crocefisso - 1 year ago 65
R Question

R : Reduce eigenvalues using cumulative percentage of total of variance

Be λ a eigenvalue, Λ a set of eigenvalues, and p the cardinal of Λ, I'm trying to make a loop in R allowing to find n so

Here is an example of input (L components always are decreasing) :

``````> L
[1] 4.903332 4.341607 3.920309 3.716966 3.554570 3.067658 2.891222 2.553127
> sumli
[1] 28.94879
> freq <- L/sumli
> freq
[1] 0.16937952 0.14997540 0.13542221 0.12839795 0.12278821 0.10596843 0.09987368 0.08819460
``````

Answer Source
``````L <- c(4.903332, 4.341607, 3.920309, 3.716966, 3.55457, 3.067658,
2.891222, 2.553127)
``````

I think we can use `cumsum` (cumulative sum).

Depending on whether you want the first `n` such that cumulative contribution is greater than 95%, or the last `n` such that cumulative contribution is less than 95%, you need

``````which(cumsum(L) / sum(L) > 0.95)
# [1] 8

which(cumsum(L) / sum(L) > 0.95) - 1
# [1] 7
``````

(Your example `L` is not very representative as eigen values do not have a sharp decay.)

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