在 Python 中使用循环组合

Question

I am trying to write a loop that can take all the different combinations of ON_PEAK, MID_PEAK and OFF_PEAK consumption combinations, which will result in the sum of 114.12 of the below-mentioned equation.

I also want those combinations to be appended to a dictionary.

In the dictionary, the key = (SUM of ON_PEAK, MID_PEAK, OFF_PEAK) and the values to be the list of combinations of On_Peak, MID_PEAK and OFF_PEAK combinations. I am new to python and struggle with loops. I appreciate whoever can provide help.

Please see the example below:

d = {}
on_kwh = range(1, 5001, 1)
mid_kwh = range(1, 5001, 1)
off_kwh = range(1, 5001, 1)


if 0.101*(on_kwh) + 0.065(off_kwh) + 0.14(mid_kwh) = 114.12:
    d[on_kwh+mid_kwh+off_kwh].append[on_kwh,off_kwh,mid_kwh] 
    
print(d)


output (Example:1 Combination)
{1000,[150,150,700]}

*Output is based on this logic: 
0.101(150) + 0.0065(150) + 0.14(700) = 114.12

Answer 1

you could try a brute force approach, testing all 5000^3 (125'000'000'000) possible combinations of on_kwh, mid_kwh and off_kwh. a better approach would be to do some pruning and to abort your loops, trying a new combination, as soon as the sum exceeds your target value. eg

d = dict()
target = 114.12
for on_kwh in range(1, 5001, 1):
    on = on_kwh * 0.101
    if on > target: # can abort here
        break
    for mid_kwh in range(1, 5001, 1):
        mid = mid_kwh * 0.14
        if on + mid > target: # can abort here
            break
        for off_kwh in range(1, 5001, 1):
            off = off_kwh * 0.065
            if on + mid + off > target: # can abort here
                break
            if on + mid + off == target:
                d.setdefault(on_kwh + mid_kwh + off_kwh, []).append([on_kwh, mid_kwh, off_kwh])

print(d) 

this will "only" test 269'348'957 possible combinations.

a yet better/faster approach - in theory - would be to drop the last loop entirely: to be a solution for on_kwh * 0.101 + mid_kwh * 0.14 + off_kwh * 0.065 = 114.12 , off_kwh must be off_kwh = (114.12 - on_kwh * 0.101 - mid_kwh * 0.14) / 0.065 for any given pair of values on_kwh and mid_kwh. if off_kwh is an integer in (1, 5000), you found a solution. that being said, the speed up is significant, but you will run into some numeric issues due to working with floating point numbers: you'd have to define some precision up to which level rounding errors are acceptable...