algorithm,math,statistics,variance,standard-deviation

Given the forward formulas Mk = Mk-1 + (xk – Mk-1) / k Sk = Sk-1 + (xk – Mk-1) * (xk – Mk), it's possible to solve for Mk-1 as a function of Mk and xk and k: Mk-1 = Mk - (xk - Mk) / (k - 1)....

sql,sql-server,tsql,sum,variance

This was successful for me. Rounding to the 1000's at a line level seemed to produce a sum that worked when rounded to the 100's. Cleanup and modify to your specific needs. IF OBJECT_ID('tempdb..#Temp') IS NOT NULL DROP TABLE #Temp GO DECLARE @Total MONEY = 5.13; ;WITH SplitData AS (...

r,variance,confidence-interval

You can calculate the standard errors of the differences among factor levels in a linear model using the "aov" function from the stats package. These can then easily be extracted for graphing: # graphing differences among factor levels (with standard errors) # require(stats) m <- lm(mpg ~ gear, data=mtcars) plot(TukeyHSD(aov(m)))...

Solution with FULL JOIN seems to be more readable, but if you search for alternative - here it is - with functions lag() and lead(): with data as ( select abp_date dt, abp_source_uid id, abp_reference_1 ref, abp_charge charge, abp_count cnt, lag(abp_source_uid) over (partition by abp_date, abp_reference_1 order by abp_source_uid) lgid,...

c#,generics,boxing,variance,unboxing

Well, there's a reference conversion between string and object, in that every string reference can be treated as an object reference. This can be done transparently, with no modification to the value at all. That's why array variance can be done cheaply - for reads, anyway. Reading from an object[]...

m = matrix(c(1001,251,200,210,10,1002,101,200,300,100,1003,251,400,190,-210,1004,251,300,300,0,1005,101,250,250,0,1006,350,200,210,10,1007,401,400,400,0),ncol = 5,nrow=7,byrow=TRUE) colnames(m) = c("transactionID","sellerID","expectedprice","actualprice","pricediff") pricing = as.data.frame(m) pricing$diffabs <- abs(pricing$pricediff) pricing transactionID sellerID expectedprice actualprice pricediff diffabs 1001 251 200 210 10 10 1002 101 200 300 100 100 1003 251 400 190 -210 210 1004 251 300 300 0 0 1005 101 250 250 0...

linux,bash,awk,statistics,variance

Standard deviation formula is described in http://www.mathsisfun.com/data/standard-deviation.html So basically you need to say: for i in items sum += [(item - average)^2]/#items Doing it in your sample input: 5 av=5/1=5 var=(5-5)/1=0 5 av=10/2=5 var=(5-5)^2+(5-5)^2/2=0 5 av=15/3=5 var=3*(5-5)^2/3=0 10 av=25/4=6.25 var=3*(5-6.25)^2+(10-6.25)^2/4=4.6875 So in awk we can say: $ awk 'BEGIN {FS=OFS=","}...

r,ggplot2,statistics,data.table,variance

Use stat_summary. Note that the documentation is wrong when it says that fun.data should "take data frame as input". ggplot(mtcars, aes(x=gear, y=qsec)) + stat_summary(fun.y = var, geom = "point") + stat_summary(fun.data = function(y) { data.frame(y = var(y), ymin = ((length(y)-1)*var(y))/qchisq(0.025,length(y)-1), ymax = ((length(y)-1)*var(y))/qchisq(0.975,length(y)-1)) }, geom = "errorbar") + ylab("var.qsec") ...

python,opencv,image-processing,computer-vision,variance

Let's take for example, the second row of variance. Since the color values are in range 0-255 per channel, we can try wrapping your values to fit into that range: >>> row = [46.664, 121.162, 326.59, 809.223, 1021.599, 5330.989] >>> wrapped = [x % 256 for x in row] >>>...

math,variance,standard-deviation,calculated

The global variation is a sum. You can compute parts of the sum in parallel trivially, and then add them together. sum(x1...x100) = sum(x1...x50) + sum(x51...x100) The same way, you can compute the global averages - compute the global sum, compute the sum of the object counts, divide (don't divide...

I am answering purely from the documentation, so I may be wrong but. Eigen does not allow Matrix - scalar, but does allow Array - scalar Try either. a2 = a1.array() - a1.mean(); Or a2.array() = a1.array() - a1.mean(); Even if neither of those work, hopefully they point you in...

r,frequency,variance,frequency-distribution

One option is using data.table. Convert the data.frame to data.table (setDT) and get the var of "Value" and sum of "Count" by "Group". library(data.table) setDT(df1)[, list(GroupVariance=var(rep(Value, Count)), TotalCount=sum(Count)) , by = Group] # Group GroupVariance TotalCount #1: A 2.7 5 #2: B 4.0 4 a similar way using dplyr is...

Your code has two problems: std only operates on double or single values (not on uint8 for example). You should cast to double within std. You should also cast X to double in order to get more precise results in the subtraction (mean line) and the division (std line). So:...

python,numpy,time-series,variance

You should take a look at pandas. For example: import pandas as pd import numpy as np # some sample data ts = pd.Series(np.random.randn(1000), index=pd.date_range('1/1/2000', periods=1000)).cumsum() #plot the time series ts.plot(style='k--') # calculate a 60 day rolling mean and plot pd.rolling_mean(ts, 60).plot(style='k') # add the 20 day rolling variance: pd.rolling_std(ts,...

While V[X] = E[X^2] - E[X]^2 is the sample variance, the var function calculates a estimator for the population variance.

I'd suggest adding columns to your data.frame using rollapply and then use ifelse to check column values. library(zoo) #data$var3<- rollapply(data$snow, 3, var, fill=0, align="left") data$var3 <- c(rollapply(data$snow, 3, var, align="left")[-1], rep(0,3)) data$snow3 <- ifelse(data$temp<0 & data$var3>0.1, 0, data$snow) temperature snow var3 snow3 1 1 3 10.3333333 3 2 2 4...

If you look at the documentation for any FunctionX class, you'll see that the return type is co-variant and the argument types are contravariant. For instance, Function2 has the signature: Function2[-T1, -T2, +R] extends AnyRef You can spot the - and + before the type parameters, where - means contravariant...

Maybe you want this (instead of your for loop): var1 = sum([(j - mean)**2 for j in magnitude])/float(len(magnitude)-1) ...

I'm approaching this with the polygon feature of ggplot, see the documentation library(ggplot2) data = rbind.data.frame(c(21.06941667, 71.07952778), c(21.06666667, 71.08158333 ), c(21.07186111, 71.08688889 ), c(21.08625 , 71.07083333 ), c(21.08719444, 71.07286111 ), c(21.08580556, 71.07686111 ), c(21.07894444, 71.08225 )) names(data) = c("Latitude", "Longitude") Your variance is quite small, I multiplied by 10 for...