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AminHP avatar AminHP commented on May 31, 2024

Hi @sadimoodi ,

I don't see the problem.

>>> round(2.545 / 0.01, 6).is_integer()
False
>>> round(2.54 / 0.01, 6).is_integer()
True

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sadimoodi avatar sadimoodi commented on May 31, 2024

then why not just use:

v =  round(v ,2)

This will ensure that the result is a mutiplication of 0.01 without the need for the above validation, you are expecting a two decimal point float in _get_modified_volume because you use:

v = round(v / si.volume_step) * si.volume_step

isnt the above approach simpler and faster?

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AminHP avatar AminHP commented on May 31, 2024

Because volume_step is not always 0.01. It is variable and I wanted to support all cases.

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sadimoodi avatar sadimoodi commented on May 31, 2024

in your pickle files all the instruments where having 0.01 as volume step but i understand where you come from.
in my case i am building a crypto trading agent which does not use contract size nor volumes to trade, rather amounts and leverage, so this step was not needed in my case, i just round to 2 decimal points (simpler)

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AminHP avatar AminHP commented on May 31, 2024

Hmm, I see. So it's not a bug. Anyway, I don't think it has a huge impact on execution speed.

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