search for: 17.26

Hi there, This seems like it should be simple. I have a data frame of climate data sampled every 10 min. I want to average the entire data frame into 30 min values (i.e., one value for each half hour). Functions like running.mean give me a moving average but I want to reduce the size of the entire frame.. Any ideas how? Cheers! Example of my data timestamp Voltage LwTempDownelling

To anyone who can help: I have two brief questions concerning sapply. Following below is the code for my example. The two problems are described at the end of the code: site <- rep(2:6, each = 12) tillage <- rep(c(1,-1), each = 6, times = 5) carbon <- c(18.23, 16.06, 17.81, 16.07, 17.26, 17.08, 14.92, 15.88, 12.11, 14.23, 16.99, 13.57, 20.34, 20.3,

I have a series of numbers I'm wanting to plot. They come from a nanodrop machine, which graphs with a specific x and y indices. X goes from 220nm to 350nm, which I can set. But the y axis should go from -5 to 65, but I'm finding it impossible to hardcode that. I've looked. I've typed ?plot at the R prompt. Google has not been my friend. _R Graphics_, if it holds the key, has not

R Help - I'm trying to run a correlation matrix with a covariate of "age" and will at some point will also want to covary other variables concurrently. I'm using the "psych" package and have tried other methods such as writing a loop to extract semi-partial correlations, but it does not seem to be working. How can I accomplish this? library(psych) > set.cor(y =

Hi Hemant, As I suspected, the code broke when I got to the line: result <- rfm_auto(df, id="user_id", payment ="subtotal_amount", date="created_at") Error in rfm_auto(df, id = "user_id", payment = "subtotal_amount", date = "cr eated_at") : could not find function "rfm_auto" It looks like you are using the hoxo-m/easyRFM

I'm trying to perform an RFM analysis on the attached dataset, I'm able to get the results using the auto_rfm function but i want to define my own breaks for RFM. as follow r <-c(30,60,90) f <-c(2,5,8) m <-c(10,20,30) but when i tried to define my own breaks i got the identical result for RFM i.e 111 for every ID. please help me with this with working R script so that i can get

With mdbox, what does dovecot lock when "pop3_lock_session(pop3): yes"? Specifically, I'm wondering if Dovecot LDA is able to deliver mail when a session is locked, if using mdbox, or if it will tempfail until the session is unlocked? Thanks, Ken -- Ken Anderson Pacific Internet - http://www.pacific.net

Dear list,I am quite new to the world of biostatistics and I encounter some issues in the precise understanding of the coxph function of the survival package.I have a set of survival data (patient who had (or died from) a breast cancer) I'd like to see which are the variables that might cause dead or not.When trying> summary(coxph(Surv(Time_to_distant_recurrence_yrs, !Distant_recurrence)~

Hello all, Although i'm aware that version 1.0.15 is rather old, that's what is used in Lenny, so... Either way, the setup is rather simple, regular dovecot install, with maildirs residing on a "local" ext3 filesystem accessed through FC to a SAN (2Gbps link). The server has 2 cores (with HT), so "almost" 4 cores and 3GB of ram. A couple weeks ago we had a major

Hi, I am implementig Dovecot as a part of my new e-mail server, which is aimed to be replacement for proprietary software I've been using sofar. Appreciating all Dovecot rich features, I lack just one. And this is the ability to customize the "quota exceeded, message rejected" message. I know I can set it's default content using quota_exceeded_message parameter, but i would

Hello all - I have an example column in a dataFrame id.name 123.45 123.45 123.45 123.45 234.56 234.56 234.56 234.56 234.56 234.56 234.56 345.67 345.67 345.67 456.78 456.78 456.78 456.78 456.78 456.78 456.78 456.78 456.78 ... [truncated] And I'd like to create a second vector of sequential values (i.e., 1:N) for each unique id.name value. In other words, I need id.name x 123.45 1 123.45

Hey, i want to define 3 ideal breaks (bin) for each variable one of those variables is attached in the previous email, i don't want to consider quartile method because quartile is not working ideally for that data set because data distribution is non normal. so i want you to suggest another method so that i can define 3 breaks with the ideal interval for Recency, frequency and monetary to

Hi Your statement about attaching data is problematic. We cannot do much with it. Instead use output from dput(yourdata) to show us what exactly your data look like. We also do not know how do you want to split your data. It would be nice if you can show also what should be the bins with respective data. Unless you provide this information you probably would not get any sensible answer. Cheers

Hi You expect us to solve your problem but you ignore advice already recieved. Your data are unreadable, use dput(yourdata) instead. see ?dput > test<-read.table("clipboard", heade=T) Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : line 115 did not have 6 elements What is ?ideal interval? can you define it? Should it be such to provide eqal

Hello, I'm working on RFM analysis and i wanted to define my own breaks but my frequency distribution is not normally distributed so when I'm using quartile its not giving the optimal results. so I'm looking for a better approach where i can define breaks dynamically because after visualization i can do it easily but i want to apply this model so that it can automatically define the