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Sign upArithmetic in aggregations #622
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grobie
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Apr 3, 2015
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That expression has unbalanced brackets. You can already mostly do that via:
What you want is a way for the filtering functions to return 1/0 instead of filtering, and then an aggregation function to group things by value. Then it'd look something like:
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See #1049 |
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We still need a |
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Actually |
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Hmm, this will need to be passed a parameter to know what the output labelname is. We don't currently have those for aggregators. Possible syntaxes:
@fabxc Thoughts? |
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In general I expect most people being happy with "value" and it would keep the language simpler. Then again So we might just as well allow parameters for
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The current reason why |
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Another option is
which makes more sense in the topk/bottomk case. |
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Makes sense and would also not make the current expressions invalid. |
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Yeah, I think the last one is probably the best. |
fabxc
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Apr 28, 2016
brian-brazil
added this to the v1.0.0 milestone
May 2, 2016
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@brian-brazil you flagged this for 1.0 – will this be breaking? |
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This isn't breaking, just a new aggregator. |
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Can we remove the 1.0 milestone then? I'd like to focus on getting that On Wed, Jun 8, 2016 at 12:16 PM Brian Brazil notifications@github.com
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Removing from the milestone – cool if we get it in but no requirement. |
fabxc
removed this from the v1.0.0 milestone
Jun 24, 2016
brian-brazil
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Jul 5, 2016
fabxc
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Jul 8, 2016
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lock
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Mar 24, 2019
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This thread has been automatically locked since there has not been any recent activity after it was closed. Please open a new issue for related bugs. |
grobie commentedApr 3, 2015
•
edited
I come across use cases from time to time where I'd love being able to do some arithmetic in by clauses of aggregations. For example:
In this example (with an exemplary syntax) I want to plot the number of services below and above a certain threshold. It's possible to construct graphs roughly showing the desired data with at least promdash, but it'd be convenient to have first class support for that. For me at least ;), what do people think?
@brian-brazil @beorn7