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Why not use a for loop?

I have seen many comments among research scientists on the Internet about how it is not recommended for cycles. However, I recently found myself in a situation where using one was useful. I would like to know if there is a better alternative for the following process (and why the alternative would be better):

I needed to start a series of repeated ANOVA measurements and approach the problem in the same way as the reproduced example, which you see below.

[I know that there are other problems associated with running several ANOVA models and that there are other options for these types of analysis, but now I just would like to hear about using the for loop

As an example, four ANOVA models with repeating measurements are four dependent variables that were measured in three cases:

set.seed(1976)
code <- seq(1:60)
time <- rep(c(0,1,2), each = 20)
DV1 <- c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 14, 2))
DV2 <- c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 10, 2))
DV3 <- c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 8, 2))
DV4 <- c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 10, 2))
dat <- data.frame(code, time, DV1, DV2, DV3, DV4)

outANOVA <- list()

for (i in names(dat)) {
  y <- dat[[i]]
  outANOVA[i] <- summary(aov(y ~ factor(time) + Error(factor(code)), 
                                  data = dat))
}

outANOVA
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2

, :

outANOVA <-
  lapply(dat,function(y)
    summary(aov(y ~ factor(time) + Error(factor(code)),data = dat)))

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outANOVA, .

microbenchmark :

n <- 100000
microbenchmark::microbenchmark(
preallocated_vec  = {x <- vector(length=n); for(i in 1:n) {x[i] <- i^2}},
preallocated_vec2 = {x <- numeric(n); for(i in 1:n) {x[i] <- i^2}},
incremented_vec   = {x <- vector(); for(i in 1:n) {x[i] <- i^2}},
preallocated_list = {x <- vector(mode = "list", length = n); for(i in 1:n) {x[i] <- i^2}},
incremented_list  = {x <- list(); for(i in 1:n) {x[i] <- i^2}},
sapply            = sapply(1:n, function(i) i^2),
lapply            = lapply(1:n, function(i) i^2),
times=20)

# Unit: milliseconds
# expr                     min         lq       mean     median         uq        max neval
# preallocated_vec    9.784237  10.100880  10.686141  10.367717  10.755598  12.839584    20
# preallocated_vec2   9.953877  10.315044  10.979043  10.514266  11.792158  12.789175    20
# incremented_vec    74.511906  79.318298  81.277439  81.640597  83.344403  85.982590    20
# preallocated_list  10.680134  11.197962  12.382082  11.416352  13.528562  18.620355    20
# incremented_list  196.759920 201.418857 212.716685 203.485940 205.441188 393.522857    20
# sapply              6.557739   6.729191   7.244242   7.063643   7.186044   9.098730    20
# lapply              6.019838   6.298750   6.835941   6.571775   6.844650   8.812273    20
+9

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Microbenchmark

set.seed(1976); code = seq(1:60); time = rep(c(0,1,2), each = 20);
DV1 = c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 14, 2)); DV2 = c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 10, 2)); DV3 = c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 8, 2)); DV4 = c(rnorm(20, 10, 2), rnorm(20, 10, 2), rnorm(20, 10, 2))
dat = data.frame(code, time, DV1, DV2, DV3, DV4)

library(microbenchmark)

microbenchmark(
    `Peter Miksza` = {
        outANOVA1 = list()
        for (i in names(dat)) {
            y = dat[[i]]
            outANOVA1[i] = summary(aov(y ~ factor(time) + Error(factor(code)), 
                data = dat))
    }},
    Moody_Mudskipper = {
        outANOVA2 =
            lapply(dat,function(y)
                summary(aov(y ~ factor(time) + Error(factor(code)),data = dat)))
    },
    `catastrophic_failure` = {
        outANOVA3 = 
            lapply(lapply(lapply(dat, function(y) y ~ factor(time) + Error(factor(code))), aov, data = dat), summary)
    },
    times = 1000L)


#Unit: milliseconds
#                 expr      min       lq     mean   median       uq       max neval cld
#         Peter Miksza 26.25641 27.63011 31.58110 29.60774 32.81374 136.84448  1000   b
#     Moody_Mudskipper 22.93190 23.86683 27.20893 25.61352 28.61729 135.58811  1000  a 
# catastrophic_failure 22.56987 23.57035 26.59955 25.15516 28.25666  68.87781  1000  a

JIT, compiler::setCompilerOptions(optimize = 0) compiler::enableJIT(0) :

#Unit: milliseconds
#                 expr      min       lq     mean   median       uq      max neval cld
#         Peter Miksza 23.10125 24.27295 28.46968 26.52559 30.45729 143.0731  1000   a
#     Moody_Mudskipper 22.82366 24.35622 28.33038 26.72574 30.27768 146.4284  1000   a
# catastrophic_failure 22.59413 24.04295 27.99147 26.23098 29.88066 120.6036  1000   a

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Moody, , , . , - . . .

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