On Dec 15, 2011, at 02:51 , Hervé Pagès wrote:
Hi Peter,
On 11-12-14 08:19 AM, peter dalgaard wrote:
On Dec 14, 2011, at 16:19 , John C Nash wrote:
Following this thread, I wondered why nobody tried cumsum to see where the
integer
overflow occurs. On the shorter xx vector in the
peter dalgaard pda...@gmail.com
on Thu, 15 Dec 2011 11:40:23 +0100 writes:
On Dec 15, 2011, at 02:51 , Hervé Pagès wrote:
Hi Peter,
On 11-12-14 08:19 AM, peter dalgaard wrote:
On Dec 14, 2011, at 16:19 , John C Nash wrote:
Following this
Hi Duncan,
On 11-12-14 03:57 AM, Duncan Murdoch wrote:
On 11-12-13 6:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-10 05:27 AM, Duncan Murdoch wrote:
On 11-12-09 4:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-09 11:39 AM, Duncan Murdoch wrote:
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
On 11-12-13 6:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-10 05:27 AM, Duncan Murdoch wrote:
On 11-12-09 4:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-09 11:39 AM, Duncan Murdoch wrote:
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
Hi,
x- c(rep(180003L, 1000), -rep(120002L,
Following this thread, I wondered why nobody tried cumsum to see where the
integer
overflow occurs. On the shorter xx vector in the little script below I get a
message:
Warning message:
Integer overflow in 'cumsum'; use 'cumsum(as.numeric(.))'
But sum() does not give such a warning, which I
On Dec 14, 2011, at 16:19 , John C Nash wrote:
Following this thread, I wondered why nobody tried cumsum to see where the
integer
overflow occurs. On the shorter xx vector in the little script below I get a
message:
Warning message:
Integer overflow in 'cumsum'; use
I agree that where the overflow occurs is not critical (one can go back to
cumsum and find
out). I am assuming that Uwe still wants to know there has been an overflow at
some point
i.e., a warning. This could become more interesting as parallel computation
causes
different summation orderings
On 14.12.2011 22:16, John C Nash wrote:
I agree that where the overflow occurs is not critical (one can go back to
cumsum and find
out). I am assuming that Uwe still wants to know there has been an overflow at
some point
i.e., a warning.
Yes, sure.
Uwe
This could become more
Hi Peter,
On 11-12-14 08:19 AM, peter dalgaard wrote:
On Dec 14, 2011, at 16:19 , John C Nash wrote:
Following this thread, I wondered why nobody tried cumsum to see where the
integer
overflow occurs. On the shorter xx vector in the little script below I get a
message:
Warning message:
[See at end]
On 13-Dec-11 23:41:12, Hervé Pagès wrote:
Hi Duncan,
On 11-12-10 05:27 AM, Duncan Murdoch wrote:
On 11-12-09 4:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-09 11:39 AM, Duncan Murdoch wrote:
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
Hi,
x- c(rep(180003L, 1000),
Hi Ted,
On 11-12-13 04:52 PM, (Ted Harding) wrote:
[...]
Now, computer programs for numerical computation can broadly
be divided into two types.
In one, arbitrary precision is available: you can tell
the program how many decimal digits you want it to work to.
An example of this is 'bc':
On 11-12-09 4:41 PM, Hervé Pagès wrote:
Hi Duncan,
On 11-12-09 11:39 AM, Duncan Murdoch wrote:
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
Hi,
x- c(rep(180003L, 1000), -rep(120002L, 1500))
This is correct:
sum(as.double(x))
[1] 0
This is not:
sum(x)
[1] 4996000
Hi,
x - c(rep(180003L, 1000), -rep(120002L, 1500))
This is correct:
sum(as.double(x))
[1] 0
This is not:
sum(x)
[1] 4996000
Returning NA (with a warning) would also be acceptable for the latter.
That would make it consistent with cumsum(x):
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
Hi,
x- c(rep(180003L, 1000), -rep(120002L, 1500))
This is correct:
sum(as.double(x))
[1] 0
This is not:
sum(x)
[1] 4996000
Returning NA (with a warning) would also be acceptable for the latter.
That would
Hi Duncan,
On 11-12-09 11:39 AM, Duncan Murdoch wrote:
On 09/12/2011 1:40 PM, Hervé Pagès wrote:
Hi,
x- c(rep(180003L, 1000), -rep(120002L, 1500))
This is correct:
sum(as.double(x))
[1] 0
This is not:
sum(x)
[1] 4996000
Returning NA (with a warning) would also be
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