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I saw (in use case and ion your code) ETA is obtained with time delta of one step, not the start time / n step * (number of step to finish)
I think it will be better, because ETA will be automatically a means not a value a t time, and more accurate. Are you agree?
I have replaced your code by: remaining=elapsed/(n-initial)*(total-n+initial) (before it was remaining = (total - n) / rate and rate= (n - initial) / elapsed, you forget to do total -(n-initial) to get the first step)
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Hi,
I saw (in use case and ion your code) ETA is obtained with time delta of one step, not the start time / n step * (number of step to finish)
I think it will be better, because ETA will be automatically a means not a value a t time, and more accurate. Are you agree?
I have replaced your code by:
remaining=elapsed/(n-initial)*(total-n+initial)
(before it wasremaining = (total - n) / rate
andrate= (n - initial) / elapsed
, you forget to dototal -(n-initial)
to get the first step)Beta Was this translation helpful? Give feedback.
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