On 7 Apr 2004, Peter Dalgaard wrote:
Robin Hankin [EMAIL PROTECTED] writes:
Hi again everybody.
Yesterday I had a problem with c-style += functions. One suggestion
was to define
R plus- - function(x,value){x+value}
Then
R a - matrix(1:9,3,3)
R plus(a[a%%2==1]) - 1000
Prof Brian Ripley [EMAIL PROTECTED] writes:
This *should* be somewhere in the R Language Definition, although I'm
not sure it is actually there. Or the blue book, although I suspect
that S v.3 actually used positional matching (and jumped through
hoops).
It did. This is in the FAQ,
Hi again everybody.
Yesterday I had a problem with c-style += functions. One suggestion
was to define
R plus- - function(x,value){x+value}
Then
R a - matrix(1:9,3,3)
R plus(a[a%%2==1]) - 1000
works as desired.
QUESTION: why does this behave differently:
R plus- - function(x,y){x+y}
R a -
Value in an internal variable of a command function,
have a look on R-help apropos function.
A.S.
Alessandro Semeria
Models and Simulations Laboratory
Montecatini Environmental Research Center (Edison Group),
Via Ciro Menotti 48,
48023 Marina di Ravenna (RA),
Robin Hankin [EMAIL PROTECTED] writes:
Hi again everybody.
Yesterday I had a problem with c-style += functions. One suggestion
was to define
R plus- - function(x,value){x+value}
Then
R a - matrix(1:9,3,3)
R plus(a[a%%2==1]) - 1000
works as desired.
QUESTION: why does this
On Tue, 6 Apr 2004 14:18:42 + (UTC), you wrote:
Robin Hankin rksh at soc.soton.ac.uk writes:
R a - matrix(1:9,3,3)
But the following caught me off-guard:
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% 1000*(1:5)
How about this way:
a - matrix(1:9,3,3)
plus- - function(a,value) a+value
Duncan Murdoch [EMAIL PROTECTED] writes:
On Tue, 6 Apr 2004 14:18:42 + (UTC), you wrote:
Robin Hankin rksh at soc.soton.ac.uk writes:
R a - matrix(1:9,3,3)
But the following caught me off-guard:
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% 1000*(1:5)
How about this way:
a -
On 07 Apr 2004 13:51:52 +0200, Peter Dalgaard
[EMAIL PROTECTED] wrote :
Duncan Murdoch [EMAIL PROTECTED] writes:
Wouldn't a clearer syntax to follow be Pascal's? I.e.
inc(x[3], 1)
But that's not kosher in functional programming languages where you
expect function calls not to modify their
Some time ago, Peter Dalgaard made the wonderful suggestion to define
%+=% - function(a,b) {eval.parent(substitute(a - a + b)) }
Which I use (R-1.8.1) as follows:
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% (1000*(1:5))
R a
[,1] [,2] [,3]
[1,] 10014 4007
[2,]2 30058
[3,] 20036
Robin Hankin [EMAIL PROTECTED] writes:
Some time ago, Peter Dalgaard made the wonderful suggestion to define
%+=% - function(a,b) {eval.parent(substitute(a - a + b)) }
...
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% 1000*(1:5)
[1] 1001 2006 3015 4028 5045
R a
[,1] [,2] [,3]
[1,] 1001
On Tue, 6 Apr 2004 13:36:32 +, Robin Hankin [EMAIL PROTECTED]
wrote :
the following caught me off-guard:
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% 1000*(1:5)
[1] 1001 2006 3015 4028 5045
R a
[,1] [,2] [,3]
[1,] 10014 1007
[2,]2 10058
[3,] 10036 1009
R
Why the
Look at Syntax on R-help apropos operators precedence
A.S.
Alessandro Semeria
Models and Simulations Laboratory
Montecatini Environmental Research Center (Edison Group),
Via Ciro Menotti 48,
48023 Marina di Ravenna (RA), Italy
Tel. +39 544 536811
Fax. +39 544
Robin Hankin rksh at soc.soton.ac.uk writes:
R a - matrix(1:9,3,3)
But the following caught me off-guard:
R a - matrix(1:9,3,3)
R a[a%%2==1] %+=% 1000*(1:5)
How about this way:
a - matrix(1:9,3,3)
plus- - function(a,value) a+value
plus(a[a%%2==1]) - 1000*(1:5)
a
[,1] [,2] [,3]
[1,]
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