search for: haj

Also, excuse me if my issue is a little bit application-specific. I need Mechanize to submit bunch of data from our server to an online reservation system. The url ishttp://hajres.iranair.com/haj/request.php" It works in my browser however, using mechanize following error appears: >> agent.get("http://hajres.iranair.com/haj/request.php") Net::HTTP::Get: /haj/request.php request-header: accept-language => en-us,en;q=0.5 request-header: accept =>...

Haj i try to perform a principal component analysis by using a tetrachoric correlation matrix as data input tetra <- tetrachoric (image_na, correct=TRUE) t_matrix <- tetra$rho pca.tetra <- principal(t_matrix, nfactors = 10, n.obs = nrow(image_na), rotate="varimax", scores=TRUE) th...

...w.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner?s personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/ D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/

...cheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner?s personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/ > D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/ >

...any-osobnich-udaju/ | Information > about processing and protection of business partner?s personal data are > available on website: > https://www.precheza.cz/en/personal-data-protection-principles/ > D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou > d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? > odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any > documents attached to it may be confidential and are subject to the legally > binding disclaimer: https://www.precheza.cz/en/01-disclaimer/ > > _________________...

...ds are 0 . Regards Petr Smetana ---------- Disclaimer ---------- Informace obsa?en? v tomto e-mailu jsou ur?eny v?lu?n? pro pot?eby jeho adres?ta. Text nebo p??lohy mohou obsahovat utajovan? informace, informace pova?ovan? spole?nost? T-Mobile za obchodn? tajemstv?, p??padn? jin? informace podl?haj?c? ochran? dle p??slu?n?ch pr?vn?ch p?edpis?. Pokud V?m tento e-mail byl omylem doru?en, zdr?te se, pros?m, jak?koli manipulace s textem ?i p??lohami, jako je kop?rov?n?, p?esm?rov?n?, zp??stupn?n? dal?? osob? a podobn?. O chybn?m doru?en? informujte odes?latele a e-mail v?etn? p??loh vyma?te ze sv...

Dear R users, I have an xlsx file (attached to this mail) that shows the values of a "der" series observed on a daily basis from January 1, 2017 to January 25, 2017. This series is strictly positive during two periods: from January 8, 2017 to January 11, 2017 and from January 16, 2017 to January 20, 2017. I would like to plot the series with two shaded areas corresponding to the

..., collapse="") [1] "1.0.1.1.0.1.1.1." If you did not want last dot, it is a bit shorter. > paste(unlist(strsplit(x,"")),collapse=".") [1] "1.0.1.1.0.1.1.1" > Cheers Petr Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/ > -----Original Message----- > From...

...X1986.01.04.10.30.00 1986-01-04 16.90191 > X1986.01.05.10.30.00 1986-01-05 16.00480 > X1986.01.06.10.30.00 1986-01-06 16.04405 > ID <NA> 3.50000 > > Cheers > Petr > > Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a > podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: > https://www.precheza.cz/01-dovetek/ | This email and any documents > attached to it may be confidential and are subject to the legally binding > disclaimer: https://www.precheza.cz/en/01-disclaimer/ > > > -----Origina...

...00 X1986.01.03.10.30.00 1986-01-03 16.51526 X1986.01.04.10.30.00 1986-01-04 16.90191 X1986.01.05.10.30.00 1986-01-05 16.00480 X1986.01.06.10.30.00 1986-01-06 16.04405 ID <NA> 3.50000 Cheers Petr Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/ > -----Original Message----- > From...

Dear all, I have daily data in wide-format from 01/01/1986 to 31/12/2016 by ID. I would like to convert this to weekly average data. The data has been generated by an algorithm. I know that I can use the lubridate package but that would require me to first convert the data to long-form (which is what I want). I am at a bit of loss of how to extract the date from the variable names and then

Hi -- I'm looking for alternatives to regex for a fairly simply 'reformatting' problem. Alternatives only because a lot of folks have trouble parsing/interpreting regex expressions, and I'm looking for suggestions for something more 'transparent'. Here is an example of what I'm trying to do. Take the following string, which I call x, and for each character in the

...; > X1986.01.05.10.30.00 1986-01-05 16.00480 > > X1986.01.06.10.30.00 1986-01-06 16.04405 > > ID <NA> 3.50000 > > > > Cheers > > Petr > > > > Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a > > podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: > > https://www.precheza.cz/01-dovetek/ | This email and any documents > > attached to it may be confidential and are subject to the legally binding > > disclaimer: https://www.precheza.cz/en/01-disclaimer/ > > &g...

Dear Rcommunity, I hope this email finds you well. I am writing to seek your assistance with a data visualization problem I am facing while working with R. Problem Description: I have a dataframe named "df" containing the following columns: "longitude", "latitude", "sea_currents_mag", and "sea_currents_direction". The dataframe includes sea

?Startup says: "Note that when the site and user profile files are sourced only the base package is loaded, so objects in other packages need to be referred to by e.g. utils::dump.frames or after explicitly loading the package concerned." So you need utils::installed.packages and utils::install.packages . Duncan Murdoch On 21/03/2023 8:04 a.m., PIKAL Petr wrote: > Dear all.

Friends, I am doing a URL classification, based on certain key words whether it contains an executive information or not. I have already went through 50K URL's and identified the key words and made it as 0, 1 ( 0 - do not have the key word and 1 - have the key word) and 0- do not contain executive information 1 - contains executive information. A sample set of data is shown below. DomainID

Hi all; I need to put 12 different plot2 into the same matrix. So my array for the matrix will be par(mfrow=c(3,4)). I am running ggplot2 to produce my 12 plots. For some reason, par(mfrow=c(3,4)) did not turn out 3*4 matrix. my basic R codes for each plot is par(mfrow=c(3,4)) library(ggplot2) p <- ggplot(a, aes(x=Genotypes, y=Plant_hight, size=Plant_hight, color=Showing_rate)) + . . Best

There are two key concepts you seem to be unaware of regarding ggplot: 1) you really need to put your data in long format to work with multiple curves, and 2) the column containing the names of the curves should be a factor with levels in the order you wish them to be presented in the legend (bottom to top). I am not in a position at the moment to give you a full reprex, but Google can probably

Hi, I'd like to graph three lines on ggplot2 and I intend the lines to be "solid", "dashed", and "dotted". The legend names are "name_b", "name_c", "name_a". I'd like to legend to present in the order: the "name_b" at the top, and "name_a" at the bottom. Could it be done by order function or its inverse?

Hi Pedro, melt() is probably working. The problem is I did not finish the copy and paste.? It would have been better if I had included the ggplot() command. Try ============================================================== library(reshape2) library(ggplot2) dat1? <- structure(list(N = c("I", "II", "III", "IV", "V", "VI",