Python实现零件分配至指定时隙的调度程序优化咨询

我目前正在开发一款可将零件分配至指定时隙的调度程序，以下是我基于while/for循环、if/else条件语句完成的初步实现代码，作为Python编程入门者，我想咨询是否存在更高效合理的实现方案，感谢各位的帮助。

import pandas as pd
from datetime import datetime, timedelta

day_start = str('2021-07-28 06:30')
break_1_start = str('2021-07-28 08:50')
lunch_start = str('2021-07-28 12:00')
break_2_start = str('2021-07-28 14:50')
day_end = str('2021-07-28 17:40')
day_start = datetime.strptime(day_start, "%Y-%m-%d %H:%M")
break_1_start = datetime.strptime(break_1_start, "%Y-%m-%d %H:%M")
break_1_end = break_1_start + timedelta(minutes=20)
lunch_start = datetime.strptime(lunch_start, "%Y-%m-%d %H:%M")
lunch_end = lunch_start + timedelta(minutes=30)
break_2_start = datetime.strptime(break_2_start, "%Y-%m-%d %H:%M")
break_2_end = break_2_start + timedelta(minutes=20)
day_end = datetime.strptime(day_end, "%Y-%m-%d %H:%M")

sched_df = pd.DataFrame({"Time": [day_start, break_1_start, break_1_end, lunch_start, lunch_end,
                                  break_2_start, break_2_end, day_end]})

print(sched_df)

demand_df = pd.DataFrame({"PartNo": ['Part1', 'Part2', 'Part3', 'Part4', 'Part5'],
                          "TTJ": [75, 100, 180, 30, 60]})

demand_df = demand_df.astype({'TTJ': float})

print(demand_df)

a = 0
b = 1
c = 2
x = sched_df.iloc[a]['Time']
y = sched_df.iloc[0]['Time']
z = 0


for n in range(0, len(demand_df)):
    while y < sched_df.iloc[1]['Time']:
        y = x + timedelta(minutes=demand_df.iloc[a]['TTJ'])
        if y > sched_df.iloc[1]['Time']:
            end = sched_df.iloc[1]['Time']
            print(demand_df.iloc[a]['PartNo'], 'time from : ', x.time(), ' to ', end.time())
            start = sched_df.iloc[1]['Time']
            end = sched_df.iloc[2]['Time']
            print('Break time from : ', start.time(), ' to ', end.time())

            xyz = sched_df.iloc[1]['Time'] - x
            delta = xyz.total_seconds()
            minutes = float(delta // 60)
            z = demand_df.iloc[a]['TTJ'] - minutes

            x = end + timedelta(minutes=z)
            counter = a

        else:
            print(demand_df.iloc[a]['PartNo'], 'time from : ', x.time(), ' to ', y.time())
            x = y
            a = a+1

if x > sched_df.iloc[3]['Time']:
    start = sched_df.iloc[2]['Time']
    end = sched_df.iloc[3]['Time']
    print(demand_df.iloc[a]['PartNo'], 'time from : ', start.time(), ' to ', end.time())
    xyz = x-end
    delta = xyz.total_seconds()
    minutes = float(delta // 60)
    z = minutes

    start = sched_df.iloc[3]['Time']
    end = sched_df.iloc[4]['Time']
    print('Lunch time from : ', start.time(), ' to ', end.time())

    start = sched_df.iloc[4]['Time']
    x = start + timedelta(minutes=z)

    print(demand_df.iloc[a]['PartNo'], 'time from : ', start.time(), ' to ', x.time())
    z = 0
    a = a + 1
else:
    start = sched_df.iloc[2]['Time']
    print(demand_df.iloc[a]['PartNo'], 'time from : ', start.time(), ' to ', x.time())
    z = 0
    a = a + 1
    y = x
    for n in range(0, len(demand_df)):
        while y < sched_df.iloc[3]['Time']:
            y = x + timedelta(minutes=demand_df.iloc[a]['TTJ'])
            if y > sched_df.iloc[3]['Time']:
                end = sched_df.iloc[3]['Time']
                print(demand_df.iloc[a]['PartNo'], 'time from : ', x.time(), ' to ', end.time())
                start = sched_df.iloc[3]['Time']
                end = sched_df.iloc[4]['Time']
                print('Lunch time from : ', start.time(), ' to ', end.time())

                xyz = sched_df.iloc[3]['Time'] - x
                delta = xyz.total_seconds()
                minutes = float(delta // 60)
                z = demand_df.iloc[a]['TTJ'] - minutes

                start = sched_df.iloc[4]['Time']
                x = start + timedelta(minutes=z)

                print(demand_df.iloc[a]['PartNo'], 'time from : ', start.time(), ' to ', x.time())
                z = 0
                a = a + 1

            else:
                print(demand_df.iloc[a]['PartNo'], 'time from : ', x.time(), ' to ', y.time())
                x = y
                a = a + 1

可参考的优化思路 #

• 硬编码逻辑解耦：现有代码的作息时间、时段索引都是固定写死的，后续调整作息、扩展多日调度很容易出错。可以将所有时段统一整理为结构化的时段列表，标记每个时段是工作还是休息，后续只需修改这个列表就能适配不同排班规则，不用单独维护一堆时间变量，也不用依赖sched_df.iloc[x]这种固定下标取值。
• 重复逻辑抽离：现在处理第一趟休息、午餐时段的逻辑几乎完全一致，只是判断的时段边界不同，可以把零件分配的逻辑抽为通用函数，输入当前零件加工时长、当前时间、可用时段列表，直接返回分配结果和下一个可用时间，大幅减少重复代码，可读性也会提升很多。
• 变量命名规范：现有代码里a、b、c、x、y、z都是无意义的变量名，后续排查问题很难快速对应逻辑，可以改成current_part_idx、current_time、remaining_process_minutes这种见名知意的命名，不用来回翻代码对应变量含义。
• 减少不必要的依赖：当前场景下pandas只是用来存储两个表格，完全可以用Python原生的列表存储零件需求、时段信息，减少不必要的依赖，运行效率也更高，如果后续需要导出调度结果再引入pandas也不迟。
• 补充边界处理逻辑：现有代码没有覆盖超过下班时间、零件加工时长超过单个工作时段的通用场景，可以加通用的循环判断逻辑：每次取当前可用的工作时段，判断剩余加工时长是否能在当前时段内完成，能就直接分配，不能就占满当前时段，减去对应加工时长后自动跳到下一个工作时段（自动跳过休息/午餐），直到当前零件加工完成即可。

内容的提问来源于stack exchange，提问作者Bajuns