问题
I'm working with some Date
columns and trying to cleanse for obviously incorrect dates. I've written a function using the safe.ifelse
function mentioned here.
Here's my toy data set:
df1 <- data.frame(id = 1:25
, month1 = seq(as.Date('2012-01-01'), as.Date('2014-01-01'), by = 'month' )
, month2 = seq(as.Date('2012-01-01'), as.Date('2014-01-01'), by = 'month' )
, month3 = seq(as.Date('2012-01-01'), as.Date('2014-01-01'), by = 'month' )
, letter1 = letters[1:25]
)
This works fine for a single column:
df1$month1 <- safe.ifelse(df1$month1 > as.Date('2013-10-01'), as.Date('2013-10-01'), df1$month1)
As I have multiple columns I'd like to use a function and apply to take care of all Date
columns at once:
capDate <- function(x){
today1 <- Sys.Date()
safe.ifelse <- function(cond, yes, no){ class.y <- class(yes)
X <- ifelse(cond,yes,no)
class(X) <-class.y; return(X)}
x <- safe.ifelse(as.Date(x) > as.Date(today1), as.Date(today1), as.Date(x))
}
However when I try to use sapply()
df1[,dateCols1] <- sapply(df1[,dateCols1], capDate)
or apply()
df1[,dateCols1] <- apply(df1[,dateCols1],2, capDate))
the Date
columns lose their Date
formatting. The only way I've found to get around this is by using lapply()
and then converting back to a data.frame()
. Can anyone explain this?
df1[,dateCols1] <- as.data.frame(lapply(df1[,dateCols1], capDate))
回答1:
Both sapply
and apply
convert the result to matrices. as.data.frame(lapply(...))
is a safe way to loop over data frame columns.
as.data.frame(
lapply(
df1,
function(column)
{
if(inherits(column, "Date"))
{
pmin(column, Sys.Date())
} else column
}
)
)
It's a little cleaner with ddply
from plyr
.
library(plyr)
ddply(
df1,
.(id),
colwise(
function(column)
{
if(inherits(column, "Date"))
{
pmin(column, Sys.Date())
} else column
}
)
)
来源:https://stackoverflow.com/questions/19515692/r-sapply-vs-apply-vs-lapply-as-data-frame