Hi,
I'm using version 0.5.1 of tm package with R 2.10.1. It looks to me
as if after the following
reuters21578 <- Corpus(DirSource(corpusDir), readerControl =
list(reader = readReut21578XMLasPlain))
reuters21578 <- tm_map(reuters21578, stripWhitespace)
reuters21578 <- tm_map(reuters
Hi,
I have a list of vectors (of varying lengths). I'd like to sort this
list by applying a function to each pair of vectors in the list and
returning information to sorting routine that let's it know which one
is larger.
To solve problems like this in Common Lisp, the sort function accepts
a fu
o pairs v1 and v3, v1 and v4, and so on?
>
> Which one is larger, mylist$v1 or mylist$v2? Longer, yes, mylist$v2 is
> longer. But larger?
>
> And ultimately you want to have the list with its elements in some other
> order,
> perhaps v4 comes first, then v3, and so on?
>
On Sat, Feb 6, 2010 at 3:22 PM, Hans W Borchers
wrote:
> David Neu davidneu.com> writes:
>
>>
>> Hi,
>>
>> I have a list of vectors (of varying lengths). I'd like to sort this
>> list by applying a function to each pair of vectors in the list and
>
On Sat, Feb 6, 2010 at 6:30 PM, Hans W Borchers
wrote:
> David Neu davidneu.com> writes:
>
>> David Neu davidneu.com> writes:
>>
>> Hi,
>>
>> I have a list of vectors (of varying lengths). I'd like to sort this
>> list by applying a function
Hi,
Is there a way to create a matrix in which the column names are not
checked to see if they are valid variable names?
I'm looking something similar to the check.names argument to
data.frame. If so, would such an approach work for the sparse matrix
classes in the Matrix package.
Many thanks!
Hi,
I have two (large) matrices A and B of dimensions (m,n) and (p,n) respectively.
I'd like to see if the is a fast way to compute a new matrix C with
dimension (m*p,n) in which each row in C is found by applying some
function f to each pair of rows (x,y) where x is a row in A and y is a
row in
Thanks to everyone who responded!
On Sun, May 23, 2010 at 1:32 AM, Jorge Ivan Velez
wrote:
> Yes, it should be. Thank you for pointing that out. Apologies for the noise.
>
> Regards,
> Jorge
>
>
> On Sun, May 23, 2010 at 1:07 AM, Berend Hasselman <> wrote:
>
>>
>>
>> Jorge Ivan Velez wrote:
>> >
Hi,
I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically. Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.
library(lattice);
x <- rnorm(100);
y <- as.factor(sapply(1:100, function(k) sample(c("
On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius wrote:
>
> On Apr 23, 2011, at 9:26 AM, David Neu wrote:
>
>> Hi,
>>
>> I'd like to change the default orientation of bwplot() and stripplot()
>> so the plots are displayed vertically. Passing horizontal=FALS
On Sat, Apr 23, 2011 at 10:23 AM, David Winsemius
wrote:
>
> On Apr 23, 2011, at 10:13 AM, David Neu wrote:
>
>> On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius
>> wrote:
>>>
>>> On Apr 23, 2011, at 9:26 AM, David Neu wrote:
>>>
>>>>
On Sat, Apr 23, 2011 at 10:37 AM, Peter Ehlers wrote:
> On 2011-04-23 07:13, David Neu wrote:
>>
>> On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius
>> wrote:
>>>
>>> On Apr 23, 2011, at 9:26 AM, David Neu wrote:
>>>
>>>> Hi,
>>
Hi,
The rows of a data frame can be sorted lexicographically as shown in
this example,
my.df <- data.frame(x=c(1,1,1,2,2), y=c(1,2,3,2,1))
my.df[order(my.df$x,my.df$y, decreasing=TRUE), ]
however, I'm wondering if it's possible to pass a variable in as the
first argument to order() so that the a
On Sat, Apr 23, 2011 at 7:06 PM, Duncan Murdoch
wrote:
> On 11-04-23 2:00 PM, David Neu wrote:
>>
>> Hi,
>>
>> The rows of a data frame can be sorted lexicographically as shown in
>> this example,
>>
>> my.df<- data.frame(x=c(1,1,1,2,2), y=c(1,2,3,
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