quantile

I was wondering how I could have R tell me the SD (as an argument in the qnorm() built in R) for a normal distribution whose 95% limit values are already known? As an example, I know the two 95% limit values for my normal are 158, and 168, respectively. So, in the below R code SD is shown as "x". If "y" (the answer of this simple qnorm() function) needs to be (158, 168), then can R tell me what should be x ? y <- qnorm(c(.025,.975), 163, x) A general procedure for Normal distribution Suppose we have a Normal distribution X ~ N(mu, sigma) , with unknown mean mu and unknown standard deviation

require(ggplot2) require(cowplot) d = iris ggplot2::ggplot(d, aes(factor(0), Sepal.Length)) + geom_violin(fill="black", alpha=0.2, draw_quantiles = c(0.25, 0.5, 0.75) , colour = "red", size = 1.5) + stat_boxplot(geom ='errorbar', width = 0.1)+ geom_boxplot(width = 0.2)+ facet_grid(. ~ Species, scales = "free_x") + xlab("") + ylab (expression(paste("Value"))) + coord_cartesian(ylim = c(3.5,9.5)) + scale_y_continuous(breaks = seq(4, 9, 1)) + theme(axis.text.x=element_blank(), axis.text.y = element_text(size = rel(1.5)), axis.ticks.x = element_blank(), strip.background=element_rect(fill="black"),

My code is based on the methods described here and here . def fraction?(number) number - number.truncate end def percentile(param_array, percentage) another_array = param_array.to_a.sort r = percentage.to_f * (param_array.size.to_f - 1) + 1 if r <= 1 then return another_array[0] elsif r >= another_array.size then return another_array[another_array.size - 1] end ir = r.truncate another_array[ir] + fraction?((another_array[ir].to_f - another_array[ir - 1].to_f).abs) end Example usage: test_array = [95.1772, 95.1567, 95.1937, 95.1959, 95.1442, 95.061, 95.1591, 95.1195, 95.1065, 95.0925, 95.199,

I am trying to calculate quantile for a column values manually, but not able to find the correct quantile value manually using the formula when compared to result output from Pandas. I looked around for different solutions, but did not find the right answer In [54]: df Out[54]: data1 data2 key1 key2 0 -0.204708 1.393406 a one 1 0.478943 0.092908 a two 2 1.965781 1.246435 a one In [55]: grouped = df.groupby('key1') In [56]: grouped['data1'].quantile(0.9) Out[56]: key1 a 1.668413 using the formula to find it manually,n is 3 as there are 3 values in data1 column quantile(n+1) applying the values

I'd like to replace all values in my relatively large R dataset which take values above the 95th and below the 5th percentile, with those percentile values respectively. My aim is to avoid simply cropping these outliers from the data entirely. Any advice would be much appreciated, I can't find any information on how to do this anywhere else. This would do it. fun <- function(x){ quantiles <- quantile( x, c(.05, .95 ) ) x[ x < quantiles[1] ] <- quantiles[1] x[ x > quantiles[2] ] <- quantiles[2] x } fun( yourdata ) You can do it in one line of code using squish() : d2 <- squish(d, quantile(d, c(

Judging from the documentation boost seems to offer quantile functions (inverse cdf functions) for both normal and gamma distributions, but its not clear for me how can I actually use them. Could someone paste an example please? The quantile calculation is implemented as a free function. Here's an example: #include <boost/math/distributions/normal.hpp> boost::math::normal dist(0.0, 1.0); // 95% of distribution is below q: double q = quantile(dist, 0.95); You can also get the complement (quantile from the right) using: // 95% of distribution is above qc: double qc = quantile(complement(dist, 0

I have a data.table and would like to compute stats by groups. R) set.seed(1) R) DT=data.table(a=rnorm(100),b=rnorm(100)) Those groups should be defined by R) quantile(DT$a,probs=seq(.1,.9,.1)) 10% 20% 30% 40% 50% 60% 70% 80% 90% -1.05265747329 -0.61386923071 -0.37534201964 -0.07670312896 0.11390916079 0.37707993057 0.58121734252 0.77125359976 1.18106507751 How can I compute say the average of b per bin, say if b=-.5 I am within [-0.61386923071,-0.37534201964] so in bin 3 How about : > DT[, mean(b), keyby=cut(a,quantile(a,probs=seq(.1,.9,.1)))] cut V1 1: NA -0.31359818 2: (-1.05,-0.614] -0