.locf convert from num to chr. Even Date is converted in chr. I'm reading the help of this function but i can't find trace about the possibility of this conversion. It seems that i can't get in anyway a numeric dataframe without NA o NaN! Ok, i admit that i'm a newbie, but i'm trying every day to gain confidence with R Can i ask you the courtesy to use na.locf function to see if also on your computer this function convert all to CHR? Thank you At 21.37 18/03/2013, David L Carlson wrote: It appears that you MUST use the colClasses= argument with read.xlsx2: Foglio1 - read.xlsx2(mydb.xlsx, 1, colClasses=c(Date, rep(numeric, 14))) However, e and n are converted to NaN not NA so you would need to convert those columns (at least, I didn't check for missing values in the other columns): Foglio1$e - ifelse(is.nan(Foglio1$e), NA, Foglio1$e) Foglio1$n - ifelse(is.nan(Foglio1$n), NA, Foglio1$n) str(Foglio1) 'data.frame': 1489 obs. of 15 variables: $ Date: Date, format: 2001-08-17 2001-08-20 ... $ a : num 202 201 202 201 202 ... $ b : num 231 230 230 230 232 ... $ c : num 177 179 181 180 182 ... $ d : num 277 277 276 276 275 ... $ e : num NA NA NA NA NA NA NA NA NA NA ... $ f : num 275 277 279 279 279 ... $ g : num 91.7 90.7 90.8 91.1 91 ... $ h : num 11446 11258 11280 11396 11127 ... $ i : num 388 389 393 392 393 ... $ l : num 93.2 94 92.4 93.4 93.1 ... $ m : num 128 127 126 129 130 ... $ n : num NA NA NA NA NA NA NA NA NA NA ... $ o : num 133 133 133 133 133 ... $ p : num 107 107 107 107 107 ... --- David -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of David L Carlson Sent: Monday, March 18, 2013 3:22 PM To: 'Pietro'; 'Berend Hasselman' Cc: r-h...@stat.math.ethz.ch Subject: Re: [R] data.frame with NA Try this Open the spreadsheet in Excel. Select all of the data click Copy. Don't close Excel. Open R and type the following command: Foglio1 - read.table(clipboard-128, header=TRUE, sep=\t) Now take a look at the structure of the data.frame str(Foglio1) 'data.frame': 1489 obs. of 15 variables: $ Date: Factor w/ 1489 levels 1/10/2002,1/10/2003,..: 1275 1291 1295 1299 1304 1309 1321 1325 1329 1337 ... $ a : num 202 201 202 201 202 ... $ b : num 231 230 230 230 232 ... $ c : num 177 179 181 180 182 ... $ d : num 277 277 276 276 275 ... $ e : num NA NA NA NA NA NA NA NA NA NA ... $ f : num 275 277 279 279 279 ... $ g : num 91.7 90.7 90.8 91.1 91 ... $ h : num 11446 11258 11280 11396 11127 ... $ i : num 388 389 393 392 393 ... $ l : num 93.2 94 92.4 93.4 93.1 ... $ m : num 128 127 126 129 130 ... $ n : num NA NA NA NA NA NA NA NA NA NA ... $ o : num 133 133 133 133 133 ... $ p : num 107 107 107 107 107 ... -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Pietro Sent: Monday, March 18, 2013 1:57 PM To: Berend Hasselman Cc: r-h...@stat.math.ethz.ch Subject: Re: [R] data.frame with NA Yes, it's true Berend! What i do is simply use read.xlsx function db - read.xlsx2(c:/mydb.xlsx,1,as.data.frame=T) This is excel file i use: http://dl.dropbox.com/u/102669/mydb.xlsx I can't find a way to import as numeric. My objective is to be able to work (in R) with my NA's At 18.46 18/03/2013, Berend Hasselman wrote: On 18-03-2013, at 16:49, Pete freeri...@gmail.com wrote: I have this little data.frame http://dl.dropbox.com/u/102669/nanotna.rdata Two column contains NA, so the best thing to do is use na.locf function (with fromLast = T) But locf function doesn't work because NA in my data.frame are not recognized as real NA. Is there a way to substitute fake NA with real NA? In this case na.locf function should work Your data are all characters. Do str(db) to see that. What is probably supposed to be numeric is also character, Somehow you have managed to read in data that R thinks is all chr. Your NA are NA in reality: a character string NA. You will have to review the method you used to get the data into R. And make sure that what you
Re: [R] data.frame with NA
Yes, it's true Berend! What i do is simply use read.xlsx function db - read.xlsx2(c:/mydb.xlsx,1,as.data.frame=T) This is excel file i use: http://dl.dropbox.com/u/102669/mydb.xlsx I can't find a way to import as numeric. My objective is to be able to work (in R) with my NA's At 18.46 18/03/2013, Berend Hasselman wrote: On 18-03-2013, at 16:49, Pete freeri...@gmail.com wrote: I have this little data.frame http://dl.dropbox.com/u/102669/nanotna.rdata Two column contains NA, so the best thing to do is use na.locf function (with fromLast = T) But locf function doesn't work because NA in my data.frame are not recognized as real NA. Is there a way to substitute fake NA with real NA? In this case na.locf function should work Your data are all characters. Do str(db) to see that. What is probably supposed to be numeric is also character, Somehow you have managed to read in data that R thinks is all chr. Your NA are NA in reality: a character string NA. You will have to review the method you used to get the data into R. And make sure that what you want to be numeric is indeed numeric. Then you can start to think about doing something about the NA's. Berend __ 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] Counting number of common elements between the rows of two different matrices
Hello
I'm trying to solve this problem without using a for loop but I have so
far failed to find a solution.
