This change is breaking, use the 'bool' modifier for such comprisons.
After this change all comparisons without 'bool' will filter, and all
comparisons with 'bool' will return 0/1. This makes the language more
consistent and orthogonal, and ultimately easier to learn and use.
If we ever figure out sane semantics for filtering scalar/scalar
comparisons we can add them in, which will most likely come out of how
the new vector() function is used.
This change is breaking, use increase() instead.
I'm not cleaning up the function in this PR, as my solution to #581 will
rewrite and simplify increase/rate/delta.
irate is a rate function that only looks at the most
recent two data points, and calucaltes a per-second value
from that. This produces much more granular graphs for
fast moving data, and works sanely across many scrape intervals.
It doesn't do so well for slowly moving data.
Currently the only way to convert a scalar to a vector is to
use absent(), which isn't very clean. This adds a vector()
function that's the inverse of scalar() and lets your optionally
set labels.
Example usage would be
vector(time() % 86400) < 3600
to filter to only the first hour of the day.
When doing comparison operations on vectors, filtering
sometimes gets in the way and you have to go to a fair bit of
effort to workaround it in order to always return a result.
The 'bool' modifier instead of filtering returns 0/1 depending
on the result of the compairson.
This is also a prerequisite to removing plain scalar/scalar comparisons,
as it maintains the current behaviour under a new syntax.
`keep_common` is more in line with the function name
`drop_common_labels()` terminology-wise, and also more in line with
`group_left`/`group_right` (no `...ing` verb suffix).
We could also go the full way and call it `keep_common_labels`. That
would have the benefit of being even more consistent with the function
`drop_common_labels()` and would be more explanatory, but it also seems
quite long.
This calculates how much a counter increases over
a given period of time, which is the area under the curve
of it's rate.
increase(x[5m]) is equivilent to rate(x[5m]) * 300.
Fabian Reinartz