将数据数据中给定的值与遵循的规则进行排列-熊猫

Question

我有一个水果计数dataframe.From的数据我有总数为7，这些水果将分发7天每果一天。

fruit  count   
apple  3  
orange 1  
lemon  1  
cherry 1  
banana 1

分割水果的规则：

1.苹果果实不应连续几天送出，苹果之间应保持至少一天的间隔。

2.橘子必须在day7上发送(即最后一天)。

3.在第6天，柠檬不应发送，在橙和柠檬之间应保持至少一天的间隔。

4.如果水果数超过7，则不要与苹果、樱桃或香蕉搭配任何其他带有柠檬和橘子的水果。

Expected_ouput

day fruit
d1   apple
d2   cherry
d3   apple
d4   lemon
d5   apple
d6   banana
d7   orange

案例1-少于7个水果

fruit  count   
apple  1  
orange 1  
lemon  1  
cherry 1  
banana 1

预期产出：

day fruit
d1   apple
d2   cherry
d4   lemon
d6   banana
d7   orange

案例2-7多个水果：

fruit  count  
apple  3  
orange 1
pomerante 1  
lemon  1  
cherry 1  
banana 1 

Expected_ouput

day fruit
d1   apple
d2   cherry
d3   apple
d3   pomegranate
d4   lemon
d5   apple
d6   banana
d7   orange

我试过：

apple = 3
orange = 1
week = 7
start_day=1
end_day = 8
days = [f"Day{i}" for i in range(start_day,end_day)]
profile = {key : [] for key in days}
profile["Day7"].append("orange")
c = 6 // apple

apple_day = 1
for i in range(apple):
    profile["Day{}".format(apple_day)].append("apple")
    apple_day += c

输出：

{'Day1': ['apple'],
 'Day2': [],
 'Day3': ['apple'],
 'Day4': [],
 'Day5': ['apple'],
 'Day6': [],
 'Day7': ['orange']}

Answer 1

这是我最先进的解决方案，希望你能接受这个答案。

它适用于任何有解决方案的极端示例：

basket_of_fruits = ['apple','apple','apple','cherry','cherry','banana','banana','orange','lemon','kiwi','kiwi']
n_days = 7
# Create day plan
plan = {}
for i in range(n_days):
    plan['Day_'+str(i+1)] = []
# Set startiong values
apple_day = n_days % 2 +1
occupy = 1
days_occupied = []
# solve 1 orange
if 'orange' in basket_of_fruits:
    plan['Day_'+str(n_days)].append('orange')
    basket_of_fruits.remove('orange')
    days_occupied.append(n_days)
# solve 1 lemon
if 'lemon' in basket_of_fruits:
    plan['Day_'+str(n_days-2)].append('lemon')
    basket_of_fruits.remove('lemon')
    days_occupied.append(n_days-2)
# solve other fruit then apple, cherry or banana
basket_of_fruits_rest = basket_of_fruits.copy()
while 'apple' in basket_of_fruits_rest:
    basket_of_fruits_rest.remove('apple')
while 'cherry' in basket_of_fruits_rest:
    basket_of_fruits_rest.remove('cherry')
while 'banana' in basket_of_fruits_rest:
    basket_of_fruits_rest.remove('banana')
for i in range(len(basket_of_fruits_rest)):
    while occupy in days_occupied:
        occupy += 1
        if occupy > n_days:
            days_occupied = [n_days,n_days-2]
            occupy -= n_days
    selected_fruit = basket_of_fruits_rest[i]
    plan['Day_'+str(occupy)].append(selected_fruit)
    basket_of_fruits.remove(selected_fruit)
    days_occupied.append(occupy)
    occupy += 1

# solve cherry or banana
basket_of_fruits_rest = basket_of_fruits.copy()
while 'apple' in basket_of_fruits_rest:
    basket_of_fruits_rest.remove('apple')
for i in range(len(basket_of_fruits_rest)):
    while occupy in days_occupied:
        occupy += 1
        if occupy > n_days:
            days_occupied = [n_days,n_days-2]
            occupy -= n_days
    selected_fruit = basket_of_fruits_rest[i]
    plan['Day_'+str(occupy)].append(selected_fruit)
    basket_of_fruits.remove(selected_fruit)
    days_occupied.append(occupy)
    occupy += 1
    
# solve apples
while 'apple' in basket_of_fruits:
    while ('lemon' in plan['Day_'+str(apple_day)]) | ('orange' in plan['Day_'+str(apple_day)]) :
        apple_day += 2
        if apple_day > n_days:
            apple_day -= n_days-1
    
        
    plan['Day_'+str(apple_day)].append('apple')
    basket_of_fruits.remove('apple')
    apple_day += 2
    if apple_day > n_days:
        apple_day -= n_days-1

返回：

{'Day_1': ['kiwi', 'banana'],
 'Day_2': ['kiwi', 'apple'],
 'Day_3': ['cherry'],
 'Day_4': ['cherry', 'apple'],
 'Day_5': ['lemon'],
 'Day_6': ['banana', 'apple'],
 'Day_7': ['orange']}