Prometheus 速率函数如何计算结果?
我看到这个关于利率的问题 我不明白考试的结果Prometheus 速率函数如何计算结果?,prometheus,Prometheus,我看到这个关于利率的问题 我不明白考试的结果 rate(counter[2s]): will get the average from the increment in 2 sec and distribute it among the seconds So in the first 2 second we got an increment of total 3 which means the average is 1.5/sec. final result: second result 1
rate(counter[2s]): will get the average from the increment in 2 sec and distribute it among the seconds
So in the first 2 second we got an increment of total 3 which means the average is 1.5/sec. final result:
second result
1 1,5
2 1,5
3 2
4 2
5 3,5
6 3,5
7 3,5
8 3,5
9 4,5
10 4,5
及
你能给我解释一下逻辑吗?我想问题中的OP只是算错了。这个问题的答案非常清楚和详细地解释了它,提供了使用的确切公式。还举了一个例子(在评论中): 如果
t1
为1,t5
为5;而t1
处的值为1,t5
处的值为10(如上所述),则(v5-v1)/(t5-t1)
为(10-1)/(5-1)
,即2.25
这显然与原始示例第5秒时的速率(计数器[5s])相矛盾,即2。还请注意,在回答中,第1节的ICH应为0
现在,我将尝试手动计算:
rate(counter[5s]): (v_y - v_x) / (t_y - t_x)
second result calculation
1 N/A (1 - 1) / (1 - 1)
2 2 (3 - 1) / (2 - 1)
3 2.5 (6 - 1) / (3 - 1)
4 2 (7 - 1) / (4 - 1)
5 2.25 (10 - 1) / (5 - 1)
6 2.75 (14 - 3) / (6 - 2)
7 2.75 (17 - 6) / (7 - 3)
8 3.5 (21 - 7) / (8 - 4)
9 3.75 (25 - 10) / (9 - 5)
10 4 (30 - 14) / (10 - 6)
rate(counter[5s]): will get the average from the increment in 5 sec and distribute it among the seconds
The same as for [2s] but we calculate the average from total increment of 5sec. final result:
second result
1 2
2 2
3 2
4 2
5 2
6 4
7 4
8 4
9 4
rate(counter[5s]): (v_y - v_x) / (t_y - t_x)
second result calculation
1 N/A (1 - 1) / (1 - 1)
2 2 (3 - 1) / (2 - 1)
3 2.5 (6 - 1) / (3 - 1)
4 2 (7 - 1) / (4 - 1)
5 2.25 (10 - 1) / (5 - 1)
6 2.75 (14 - 3) / (6 - 2)
7 2.75 (17 - 6) / (7 - 3)
8 3.5 (21 - 7) / (8 - 4)
9 3.75 (25 - 10) / (9 - 5)
10 4 (30 - 14) / (10 - 6)