如何在Python中不循环获取键的值

Question

我有一个 for 循环，我需要在其中分配从另一个字典中获取的新键的值。我试图避免 for 循环中的 for 循环。我的第一本字典如下所示：ids = [{'1020': 'ID-2522'}, {'1030': 'ID-2523'}, {'1040': 'ID-2524'}]我正在循环的字典列表如下所示：data = [{'sf_id': '1020', TotalPrice': '504'}, {'sf_id': '1030', TotalPrice': '400'}, {'sf_id': '1040', TotalPrice': '500'}]这是我的for循环：for index, my_dict in enumerate(data):    for key, value in my_dict.items():         new_id = my_dict["sf_id"]         opportunity = ids[new_id]以便它获取对应的值。期望的输出是：print(opportunity)ID-2522ID-2523ID-2524

Answer 1

假设整个集合中没有重复键，您可以使用reduce将每个 id 字典合并到一个 id 字典中，因此您只需迭代两个集合一次

from functools import reduce

ids = [{'1020': 'ID-2522'}, {'1030': 'ID-2523'}, {'1040': 'ID-2524'}]

data = [{'sf_id': '1020', 'TotalPrice': '504'}, {'sf_id': '1030', 'TotalPrice': '400'}, {'sf_id': '1040', 'TotalPrice': '500'}]

ids_map = reduce(lambda x, y: x.update(y) or x , ids, {})

for my_dict in data:

  new_id = my_dict["sf_id"]

  opportunity = ids_map[new_id]

  print(opportunity)

Answer 2

ids是字典列表，而不是字典。这使得查找其中一个键的效率非常低，因为您必须遍历整个列表才能找到它。

因此，第一步是将所有这些小字典合并成一个大字典。

然后，我们可以使用您的预期输出有效地构建一个列表：

ids = [{'1020': 'ID-2522'}, {'1030': 'ID-2523'}, {'1040': 'ID-2524'}]

data = [{'sf_id': '1020', 'TotalPrice': '504'}, 

        {'sf_id': '1030', 'TotalPrice': '400'}, 

        {'sf_id': '1040', 'TotalPrice': '500'}]

ids_dict = {k:v for dct in ids for k, v in dct.items()}

opportunity = []

for my_dict in data:

    new_id = my_dict["sf_id"]

    opportunity.append(ids_dict[new_id])

print(opportunity)

# ['ID-2522', 'ID-2523', 'ID-2524']

（注意enumerate如果不使用索引就不需要）

或者，更短的，使用列表理解：

opportunity = [ids_dict[my_dict["sf_id"]] for my_dict in data]

print(opportunity)

#  ['ID-2522', 'ID-2523', 'ID-2524']

Answer 3

# Create a new variable and save sftoap in it

new_ids = ids

# pop the items from original list of dicts and save it in a new dict 

sftoap_dict = {}

for d in new_ids:

    key, value = d.popitem()

    sftoap_dict[key] = value