[R] P-values Kolmogorov–Smirnov test
Boo G.
g|@n|uc@@boo @end|ng |rom @oton@@c@uk
Thu Sep 5 20:42:36 CEST 2019
Thanks for your reply, Rui.
I don’t think that I can use directly the ks.test because I have a weighted sample (see m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]) and I want to account for that. That’s why I am trying to compute everything manually.
Also, if you look at the results of the ks.test in your simulation, you will notice that the p-value always implies that the sample is always (even with same size = 1) drawn form the same distribution. This looks suspicious to me.
What are your thoughts?
>
> On 5 Sep 2019, at 20:29, Rui Barradas <ruipbarradas using sapo.pt> wrote:
>
> Hello,
>
> I don't have the algorithms at hand but the KS statistic calculation is more complicated than your max/abs difference.
>
> Anyway, why not use ks.test? it's not that difficult:
>
>
> set.seed(1234)
> #reference distribution
> d_1 <- sort(rpois(1000, 500))
> p_1 <- d_1/sum(d_1)
> m_1 <- data.frame(d_1, p_1)
>
> #data frame to store the values of the simulation
> d_stat <- data.frame(1:1000, NA, NA)
> names(d_stat) <- c("sample_size", "ks_distance", "p_value")
>
> #simulation
> for (i in 1:1000) {
> #sample from the reference distribution
> m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
> d_2 <- m_2$d_1
>
> ht <- ks.test(d_1, d_2)
> #kolmogorov-smirnov distance
> d_stat[i, 2] <- ht$statistic
> d_stat[i, 3] <- ht$p.value
> }
>
> hist(d_stat[, 2])
> hist(d_stat[, 3])
>
>
> Note that d_2 is not sorted, but the results are equal in the sense of function identical(), meaning they are *exactly* the same. Why shouldn't they?
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 17:06 de 05/09/19, Boo G. escreveu:
>> Hello,
>> I am trying to perform a Kolmogorov–Smirnov test to assess the difference between a distribution and samples drawn proportionally to size of different sizes. I managed to compute the Kolmogorov–Smirnov distance but I am lost with the p-value. I have looked into the ks.test function unsuccessfully. Can anyone help me with computing p-values for a two-tailed test?
>> Below a simplified version of my code.
>> Thanks in advance.
>> Gianluca
>> library(spatstat)
>> #reference distribution
>> d_1 <- sort(rpois(1000, 500))
>> p_1 <- d_1/sum(d_1)
>> m_1 <- data.frame(d_1, p_1)
>> #data frame to store the values of the siumation
>> d_stat <- data.frame(1:1000, NA, NA)
>> names(d_stat) <- c("sample_size", "ks_distance", "p_value")
>> #simulation
>> for (i in 1:1000) {
>> #sample from the reference distribution
>> m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
>> m_2 <-m_2[order(m_2$d_1),]
>> d_2 <- m_2$d_1
>> p_2 <- m_2$p_1
>> #weighted ecdf for the reference distribution and the sample
>> f_d_1 <- ewcdf(d_1, normalise=F)
>> f_d_2 <- ewcdf(d_2, 1/p_2, normalise=F, adjust=1/length(d_2))
>> #kolmogorov-smirnov distance
>> d_stat[i,2] <- max(abs(f_d_1(d_2) - f_d_2(d_2)))
>> }
>> [[alternative HTML version deleted]]
>> ______________________________________________
>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://eur03.safelinks.protection.outlook.com/?url=https%3A%2F%2Fstat.ethz.ch%2Fmailman%2Flistinfo%2Fr-help&data=01%7C01%7Cgianluca.boo%40soton.ac.uk%7C0c709068527c41e062dd08d7322f0d72%7C4a5378f929f44d3ebe89669d03ada9d8%7C0&sdata=y9jfixyNiroKwKZEJj0owuCcWoeFQKZdaG9WLe2xHQ8%3D&reserved=0
>> PLEASE do read the posting guide https://eur03.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.R-project.org%2Fposting-guide.html&data=01%7C01%7Cgianluca.boo%40soton.ac.uk%7C0c709068527c41e062dd08d7322f0d72%7C4a5378f929f44d3ebe89669d03ada9d8%7C0&sdata=7n0doy4P1S1TpApX1zpUborAnUnxuOxYtn%2FQ%2BtVztGM%3D&reserved=0
>> and provide commented, minimal, self-contained, reproducible code.
More information about the R-help
mailing list