On 5/17/07, kamakura <[EMAIL PROTECTED]> wrote:
Then, is the following syntactics wrong?
mean"3 u
In this case,
#$u
3
mean"3 u and mean u are the same.
if mean"1&.|:u is true syntactics. It seems to give the same result.
When we want to operate user defined function f, is f"1&.|:u correct
rule?
mean"1 means take the median of rank 1 values.
mean"1&.|: means, in essence, use the first dimension for these
rank 1 values, rather than the last.
In other words, the definition of "correct rule" depends on what
you want.
That said,
(mean -: mean"1&.|:) u
1
...given the definition of mean, this is not an issue. This is because
mean u is
(+/ % #) u
which is
(+/u) % #u
where
#u
3
and
+/u
is
(0{u) + (1{u) + (2{u) + 3{u
or
+/u
8 15 12 15 15
12 7 13 13 15
14 23 14 22 8
10 24 12 15 14
In other words, you are already adding atoms along the first
dimension, and dividing by the value of that dimension.
It's only for something like median where you get different results:
(median -: median"1&.|:)u
0
When you look at how median works, this should become obvious:
midpt u
1
1 { /:~ u
1 6 0 4 4
1 3 3 6 6
5 7 8 8 2
1 9 7 9 9
(and the rest of median's definition makes no difference in this
case.)
Basically, median is picking the median item and median item
is not what you are looking for.
--
Raul
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