I have two matrices of K columns each, e.g. (K=5), and with numbers of
row N_A and N_B respectively
A = (1 5 3 8 15;
2 7 20 11 13;
12 19 20 21 43)
B = (2 6 30 8 16;
3 8 19 11 13)
(the actual matrices have hundreds of thousands of entry, that's why I'm
keen to avoid for loops)
And what I need to do is to apply a function which counts the number of
common elements between ANY row of A and ANY row of B, giving a result
like this:
A1 vs B1: 1 # (8 is a common element)
A1 vs B2: 1 # (8 is a common element)
A2 vs B1: 1 # (2 is a common element)
A2 vs B2: 1 # 11, 13 are common elements
Etc.
I've built a function that counts the number of common elements between
two vectors, based on the intersect function in the R manual
common_elements - function(x,y) length(y[match(x,y,nomatch=0)])
And a double loop who solves my problem would be something like
(pseudo-code)
For(i in 1:N_A){
for(j in 1:N_B){
ce(i,j)=common_elements(a(i),b(j))
}
}
Is there an efficient, clean way to do the same job and give as an
output a matrix N_A x N_B such as that above?
Thanks a lot for your help
Regards
Pietro
__
For information pertaining to Willis' email confidentiality and monitoring
policy, usage restrictions, or for specific company registration and regulatory
status information, please visit http://www.willis.com/email_trailer.aspx
We are now able to offer our clients an encrypted email capability for secure
communication purposes. If you wish to take advantage of this service or learn
more about it, please let me know or contact your Client Advocate for full
details. ~W67897
__
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] Counting number of common elements between the rows of two different matrices
Jim
I tried that and it works. Thank you very much for your help!
Regards
Pietro
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: 04 November 2011 13:38
To: Parodi, Pietro
Cc: r-help@r-project.org
Subject: Re: [R] Counting number of common elements between the rows of two
different matrices
Try this:
# create dummy data
a - matrix(sample(20, 50, TRUE), ncol = 5)
b - matrix(sample(20, 50, TRUE), ncol = 5)
# create combinations to test
x - expand.grid(seq(nrow(a)), seq(nrow(b)))
# test
result - mapply(function(m1, m2) any(a[m1, ] %in% b[m2, ])
, x[, 1]
, x[, 2]
)
# create the output matrix
result.m - matrix(result, nrow = nrow(a), ncol = nrow(b))
On Fri, Nov 4, 2011 at 8:51 AM, Parodi, Pietro pietro.par...@willis.com wrote:
Hello
I'm trying to solve this problem without using a for loop but I have so
far failed to find a solution.
I have two matrices of K columns each, e.g. (K=5), and with numbers of
row N_A and N_B respectively
A = (1 5 3 8 15;
2 7 20 11 13;
12 19 20 21 43)
B = (2 6 30 8 16;
3 8 19 11 13)
(the actual matrices have hundreds of thousands of entry, that's why I'm
keen to avoid for loops)
And what I need to do is to apply a function which counts the number of
common elements between ANY row of A and ANY row of B, giving a result
like this:
A1 vs B1: 1 # (8 is a common element)
A1 vs B2: 1 # (8 is a common element)
A2 vs B1: 1 # (2 is a common element)
A2 vs B2: 1 # 11, 13 are common elements
Etc.
