[R] Take the maximum of every 12 columns

Tue Feb 20 17:58:52 CET 2018

Ista, et. al: efficiency?
(Note: I needed to correct my previous post: do.call() is required for
pmax() over the data frame)

> x <- data.frame(matrix(runif(12e6), ncol=12))

> system.time(r1 <- do.call(pmax,x))
   user  system elapsed
  0.049   0.000   0.049

> identical(r1,r2)
[1] FALSE
> system.time(r2 <- apply(x,1,max))
   user  system elapsed
  2.162   0.045   2.207

## 150 times slower!

> identical(r1,r2)
[1] TRUE

pmax() is there for a reason.
Or is there something I am missing?

Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Tue, Feb 20, 2018 at 8:16 AM, Miluji Sb <milujisb at gmail.com> wrote:

> This is what I was looking for. Thank you everyone!
>
> Sincerely,
>
> Milu
>
>
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> On Tue, Feb 20, 2018 at 5:10 PM, Ista Zahn <istazahn at gmail.com> wrote:
>
>> Hi Milu,
>>
>> byapply(df, 12, function(x) apply(x, 1, max))
>>
>> You might also be interested in the matrixStats package.
>>
>> Best,
>> Ista
>>
>> On Tue, Feb 20, 2018 at 9:55 AM, Miluji Sb <milujisb at gmail.com> wrote:
>> >  Dear all,
>> >
>> > I have monthly data in wide format, I am only providing data (at the
>> bottom
>> > of the email) for the first 24 columns but I have 2880 columns in total.
>> >
>> > I would like to take max of every 12 columns. I have taken the mean of
>> > every 12 columns with the following code:
>> >
>> > byapply <- function(x, by, fun, ...)
>> > {
>> >   # Create index list
>> >   if (length(by) == 1)
>> >   {
>> >     nc <- ncol(x)
>> >     split.index <- rep(1:ceiling(nc / by), each = by, length.out = nc)
>> >   } else # 'by' is a vector of groups
>> >   {
>> >     nc <- length(by)
>> >     split.index <- by
>> >   }
>> >   index.list <- split(seq(from = 1, to = nc), split.index)
>> >
>> >   # Pass index list to fun using sapply() and return object
>> >   sapply(index.list, function(i)
>> >   {
>> >     do.call(fun, list(x[, i], ...))
>> >   })
>> > }
>> >
>> > ## Compute annual means
>> > y <- byapply(df, 12, rowMeans)
>> >
>> > How can I switch rowMeans with a command that takes the maximum? I am a
>> bit
>> > baffled. Any help will be appreciated. Thank you.
>> >
>> > Sincerely,
>> >
>> > Milu
>> >
>> > ###
>> > dput(droplevels(head(x, 5)))
>> > structure(list(X0 = c(295.812103271484, 297.672424316406,
>> 299.006805419922,
>> > 297.631500244141, 298.372741699219), X1 = c(295.361328125,
>> > 297.345092773438,
>> > 298.067504882812, 297.285339355469, 298.275268554688), X2 =
>> > c(294.279602050781,
>> > 296.401550292969, 296.777984619141, 296.089111328125, 297.540374755859
>> > ), X3 = c(292.103118896484, 294.253601074219, 293.773803710938,
>> > 293.916229248047, 296.129943847656), X4 = c(288.525024414062,
>> > 291.274505615234, 289.502777099609, 290.723388671875, 293.615112304688
>> > ), X5 = c(286.018371582031, 288.748565673828, 286.463134765625,
>> > 288.393951416016, 291.710266113281), X6 = c(285.550537109375,
>> > 288.159149169922, 285.976501464844, 287.999816894531, 291.228271484375
>> > ), X7 = c(289.136962890625, 290.751159667969, 290.170257568359,
>> > 291.796203613281, 293.423248291016), X8 = c(292.410003662109,
>> > 292.701263427734, 294.25244140625, 295.320404052734, 295.248199462891
>> > ), X9 = c(293.821655273438, 294.139068603516, 296.630157470703,
>> > 296.963531494141, 296.036224365234), X10 = c(294.532531738281,
>> > 295.366607666016, 297.677551269531, 296.715911865234, 296.564178466797
>> > ), X11 = c(295.869476318359, 297.010070800781, 299.330169677734,
>> > 297.627593994141, 297.964935302734), X12 = c(295.986236572266,
>> > 297.675567626953, 299.056671142578, 297.598907470703, 298.481842041016
>> > ), X13 = c(295.947601318359, 297.934448242188, 298.745391845703,
>> > 297.704925537109, 298.819091796875), X14 = c(294.654327392578,
>> > 296.722717285156, 297.0986328125, 296.508239746094, 297.822021484375
>> > ), X15 = c(292.176361083984, 294.49658203125, 293.888305664062,
>> > 294.172149658203, 296.117095947266 <(709)%20594-7266>), X16 =
>> c(288.400726318359,
>> > 291.029113769531, 289.361907958984, 290.566772460938, 293.554016113281
>> > ), X17 = c(285.665222167969, 288.293029785156, 286.118957519531,
>> > 288.105285644531, 291.429382324219), X18 = c(285.971252441406,
>> > 288.3798828125, 286.444580078125, 288.495880126953, 291.447326660156
>> > ), X19 = c(288.79296875, 290.357543945312, 289.657928466797,
>> > 291.449066162109, 293.095275878906), X20 = c(291.999877929688,
>> > 292.838348388672, 293.840362548828, 294.412322998047, 294.941253662109
>> > ), X21 = c(293.615447998047, 294.028106689453, 296.072296142578,
>> > 296.447387695312, 295.824615478516), X22 = c(294.719848632812,
>> > 295.392028808594, 297.453216552734, 297.114288330078, 296.883209228516
>> > ), X23 = c(295.634429931641, 296.783294677734, 298.592346191406,
>> > 297.469512939453, 297.832122802734)), .Names = c("X0", "X1",
>> > "X2", "X3", "X4", "X5", "X6", "X7", "X8", "X9", "X10", "X11",
>> > "X12", "X13", "X14", "X15", "X16", "X17", "X18", "X19", "X20",
>> > "X21", "X22", "X23"), row.names = c(NA, 5L), class = "data.frame")
>> >
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>> > priva di virus. www.avast.com
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>> >
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>> >
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