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2 weeks ago

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. 🙂

1 ACCEPTED SOLUTION

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2 weeks ago

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
```

5 REPLIES 5

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2 weeks ago

Is the `/`

for of `do`

what you are looking for?

https://code.kx.com/q/ref/accumulators/#do

```
q)750 +/2 2
754
```

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2 weeks ago - last edited 2 weeks ago

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

1500

I'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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2 weeks ago

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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2 weeks ago

The very one. 🙂

Fair play and thanks!

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2 weeks ago

See article on the accumulator iteration operators:

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