递归函数嵌套字典python答案

【问题标题】：Recursive function nested dictionary python递归函数嵌套字典python
【发布时间】：2021-12-04 07:20:47
【问题描述】：

我需要调用一个递归函数来打印嵌套字典的内容，如下所示：

lunch : {"sandwich": 2, "chips": 1, "drink": 1}

dinner : {"ribs": 6, "mac_cheese": 1, "roll": 1, "pie": 1}

我尝试了以下代码，但它只是迭代嵌套字典的键

# recursive function for printing out meal plans 
def print_meals(meal_plan):
    for key, value in meal_plan.items():
        if type(value) is dict:
            print_meals(value)
        else:
            print(key, ":", value)

    # create an empty dictionary
    meal_plan = {}

    # create individual dictionaries for each meal
    lunch = {"sandwich": 2, "chips": 1, "drink": 1}
    dinner = {"ribs": 6, "mac_cheese": 1, "roll": 1, "pie": 1}

    #create a nested dictionary
    meal_plan["lunch"] = lunch 
    meal_plan["dinner"] = dinner 

    # call recursive function 'print_meals'
    print_meals(meal_plan)

【问题讨论】：

输入中的嵌套字典在哪里？我可以看到，都是普通的字典。
@JohnByro 我正在创建嵌套字典的注释下创建嵌套字典。它被存储在字典 meal_plan
是否保证嵌套字典中只有一层深度？或者你可以有例如meal_plan = {'monday': {'lunch': {...}, 'dinner': {...}}, 'tuesday': {...}}
我建议使用isinstance(value, dict) 而不是type(value) is dict。例如，如果value 是dict 的子类，那么isinstance(value, dict) 仍将返回True，但type(value) is dict 将返回False。
如果没有任意级别的嵌套，则不需要递归。

标签： python dictionary recursion

【解决方案1】：

你可以像下面这样定义你的function：

>>> def print_meals(dct):
...    for k,v in dct.items():
...        print(f'{k} : {v}');print()

输出：

>>> meal_plan
{'lunch': {'sandwich': 2, 'chips': 1, 'drink': 1},
 'dinner': {'ribs': 6, 'mac_cheese': 1, 'roll': 1, 'pie': 1}}

>>> print_meals(meal_plan)
lunch : {'sandwich': 2, 'chips': 1, 'drink': 1}

dinner : {'ribs': 6, 'mac_cheese': 1, 'roll': 1, 'pie': 1}

【讨论】：

【解决方案2】：

这是一个想法：

def print_meals(meal_plan, depth=0):
    if isinstance(meal_plan, dict):
        for k, v in meal_plan.items():
            print('  '*depth, k, ':')
            print_meals(v, depth+1)
    else:
        print('  '*depth, meal_plan)

print_meals('sandwich')
#  sandwich

print_meals({'sandwich': 1})
#  sandwich :
#    1

print_meals({'lunch': {'sandwich': 2, 'chips': 1, 'drink': 1}, 'dinner': {'ribs': 6, 'mac_cheese': 1, 'roll': 1, 'pie': 1}})
#  lunch :
#    sandwich :
#      2
#    chips :
#      1
#    drink :
#      1
#  dinner :
#    ribs :
#      6
#    mac_cheese :
#      1
#    roll :
#      1
#    pie :
#      1

print_meals({'monday': {'lunch': {'sandwich': 1, 'apple': 2}, 'dinner': {'soup': 1, 'yaourt': 1}}, 'tuesday': {'breakfast': 'pain au chocolat', 'lunch': {'salad': 2, 'lasagna': 1}, 'dinner': {'beans':3}}})
#  monday :
#    lunch :
#      sandwich :
#        1
#      apple :
#        2
#    dinner :
#      soup :
#        1
#      yaourt :
#        1
#  tuesday :
#    breakfast :
#      pain au chocolat
#    lunch :
#      salad :
#        2
#      lasagna :
#        1
#    dinner :
#      beans :
#        3

或者，为了减少换行符，但假设膳食计划是一个字典：

def print_meals(meal_plan, depth=0):
  for k,v in meal_plan.items():
    if isinstance(v, dict):
      print('  '*depth, k, ':')
      print_meals(v, depth+1)
    else:
      print('  '*depth, k, ':', v)

print_meals({'sandwich': 1})
#  sandwich : 1

print_meals({'lunch': {'sandwich': 2, 'chips': 1, 'drink': 1}, 'dinner': {'ribs': 6, 'mac_cheese': 1, 'roll': 1, 'pie': 1}})
#  lunch :
#    sandwich : 2
#    chips : 1
#    drink : 1
#  dinner :
#    ribs : 6
#    mac_cheese : 1
#    roll : 1
#    pie : 1

print_meals({'monday': {'lunch': {'sandwich': 1, 'apple': 2}, 'dinner': {'soup': 1, 'yaourt': 1}}, 'tuesday': {'breakfast': 'pain au chocolat', 'lunch': {'salad': 2, 'lasagna': 1}, 'dinner': {'beans':3}}})
#  monday :
#    lunch :
#      sandwich : 1
#      apple : 2
#    dinner :
#      soup : 1
#      yaourt : 1
#  tuesday :
#    breakfast : pain au chocolat
#    lunch :
#      salad : 2
#      lasagna : 1
#    dinner :
#      beans : 3

【讨论】：