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Updating input parameter over iteration
Hi,
I was wondering if there’s a way to run a function over x number of iterations and also update a second value based on the function’s output.
For e.g.
summer:{[ID;y;z] y:y+z; ID:ID+1}
{summer[x;y;z]}[;2;2]/[{x<750};0]Currently this runs for 750 iterations however y and z are set to 2 on input so y always equals 4, I have checked this using namespaces.
Is there a way to include another over somewhere so that y will run like 2+2=4, 4+2=6… for 750 iterations?
Any help appreciated!!
PS. The code is being used within peach, that’s why I can’t just use namespaces or a global variable to upsert into.
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5 Replies
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Something more like:
q)750{(x[0]+1;x[1]+x[2];x[2])}/0
0 2 750 1500 2
Or use a dictionary for readability:
q)750{x[`ID]+:1;x[`y]:sum x`y`z;x}/`ID`y`z!0 0 2
ID| 750
y | 1500
z | 2
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Is the
/for ofdowhat you are looking for?https://code.kx.com/q/ref/accumulators/#do
q)750 +/2 2 754
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Hey, thanks for the reply.
So I think the above is quite close however it only allows you run sum, not appending to iteration id as required.
Following similar to above:
q)summer1:{[x;y] .m.x:x; x:x+y}q)750 summer1[;2]/2;
q).m.x
1500I’m probably missing something obvious but is there a way to manipulate this to append to ID per iteration so that:
q)summer:{[ID;y;z] y:y+z;.m.id:.m.id,ID;.m.y:y; ID:ID+1}q).m.y
1500 (as above)
q).m.ID
750 (missing piece)
Thanks again!
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The very one.
Fair play and thanks!
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See article on the accumulator iteration operators:
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