Open lalala215 opened 2 years ago
Yes, the reverse should be False. Thank you for pointing out the error. I will check the code recently.
OK! And I also have other two questions:
this line 38 no need for indentation https://github.com/wsyCUHK/Reinforcement-Learning-for-Real-time-Pricing-and-Scheduling-Control-in-EV-Charging-Stations/blob/cbd4c8b3be01f779bb2c7e9fad9cc2c822b49442/code/HSA.py#L36-L43
In the paper feature2(Eq.(15)) is electricity price multiply total charging rate f{2}\left(S{t}, A{t}\right)=-c{t} e_{t} However in line 115 the feature2 is action[1], i.e. the total charging rate. https://github.com/wsyCUHK/Reinforcement-Learning-for-Real-time-Pricing-and-Scheduling-Control-in-EV-Charging-Stations/blob/cbd4c8b3be01f779bb2c7e9fad9cc2c822b49442/code/HSA.py#L115
https://github.com/wsyCUHK/Reinforcement-Learning-for-Real-time-Pricing-and-Scheduling-Control-in-EV-Charging-Stations/blob/cbd4c8b3be01f779bb2c7e9fad9cc2c822b49442/code/HSA.py#L77-L83
Dear Author I am wondering If this order is right?
order=[operator.itemgetter(0)(t)-1 for t in sorted(enumerate(least,1), key=operator.itemgetter(1), reverse=True)]
residual_demand[:,1] is parking time residual_demand[:,0] is charging demand
order put the biggest in least as the most need to be charged. But the biggest in least means its parking time is much larger than charging demand, I thought in this case the biggest in least means it can wait more time, so cannot put it as the most urgent.
maybe the reserse should be False, such as:
order=[operator.itemgetter(0)(t)-1 for t in sorted(enumerate(least,1), key=operator.itemgetter(1), reverse=False)]