Santosh, Ivan,
This is also what I was looking for. Thanks. Looking at the source of
dataFrame.default is seems that it uses the same approach as I did:
first create a list then a data.frame from that list. I think I'll stick
with the code I already had as I don't want another dependency (mult
Forget this last email, I oversaw the implementation in the examples...
Ivan
Le 5/16/2011 11:35, Ivan Calandra a écrit :
Actually, what would be even better would be an extra argument to
specify the column names.
I don't think it's very difficult to implement and it would make
things even easi
Actually, what would be even better would be an extra argument to
specify the column names.
I don't think it's very difficult to implement and it would make things
even easier.
Ivan
Le 5/16/2011 11:25, Ivan Calandra a écrit :
Thanks Santosh!
The more I learn about R.utils, the more I think tha
Thanks Santosh!
The more I learn about R.utils, the more I think that many of its
functions should be included in the base distribution.
Ivan
Le 5/16/2011 10:42, Santosh Srinivas a écrit :
Hi Ivan, Take a look dataFrame in R.utils ... is that what you want?
from the help file:
Examples
d
Hi Ivan, Take a look dataFrame in R.utils ... is that what you want?
from the help file:
Examples
df <- dataFrame(colClasses=c(a="integer", b="double"), nrow=10)
df[,1] <- sample(1:nrow(df))
df[,2] <- rnorm(nrow(df))
print(df)
Thanks,
Santosh
On Mon, May 16, 2011 at 1:42 PM, Ivan Calan
I feel like I'm always asking this type of questions, but is it possible
to add a base function that allows creating an empty data.frame, as
matrix() does?
What I mean would be something like:
create.data.frame(number_of_columns, mode_of_columns).
I think it would make things easier than creati
Inline below.
On Sun, May 15, 2011 at 11:11 AM, Jan van der Laan wrote:
> Thanks. I also noticed myself minutes after sending my message to the list.
> My 'please ignore my question it was just a stupid typo' message was sent
> with the wrong account and is now awaiting moderation.
>
> However, m
Thanks. I also noticed myself minutes after sending my message to the
list. My 'please ignore my question it was just a stupid typo' message
was sent with the wrong account and is now awaiting moderation.
However, my other question still stands: what is the
preferred/fastest/simplest way to cr
In your post, you're missing the final "s" on the stringsAsFactors
argument in the d1 assignment. When I typed it correctly, it works as
expected.
-- Bert
On Sun, May 15, 2011 at 4:25 AM, Jan van der Laan wrote:
> I use the following code to create two data.frames d1 and d2 from a list:
> types
I use the following code to create two data.frames d1 and d2 from a list:
types <- c("integer", "character", "double")
nlines <- 10
d1 <- as.data.frame(lapply(types, do.call, list(nlines)),
stringsAsFactor=FALSE)
l2 <- lapply(types, do.call, list(nlines))
d2 <- as.data.frame(l2, s
Forget I asked. There was a typo in my example (stringsAsFactor
instead of stringAsFactors) which explained the difference. My
apologies.
My second question however still stands: How does on create a
data.frame with given column types and given dimensions? Thanks.
Regards,
Jan
Quoting J
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