I’ve been trying to grok mmin:
q)3 mmin til 6
0 0 0 1 2 3
q)mmin
k){(x-1)&':/y}
q)k)2&':/!6
0 0 0 1 2 3
which i translate to q:
q)2&':/til 6
so it’s &': with initial value 2 over til 6…
but i don’t see how 2 is a window?
what’s the key to understanding this code?
ta, jack
f/[2;x] means repeat f 2 times over x that is calculate min between every pair two times.
WBR, Andrey.
???, 1 ??? 2015 ?., 11:40:52 UTC+3 ??? effbiae ???:
I’ve been trying to grok mmin:
q)3 mmin til 6
0 0 0 1 2 3
q)mmin
k){(x-1)&':/y}
q)k)2&':/!6
0 0 0 1 2 3
which i translate to q:
q)2&':/til 6
so it’s &': with initial value 2 over til 6…
but i don’t see how 2 is a window?
what’s the key to understanding this code?
ta, jack
Hi Jack,
Repeating the mmin function:
q)mmin
k){(x-1)&':/y}
Lets say we try 3 mmin 1 5 3
If we break the expression down into two parts:
the min operator combined with the each prior adverb
the x f/y expression, meaning iterate y using f, by x times
For (1), we can combine verbs in brackets to make an adverb. In this case, combining min (&) and each priot (':)
q) minprev:(&':)
&: is taking each item of the right operand and applying it to its predecessor
Now for part (2), using our adverb minprev as the function, f, in the x f/y expression.
q)2 minprev\ 1 5 3
1 1 1
Iterate the list1 5 3using minprev 2 times
And if we change Over into Scan (\ into /) to get the intermediate results as well:
q)2 minprev\ 1 5 3
1 5 3
1 1 3
1 1 1
The original list is returned first: 1 5 3
The first iteration of the function returns
(min 0N 1; min 1 5; min 5 3)
113
The second iteration of the function returns
(min 0N 1; min 1 1; min 1 3)
111
For your original example, using 3 mmin til 6, as you say correctly translates to 2&:/til 6
Or in terms of the minprev Ive defined:
3 mmin til 6
becomes
q)2 minprev\ 0 1 2 3 4 5
0 1 2 3 4 5
0 0 1 2 3 4/ 1 iteration
0 0 0 1 2 3/ 2 iterations
Looking just at the final result with Over:
q)2 minprev/ 0 1 2 3 4 5
0 0 0 1 2 3
To calculate the 3, we look at the 3 4 5 (window size 3)
To calculate the 2, we look at the 2 3 4 (window size 3)
etc
The result only requires n-1 iterations for n sized window.
Thanks,
Matthew McAuley
KDB+ Developer, AquaQ Analytics Ltd
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So I thought the example of the k code for mmin/mmax might lead to finding the indexes of the moving maximum by the same pattern.
So I found:
q)immax:{(x-1){(y;z)(<). x@y,z}[y]':/til count y}
q)x:7 2 3 3 7
q)immax[3;x]
0 0 0 3 4
but it’s much slower than mmax:
q)\ts mmax[10;til 10000]
1 524576
q)\ts immax[10;til 10000]
84 750576
I then discovered that tampering with mmax - replacing | with {x|y} - has quite an effect on time:
q)\ts k){(x-1)|':/y}[10;!10000]
1 524816
q)\ts k){(x-1){x|y}':/y}[10;!10000]
29 619456
Can immax be relatively fast using the pattern of mmax?
[then the problem of finding first moving maximum index can be reduced to two reverses and immax]
Thanks,
Jack