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Re: [PATCH v6 14/15] scripts/simplebench: improve ascii table: add diffe
From: |
Vladimir Sementsov-Ogievskiy |
Subject: |
Re: [PATCH v6 14/15] scripts/simplebench: improve ascii table: add difference line |
Date: |
Fri, 25 Sep 2020 20:13:36 +0300 |
User-agent: |
Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Thunderbird/78.2.2 |
25.09.2020 13:24, Max Reitz wrote:
On 18.09.20 20:19, Vladimir Sementsov-Ogievskiy wrote:
Performance improvements / degradations are usually discussed in
percentage. Let's make the script calculate it for us.
Signed-off-by: Vladimir Sementsov-Ogievskiy <vsementsov@virtuozzo.com>
---
scripts/simplebench/simplebench.py | 46 +++++++++++++++++++++++++++---
1 file changed, 42 insertions(+), 4 deletions(-)
diff --git a/scripts/simplebench/simplebench.py
b/scripts/simplebench/simplebench.py
index 56d3a91ea2..0ff05a38b8 100644
--- a/scripts/simplebench/simplebench.py
+++ b/scripts/simplebench/simplebench.py
[...]
+ for j in range(0, i):
+ env_j = results['envs'][j]
+ res_j = case_results[env_j['id']]
+
+ if 'average' not in res_j:
+ # Failed result
+ cell += ' --'
+ continue
+
+ col_j = chr(ord('A') + j)
+ avg_j = res_j['average']
+ delta = (res['average'] - avg_j) / avg_j * 100
I was wondering why you’d subtract, when percentage differences usually
mean a quotient. Then I realized that this would usually be written as:
(res['average'] / avg_j - 1) * 100
+ delta_delta = (res['delta'] + res_j['delta']) / avg_j * 100
Why not use the new format_percent for both cases?
because I want less precision here
+ cell += f' {col_j}{round(delta):+}±{round(delta_delta)}%'
I don’t know what I should think about ±delta_delta. If I saw “Compared
to run A, this is +42.1%±2.0%”, I would think that you calculated the
difference between each run result, and then based on that array
calculated average and standard deviation.
Furthermore, I don’t even know what the delta_delta is supposed to tell
you. It isn’t even a delta_delta, it’s an average_delta.
not avarage, but sum of errors. And it shows the error for the delta
The delta_delta would be (res['delta'] / res_j['delta'] - 1) * 100.0.
and this shows nothing.
Assume we have = A = 10+-2 and B = 15+-2
The difference is (15-10)+-(2+2) = 5+-4.
And your formula will give (2/2 - 1) *100 = 0, which is wrong.
Anyway, my code is mess)
And that might be presented perhaps like “+42.1% Δ± +2.0%” (if delta
were the SD, “Δx̅=+42.1% Δσ=+2.0%” would also work; although, again, I do
interpret ± as the SD anyway).
I feel that I'm bad in statistics :( I'll learn a little and make a new version.
--
Best regards,
Vladimir
- Re: [PATCH v6 10/15] iotests: qemu_io_silent: support --image-opts, (continued)