Calculating the mean of every replication

a 夏天 提交于 2019-12-04 07:13:23

问题


I have the following code

set.seed(30)
nsim <- 50    ## NUMBER OF REPLICATIONS
demand <- c(12,13,24,12,13,12,14,10,11,10)

res <- replicate(nsim, {
    load <- runif(10,11,14)
    diff <- load - demand    ## DIFFERENCE BETWEEN DEMAND AND LOAD 
    return(sum(diff < 0))
})
res
[1] 6 5 7 4 4 5 4 3 6 4 5 5 5 4 2 5 3 3 3 5 3 2 4 6 5 4 4 3 5 6 4 4 3 6 5 3 5 5 4 3 3
[42] 6 4 4 4 6 6 5 4 5

I have a huge data set and the question is what is the fastest way of calculating the mean for every replication. For example the res in first replication is 6 so the result should be 6/1=6 for the second 6+5/2=5.5 for the third 6+5+7/3=6 and for the last replication is sum(res)/nsim=4.38


回答1:


In the edited version of the question (edit of Feb 11 at 5:53), the OP has specified the expected result. These indicate that the OP might be looking for a cumulative mean of the result vector res:

cumsum(res)/seq_along(res)
# [1] 6.000000 5.500000 6.000000 5.500000 5.200000 5.166667 5.000000 4.750000 4.888889
#[10] 4.800000 4.818182 4.833333 4.846154 4.785714 4.600000 4.625000 4.529412 4.444444
#[19] 4.368421 4.400000 4.333333 4.227273 4.217391 4.291667 4.320000 4.307692 4.296296
#[28] 4.250000 4.275862 4.333333 4.322581 4.312500 4.272727 4.323529 4.342857 4.305556
#[37] 4.324324 4.342105 4.333333 4.300000 4.268293 4.309524 4.302326 4.295455 4.288889
#[46] 4.326087 4.361702 4.375000 4.367347 4.380000

Alternatively, dplyr::cummean(res) can be used.




回答2:


To illustrate my comment, you can generate a matrix where columns (or rows, if you prefer) represent replications, after which you can use R's matrix operations capabilities:

set.seed(47)    # make reproducible

nsim <- 50    ## NUMBER OF REPLICATIONS
demand <- c(12,13,24,12,13,12,14,10,11,10)

loads <- matrix(runif(10 * nsim, 11, 14), ncol = nsim)

diffs <- loads - demand    # with vector recycling
# or: diffs <- apply(loads, 2, `-`, demand)    
# or: diffs <- apply(loads, 2, function(x){x - demand})

res <- colSums(diffs > 0)
LOLE <- sum(res) / nsim

LOLE
#> [1] 5.7


来源:https://stackoverflow.com/questions/42145971/calculating-the-mean-of-every-replication

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