I've built a function that counts the number of common elements between
two vectors, based on the intersect function in the R manual
common_elements - function(x,y) length(y[match(x,y,nomatch=0)])
And a double loop who solves my problem would be something like
(pseudo-code)
For(i in 1:N_A){
for(j in 1:N_B){
ce(i,j)=common_elements(a(i),b(j))
}
}
Is there an efficient, clean way to do the same job and give as an
output a matrix N_A x N_B such as that above?
Thanks a lot for your help
Regards
Pietro
__
For information pertaining to Willis' email confidentiality and monitoring
policy, usage restrictions, or for specific company registration and
regulatory status information, please visit
http://www.willis.com/email_trailer.aspx
We are now able to offer our clients an encrypted email capability for secure
communication purposes. If you wish to take advantage of this service or
learn more about it, please let me know or contact your Client Advocate for
full details. ~W67897
__
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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
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] Converting from density to cumulative distribution
Hello I'm trying to do the following vector operation: given vector x = c(x1,x2,x3,x4...xn), produce vector y = c(x1,x1+x2,x1+x2+x3,...x1+...+xn). E.g., from x = c(1,3,2,2,5), produce y = c(1,4,6,8,13). The underlying problem is finding the cumulative distribution function given the empirical density distribution function. I have done some research on this but the only relevant thing I've found is the function integrate, which however gives me the sum of the whole vector not the step-wise sum. Thanks for your help Pietro __ For information pertaining to Willis' email confidentiality and monitoring policy, usage restrictions, or for specific company registration and regulatory status information, please visit http://www.willis.com/email_trailer.aspx We are now able to offer our clients an encrypted email capability for secure communication purposes. If you wish to take advantage of this service or learn more about it, please let me know or contact your Client Advocate for full details. ~W67897 __ [[alternative HTML version deleted]] __ 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] rcorrp.cens and U statistics
I have two alternative Cox models with C-statistics 0.72 and 0.78. My question is if 0.78 is significantly greater than 0.72. I'm using rcorrp.cens. I cannot find the U statistics in the output of the function. This is the output of the help example: x1 - rnorm(400) x2 - x1 + rnorm(400) d.time - rexp(400) + (x1 - min(x1)) cens - runif(400,.5,2) death - d.time = cens d.time - pmin(d.time, cens) rcorrp.cens(x1, x2, Surv(d.time, death)) Dxy S.D. x1 more concordant x2 more concordant nmissing uncensored Relevant Pairs -8.902077e-03 2.649712e-01 4.955490e-01 5.044510e-01 4.00e+02 0.00e+00 4.00e+00 1.348000e+03 Uncertain C X1 C X2 Dxy X1 Dxy X2 1.582520e+05 9.955490e-01 9.762611e-01 9.910979e-01 9.525223e-01 Should I manually calculate the U statistics interpreting the fraction x1 more concordant and x2 more concordant of the Relevant Pairs as Uxy and Uyx? many thanks for any help or suggestion Pietro Bulian __ 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] pivot table in R
Hello, I'm struggling with an elementary problem with R. I have a simple data frame such as this one giving the number of accidents subdivided by sex, age and region. sex age region no_of_accidents F young north 10 F young south 12 F old north 5 F old south 7 M young north 24 M young south 30 M old north 12 M old south 17 and I would like to build a pivot table, e.g. obtaining the sum of the number of accidents for each sex: sex age region no_of_accidents F (any) (any) 34 M (any) (any) 83 but I can't seem to obtain this result simply in R, except by embarking in complicated for loops. I have checked the documentation on such functions as table() and the documentation on An introduction to R but couldn't solve the problem. Could you please help me with this? Cheers Pete PLEASE CONSIDER THE ENVIRONMENT BEFORE PRINTING THIS E-MAIL For Aons standard conditions associated with this e-mail please visit http://www.aon.com/uk/en/email-footer/aon-limited.jsp Aon Limited Registered Office: 8 Devonshire Square, London EC2M 4PL Registered in London No. 210725 . VAT Registration No. 480 8401 48 Aon Limited is authorised and regulated by the Financial Services Authority in respect of insurance mediation activities only. [[alternative HTML version deleted]] __ 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.
