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## mmin

New Contributor
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
3 REPLIES 3
New Contributor
f/[2;x] means repeat f 2 times over x that is calculate min between every pair two times.

WBR, Andrey.

вторник, 1 декабря 2015 г., 11:40:52 UTC+3 пользователь effbiae написал:
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
New Contributor

Hi Jack,

Repeating the mmin function:

q)mmin

k){(x-1)&':/y}

Let’s say we try 3 mmin 1 5 3

If we break the expression down into two parts:

1) the min operator combined with the each prior adverb

2) 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 list 1 5 3 using    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)

1 1 3

The second iteration of the function returns
(min 0N 1; min 1 1; min 1 3)
1 1 1

For your original example, using 3 mmin til 6, as you say correctly translates to 2&’:/til 6

Or in terms of the minprev I’ve 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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New Contributor
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