Turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- KX Community
- :
- Discussion Forums
- :
- kdb+ and q
- :
- Updating input parameter over iteration

Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Mute
- Printer Friendly Page

Options

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 06:33 AM

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

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 08:16 AM

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

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 07:23 AM

Is the `/`

for of `do`

what you are looking for?

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

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

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 07:48 AM - edited 2022.05.11 07:49 AM

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!

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 08:16 AM

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
```

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.11 08:22 AM

The very one. 🙂

Fair play and thanks!

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Report Inappropriate Content

2022.05.12 04:03 AM

See article on the accumulator iteration operators:

Main Office Contacts

**EMEA**

Tel: +44 (0)28 3025 2242

**AMERICAS**

Tel: +1 (212) 447 6700

**APAC**

Tel: +61 (0)2 9236 5700

Useful Information

Resources

Popular Links

Follow Us

KX. All Rights Reserved.

KX and kdb+ are registered trademarks of KX Systems, Inc., a subsidiary of FD Technologies plc.