Apply a seq() in R function

Question

I have written a function, but I was wondering how I could have the function repeat by applying the seq(0, 1.5, by =.1) in place of a?

Here is my R code:

n1 = 20 n2 = 20 T1 = -1.26491106 df = 38 a = "wide" nb <- (n1*n2)/(n1+n2) a = c(medium = 1/2, wide = sqrt(2)/2, verywide = 1)[[a]] OO <- function(t,nb,df,a){ integrand <- function(g){ (1+nb*g)^(-1/2)*(1+t^2/(df*(1+nb*g)))^(-(df+1)/2)*(2*pi)^(-1/2)*g^(-3/2)*exp(-a^2/(2*g)) } num <- a*integrate(integrand,0,Inf)$value denom <- (1+t^2/df)^(-(df+1)/2) return(num/denom) } OO(t=T1,nb=nb,df=df,a=a) ## HERE IS THE FUNCTION ##

Possible duplicate of passing several arguments to FUN of lapply (and others *apply) — Jake Kaupp
– Jake Kaupp, Commented Feb 2, 2017 at 21:38

gsun · Accepted Answer · 2017-02-02 21:39:50Z

1

Not sure what is the meaning of initial setting on a, but here is how to apply function on a sequence,

sapply(seq(0.1, 1.5, by =.1), function(a) OO(t=T1,nb=nb,df=df,a=a))

answered Feb 2, 2017 at 21:39

gsun

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2 Comments

rnorouzian Over a year ago

gsun, has now throgh sapply the OO() produced a vector each of whose elements are the answers of OO() if we individually had input seq(0.1, 1.5, by =.1) one number at a time?

gsun Over a year ago

Yes. It's same with input one at a time in a loop. I believe apply function will be faster than loop, and one line code is cleaner, from my perspective.

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