search for: quantilization

The attached patch adds some sanity checks to the "type" argument of quantile(). Output from the following commands show the change of behavior with the current patch: vec <- 1:10 quantile(vec, type = c(1, 2)) quantile(vec, type = 10) quantile(vec, type = "aaa") quantile(vec, type = NA_real_) quantile(vec, type = 4.3) quantile(vec, type = -1) Current behavior

On Fri, May 31, 2019 at 01:28:55AM -0400, Scott Kostyshak wrote: > The attached patch adds some sanity checks to the "type" argument of > quantile(). Output from the following commands show the change of > behavior with the current patch: > > vec <- 1:10 > quantile(vec, type = c(1, 2)) > quantile(vec, type = 10) > quantile(vec, type = "aaa")

I'm trying to work out the simplest way to remove the upper and lower quantiles, in this case upper and lower 25% from an array. I can do it in two steps but when I try it in one, it fails. Is there something simple missing from my syntax or are there other simple elegant way to accomplish this? Thanks J > i <-1:20 > i2 <- i[i<quantile(i,.75)] > i3 <-

Guys: 1).When I using the 'quantile' function, I get really confused. Here is what I met: > x<-zoo(rnorm(500,0,1)) > quantile(x,0.8) 400 1.060258 > c=rnorm(500,0,1) > quantile(c,0.8) 80% 0.9986075 why do the results display different? Is that because of the different type of the class? 2).And I want to use the 'rollapply' function to compute a

I'm not sure who is responsible for quantile(), but I assume they read this list. Ivan Frohne and I have produced a revision of the quantile.default() function which enables the computation of alternative sample quantile definitions. The code and .Rd file are attached. This enables the user to produce quantiles that are equivalent to those in various statistics package. There is a type

Hi, so I don't speak computer and I have no idea what this code is telling the program to do, but I apparently need to be able to find and isolate influencial observations. Problem, I have no idea what the error means and where it may be from in the code. error I get is below the code { ## OLS results NameC<- lm(gpanew~female+female:lastinit+agenew+canadian+mom_ed+yearstudy) ## default:

All - I have an example data frame x l.c.1 43.38812035 085 47.55710661 085 47.55710661 085 51.99211429 085 51.99211429 095 54.78449958 095 54.78449958 095 56.70201864 095 56.70201864 105 59.66361903 105 61.69573564 105 61.69573564 105 63.77469479 115 64.83191994 115 64.83191994 115 66.98222118 115 66.98222118 125 66.98222118 125 66.98222118 125 66.98222118 125 and I'd like to get the 3rd

I am trying to replicate the SAS proc univariate in R. I got most of the stats I needed for a by grouping in a data frame using: all1 <- ddply(all,"ACT_NAME", summarise, mean=mean(COUNTS), sd=sd(COUNTS), q25=quantile(COUNTS,.25),median=quantile(COUNTS,.50), q75=quantile(COUNTS,.75), q90=quantile(COUNTS,.90), q95=quantile(COUNTS,.95), q99=quantile(COUNTS,.99) )

Dear R-users, I don't know if there is a problem in wtd.quantile (from library "Hmisc"): -------------------------------- x <- c(1,2,3,4,5) w <- c(0.5,0.4,0.3,0.2,0.1) wtd.quantile(x,weights=w) ------------------------------- The output is: 0% 25% 50% 75% 100% 3.00 3.25 3.50 3.75 4.00 The version of R I am using is: 2.1.0 Best,Jing

Hi, In R-devel (2020-05-17 r78478) quantile() type 1 seems to behave a little bit strange for some ordered factors: quantile(factor(1:3, ordered = TRUE), 0.5, type = 1) returns ?2? as expected. But quantile(factor(2:4, ordered = TRUE), 0.5, type = 1) returns ?4? and quantile(factor(3:5, ordered = TRUE), 0.5, type = 1) returns ?NA?. Furthermore, the function returns ?NA? for calls like

Is it a bug that quantile() can return a lower sample quantile for a higher probability? > ##### quantile returns decreasing results with increasing probs (data at the end of the message) > quantile(x2, (0:5)/5) 0% 20% 40% 60% 80% -0.0014141174 -0.0009041968 -0.0009041968 -0.0007315023 -0.0005746115 100% 0.2905596324 >

Dear all, I am searching for a way to compute a test comparable to Chuang et al. ("Causality in Quantiles and Dynamic Stock Return-Volume Relations"). The aim of this test is to check wheter the coefficient of a quantile regression granger-causes Y in a quantile range. I have nearly computed everything but I am searching for an estimator of the density of the distribution at several

To calculate Percentile for a set of observations Excel has percentile() function. R function quantile() does the same thing. Is there any significant difference btw percentile and quantile? Regrads, -- View this message in context: http://www.nabble.com/Diff-btw-percentile-and-quantile-tp22328375p22328375.html Sent from the R help mailing list archive at Nabble.com.

I am calling quantiles as follows. I don't understand why sometimes the columns (data values) above 95% are returned as "NULL"!! When I drop the percentile down to 92%, I see colums appearing. Why would any quantile be empty? I see sometimes that 95% percentile is being chosen as "1" for my data between 0 and 1, where obviously there's no column value equal to 1. But

Hi All, I am looking at applications of percentiles to time sequenced data. I had just been using the quantile function to get percentiles over various periods, but am more interested in if there is an accepted (and/or R-implemented) method to apply weighting to the data so as to weigh recent data more heavily. I wrote the following function, but it seems quite inefficient, and not really very

bradleyd wrote > Excuse the request from an R novice! I have a data frame (DATA) that has > two numeric columns (YEAR and DAY) and 4000 rows. For each YEAR I need to > determine the 10% and 90% quantiles of DAY. I'm sure this is easy enough, > but I am a new to this. > >> quantile(DATA$DAY,c(0.1,0.9)) > 10% 90% > 12 29 > > But this is for the entire

Hi, A simple quantile question: I need to calculate the 95% and 5% quantiles (aka percentiles) for the following data: 67.12 64.51 62.06 55.45 51.41 43.78 10.74 10.14 if I use the formula: 95% quantile point= 95 (8+1)/100, I get the 8.55th point as the 95% quantile. Which does not make too much sense as I have only 8 data points. The other option is to use (95*8)/100 = 7.6th data point (which can

Hi, I'm working on a data set that contains a couple of factors and a number of dependent variables. From all of these dependent variables I would like to calculate mean, standard deviation and quantiles. With the function FUN I get all the means and stdev that I want but quantiles are only calculated for the first of the dependent variables (column 8 in the summarize command). What do I

Hi All, I want to create a categorical variable, cat.pfoa, in the file of pfas.pheno (a data frame) based on log2pfoa values. I can do it using the following code. pfas.pheno <-within(pfas.pheno, {cat.pfoa<-NA cat.pfoa[pfas.pheno$log2pfoa <=quantile(pfas.pheno$log2pfoa,0.25, na.rm =T)]<-0 cat.pfoa[pfas.pheno$log2pfoa >=quantile(pfas.pheno$log2pfoa,0.75, na.rm =T)]<-2

Habría que buscar la vuelta, yo no lo se, pero posiblemente lo siguiente da una pista. Nota: al mismo código le sume una línea al final datos<-c(2,3,4,5,6,7,8) quantile(datos) quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95)) as.matrix(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) as.data.frame(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) # ¿ y si solo solicita