?s 04:30 de 27/03/2024, Ogbos Okike escreveu:> Warm greetings to you all.
>
> Using the tapply function below:
> data<-read.table("FD1month",col.names =
c("Dates","count"))
> x=data$count
> f<-factor(data$Dates)
> AB<- tapply(x,f,mean)
>
>
> I made a simple calculation. The result, stored in AB, is of the form
> below. But an effort to write AB to a file as a data frame fails. When I
> use the write table, it only produces the count column and strip of the
> first column (date).
>
> 2005-11-01 2005-12-01 2006-01-01 2006-02-01 2006-03-01 2006-04-01
> 2006-05-01
> -4.106887 -4.259154 -5.836090 -4.756757 -4.118011 -4.487942
> -4.430705
> 2006-06-01 2006-07-01 2006-08-01 2006-09-01 2006-10-01 2006-11-01
> 2006-12-01
> -3.856727 -6.067103 -6.418767 -4.383031 -3.985805 -4.768196
> -10.072579
> 2007-01-01 2007-02-01 2007-03-01 2007-04-01 2007-05-01 2007-06-01
> 2007-07-01
> -5.342338 -4.653128 -4.325094 -4.525373 -4.574783 -3.915600
> -4.127980
> 2007-08-01 2007-09-01 2007-10-01 2007-11-01 2007-12-01 2008-01-01
> 2008-02-01
> -3.952150 -4.033518 -4.532878 -4.522941 -4.485693 -3.922155
> -4.183578
> 2008-03-01 2008-04-01 2008-05-01 2008-06-01 2008-07-01 2008-08-01
> 2008-09-01
> -4.336969 -3.813306 -4.296579 -4.575095 -4.036036 -4.727994
> -4.347428
> 2008-10-01 2008-11-01 2008-12-01
> -4.029918 -4.260326 -4.454224
>
> But the normal format I wish to display only appears on the terminal,
> leading me to copy it and paste into a text file. That is, when I enter AB
> on the terminal, it returns a format in the form:
>
> 008-02-01 -4.183578
> 2008-03-01 -4.336969
> 2008-04-01 -3.813306
> 2008-05-01 -4.296579
> 2008-06-01 -4.575095
> 2008-07-01 -4.036036
> 2008-08-01 -4.727994
> 2008-09-01 -4.347428
> 2008-10-01 -4.029918
> 2008-11-01 -4.260326
> 2008-12-01 -4.454224
>
> Now, my question: How do I write out two columns displayed by AB on the
> terminal to a file?
>
> I have tried using AB<-data.frame(AB) but it doesn't work either.
>
> Many thanks for your time.
> Ogbos
>
> [[alternative HTML version deleted]]
>
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Hello,
The main trick is to pipe to as.data.frame. But the result will have one
column only, you must assign the dates from the df's row names.
I also include an aggregate solution.
# create a test data set
set.seed(2024)
data <- data.frame(
Date = sample(seq(Sys.Date() - 5, Sys.Date(), by = "1 days"), 100L,
TRUE),
count = sample(10L, 100L, TRUE)
)
# coerce tapply's result to class "data.frame"
res <- with(data, tapply(count, Date, mean)) |> as.data.frame()
# assign a dates column from the row names
res$Date <- row.names(res)
# cosmetics
names(res)[2:1] <- names(data)
# note that the row names are still tapply's names vector
# and that the columns order is not Date/count. Both are fixed
# after the calculations.
res
#> count Date
#> 2024-03-22 5.416667 2024-03-22
#> 2024-03-23 5.500000 2024-03-23
#> 2024-03-24 6.000000 2024-03-24
#> 2024-03-25 4.476190 2024-03-25
#> 2024-03-26 6.538462 2024-03-26
#> 2024-03-27 5.200000 2024-03-27
# fix the columns' order
res <- res[2:1]
# better all in one instruction
aggregate(count ~ Date, data, mean)
#> Date count
#> 1 2024-03-22 5.416667
#> 2 2024-03-23 5.500000
#> 3 2024-03-24 6.000000
#> 4 2024-03-25 4.476190
#> 5 2024-03-26 6.538462
#> 6 2024-03-27 5.200000
Also,
I'm glad to help as always but Ogbos, you have been an R-Help
contributor for quite a while, please post data in dput format. Given
the problem the output of the following is more than enough.
dput(head(data, 20L))
Hope this helps,
Rui Barradas
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