[R] Take the maximum of every 12 columns
Bert Gunter
bgunter.4567 at gmail.com
Tue Feb 20 17:10:52 CET 2018
Do you want the max of all the data in all 12 columns -- 1 number -- or
the parallel max of the 12 columns -- a vector of length the number of rows.
In the first case,
max(df[,1:12])
will do (R uses 1-based indexing,not 0-based) ;
in the second case,
pmax(df[, 1:12])
will do.
If either of these is what you want, all your code is nonsense. Of course
you can change the indexing to march through the data frame in whatever way
you like. If this is not what you want, maybe someone wiser than I can
interpret your query.
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 7:59 AM, Miluji Sb <milujisb at gmail.com> wrote:
> Thank you for your kind replies. Maybe I was not clear with my question (I
> apologize) or I did not understand...
>
> I would like to take the max for X0...X11 and X12...X24 in my dataset.
> When I use pmax with the function byapply as in
>
> byapply(df, 12, pmax)
>
> I get back a list which I cannot convert to a dataframe. Am I missing
> something? Thanks again!
>
> Sincerely,
>
> Milu
>
>
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> On Tue, Feb 20, 2018 at 4:38 PM, Bert Gunter <bgunter.4567 at gmail.com>
> wrote:
>
>> Don't do this (sorry Thierry)! max() already does this -- see ?max
>>
>> > x <- data.frame(a =rnorm(10), b = rnorm(10))
>> > max(x)
>> [1] 1.799644
>>
>> > max(sapply(x,max))
>> [1] 1.799644
>>
>> 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 7:05 AM, Thierry Onkelinx <
>> thierry.onkelinx at inbo.be> wrote:
>>
>>> 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
>>>
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>>> not ensure that a reasonable answer can be extracted from a given body
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>>>
>>>
>>>
>>>
>>> 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 <(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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