On 10/12/2015 6:12 AM, arnaud gaboury wrote:
Here is a sample of my data frame, obtained with read_csv2 from readr package.
myDf <- structure(list(X15 = c("30.09.2015", "05.10.2015", "30.09.2015",
"29.09.2015", "10.10.2015"), X16 = c("02.10.2015", "06.10.2015",
"01.10.2015", "01.10.2015", "13.10.2015"), X17 = c("Grains",
"Grains", "Grains", "Grains", "Grains"), X18 = c("Soyabeans",
"Soyabeans", "Soyabeans", "Soyabeans", "Soyabeans"), X19 = c("20,000",
"20,000", "20,000", "29,930", "26,000")), .Names = c("X15", "X16",
"X17", "X18", "X19"), class = c("tbl_df", "data.frame"), row.names = c(NA,
-5L))
gabx@hortensia [R] str(myDf)
Classes ‘tbl_df’ and 'data.frame': 5 obs. of 5 variables:
$ X15: chr "30.09.2015" "05.10.2015" "30.09.2015" "29.09.2015" ...
$ X16: chr "02.10.2015" "06.10.2015" "01.10.2015" "01.10.2015" ...
$ X17: chr "Grains" "Grains" "Grains" "Grains" ...
$ X18: chr "Soyabeans" "Soyabeans" "Soyabeans" "Soyabeans" ...
$ X19: chr "20,000" "20,000" "20,000" "29,930" ...
I want to change date to date class and numbers (X19) to numeric, and
keep the class of my object.
This code works:
myDf$X19 <- as.numeric(gsub(",", "", myDf$X19))
myDf$X15 <- as.Date(myDf$X15, format = "%d.%m.%Y"))
myDf$X16 <- as.Date(myDf$X16, format = "%d.%m.%Y"))
Now, as I have more than 5 columns, this can be fastidious and slowing
code (?), even if I can group by type. Columns are only types of char,
num and Date, so it could be OK.
I tried with lapply for the Date columns. It works BUT will place NA
in any columns with numbers as characters.
The reuslt will be this for X19: num NA NA NA NA NA NA NA NA NA NA ..
How can I target my goal with something else than lapply or writing a
line for each type ?
I don't see how a function could reliably detect the types, but it might
be good enough to use a regular expression, possibly just on the first
line of the result. Once you've identified columns, e.g.
numcols <- 19
datecols <- c(15:16)
etc, you can use lapply:
myDf[,numcols] <- lapply(myDf[, numcools, drop=FALSE], function(x)
as.numeric(gsub(",", "", x)))
You can simplify myDf[,numcols] to myDf[numcols] if you want, but I
think it makes it less clear.
Duncan Murdoch
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