动态填充字典

Question

尝试使用嵌套词典....从用户那里获取关于他有多少个输入项目的输入（他输入了一个否），对于每个项目，代码要求他输入两个数字，从中, 代码将告诉输入的 nos 范围内的最大和最小数字...

PFB代码：

def how_many_sets():
    global set_nos_list
    set_nos_list = []
    defaults = {"range": [], "items": [], "largest": 0, "lowest": 0, "difference": 0}
    global set_no
    set_no = int(input('How many sets?'))
    if set_no < 1 or set_no > 10:
        print('error')
    else:
        for num in range(set_no):
            set_nos_list.append(num)
        print(set_nos_list)
        global set_dict
        set_dict = dict.fromkeys(set_nos_list, defaults)
        return set_no


def take_sets():
    for item in set_nos_list:
        list_per_pair = list(map(int, input().split()))
        if len(list_per_pair) != 2:
            print('There is an error in the no of items, For finding the range two items should be there in a list')
            return
        elif (list_per_pair[0] < 2) or (list_per_pair[1] > 1000000) or (list_per_pair[0] > list_per_pair[1]):
            print(' error')
            return
        else:
            print(item)
            set_dict[item]["range"] = list_per_pair
            temp_item_list = []
            for value in range(list_per_pair[0], list_per_pair[1] + 1):
                # print(value)
                temp_item_list.append(value)
                print(temp_item_list)
            #set_dict[num]['items'] = temp_item_list

    print(set_dict)


def main():
    how_many_sets()
    take_sets()


# Write code here

main()

第 set_dict[item]["range"] = list_per_pair 行给我一个错误，set_dict 只更新了最后一个项目值，尽管循环遍历了所有值....

How many sets?2
[0, 1]
3 6
0
[3, 4, 5, 6]
4 7
1
[4, 5, 6, 7]
{0: {'range': [3, 6], 'items': [], 'largest': 0, 'lowest': 0, 'difference': 0}, 1: {'range': [4, 7], 'items': [], 'largest': 0, 'lowest': 0, 'difference': 0}}

我想要这样输出的地方：

{0: {'range': [4, 7], 'items': [], 'largest': 0, 'lowest': 0, 'difference': 0}, 1: {'range': [4, 7], 'items': [], 'largest': 0, 'lowest': 0, 'difference': 0}}

Answer 1

从 Initialize dict with values 开始，您可以更改为

set_dict = {num: {**defaults} for num in range(set_no)}

然后尽可能避免global变量，更喜欢返回值frmo方法和传递参数

def how_many_sets():
    defaults = {"range": [], "items": [], "largest": 0, "lowest": 0, "difference": 0}
    set_no = int(input('How many sets? '))
    while set_no < 1 or set_no > 10:
        print('Amount set error')
        set_no = int(input('How many sets? '))
    return {num: {**defaults} for num in range(set_no)}


def take_sets(dict_structure):
    for item in dict_structure:
        list_per_pair = list(map(int, input("Gives values: ").split()))
        if len(list_per_pair) != 2:
            print('There is an error in the no of items, For finding the range two items should be there in a list')
            return
        elif (list_per_pair[0] < 2) or (list_per_pair[1] > 1000000) or (list_per_pair[0] > list_per_pair[1]):
            print('Bound Error')
            return
        else:
            dict_structure[item]["range"] = list_per_pair
            dict_structure[item]['items'] = list(range(list_per_pair[0], list_per_pair[1] + 1))

    print(dict_structure)


def main():
    dict_structure = how_many_sets()
    take_sets(dict_structure)


main()

例子

How many sets? 4
Gives values: 4 6
Gives values: 3 7
Gives values: 7 8
Gives values: 4 6

{0: {'range': [4, 6], 'items': [4, 5, 6], 'largest': 0, 'lowest': 0, 'difference': 0}, 
 1: {'range': [3, 7], 'items': [3, 4, 5, 6, 7], 'largest': 0, 'lowest': 0, 'difference': 0}, 
 2: {'range': [7, 8], 'items': [7, 8], 'largest': 0, 'lowest': 0, 'difference': 0}, 
 3: {'range': [4, 6], 'items': [4, 5, 6], 'largest': 0, 'lowest': 0, 'difference': 0}}

动态填充字典

dynamically populating a dictionary

python

dictionary

for-loop

nested-lists

nested-loops