[R] Take the maximum of every 12 columns

Thierry Onkelinx thierry.onkelinx at inbo.be
Tue Feb 20 16:05:29 CET 2018

```The maximum over twelve columns is the maximum of the twelve maxima of
each of the columns.

single_col_max <- apply(x, 2, max)
twelve_col_max <- apply(
matrix(single_col_max, nrow = 12),
2,
max
)

ir. Thierry Onkelinx
Statisticus / Statistician

Vlaamse Overheid / Government of Flanders
INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE
AND FOREST
Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance
thierry.onkelinx at inbo.be
Havenlaan 88 bus 73, 1000 Brussel
www.inbo.be

///////////////////////////////////////////////////////////////////////////////////////////
To call in the statistician after the experiment is done may be no
more than asking him to perform a post-mortem examination: he may be
able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given body
of data. ~ John Tukey
///////////////////////////////////////////////////////////////////////////////////////////

2018-02-20 15:55 GMT+01:00 Miluji Sb <milujisb at gmail.com>:
>  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), 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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```

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