根据字典中的键值检查列表中的值 Python

Question

我有两个数据源，一个是列表，一个是字典列表。我的数据如下所示：

need_placeholder = ['1200', '1300', '1400']
ad_dict = [{"Name": "A", "ID": "1999"}, {"Name": "B", "ID": "1299"}, 
           {"Name": "C", "ID": "1400"}]

我需要检查 need_placeholders 项是否等于 ad_dict 中的 ID 值。这是我的脚本：

for item in need_placeholder:
    adpoint_key = item 
    for index, my_dict in enumerate(ad_dict):
        if my_dict["ID"] == adpoint_key:
            continue
        else:    
            print(f'No key exists for {adpoint_key}')

输出为：

No key exists for 1200
No key exists for 1200
No key exists for 1200
No key exists for 1300
No key exists for 1300
No key exists for 1300

我想要的输出是：

No key exists for 1200
No key exists for 1300

如何在不循环遍历字典或列表的情况下比较这些值？

Answer 1

您的 else 在循环中的错误位置。该内部循环为每个循环运行多次。如果您首先将 ID 值提取到一个集合中，则可以完全避免该循环：

need_placeholder = ['1200', '1300', '1400']
ad_dict = [{"Name": "A", "ID": "1999"}, {"Name": "B", "ID": "1299"}, {"Name": "C", "ID": "1400"}]

ad_values = set(d['ID'] for d in ad_dict)

for v in need_placeholder:
    if v not in ad_values:
        print(f'no key exits for {v}')

打印：

no key exits for 1200
no key exits for 1300

如果顺序不重要，您可以将整个操作作为一个集合操作完成：

for v in set(need_placeholder) - ad_values:
    print(f'no key exits for {v}')

Answer 2

你可以试试这个。

>>> need_placeholder = ['1200', '1300', '1400']
>>> ad_dict = [{"Name": "A", "ID": "1999"}, {"Name": "B", "ID": "1299"}, 
               {"Name": "C", "ID": "1400"}]
>>> keys={d['ID'] for d in ad_dict} # Set of all unique ID values
>>> [key for key in need_placeholder if not key in keys]
# ['1200', '1300']

您可以使用itertools.filterfalse

list(filterfalse(lambda x:x in keys, need_placeholder))
# ['1200', '1300']

如果你不关心顺序

set(need_placeholder)-keys
# {'1300', '1200'}

使用all

>>> for key in need_placeholder:
...     if all(key != d['ID'] for d in ad_dict):
...         print(f'No key exists for {key}')
...
No key exists for 1200
No key exists for 1300

使用for-else

>>> for key in need_placeholder:
...     for d in ad_dict:
...             if key==d['ID']:
...                     break
...     else:
...             print(f'No key {key}')
...
No key 1200
No key 1300

Answer 3

一个快速的方法是使用嵌套列表理解

[print('No key for {}'.format(x)) for x in need_placeholder if x not in [y['ID'] for y in ad_dict]]

Answer 4

allowed_values = {dic["ID"] for dic in ad_dict}

for item in need_placeholder:
    if item not in allowed_values:
        print(f'No key exists for {item}')

Answer 5

for item in need_placeholder:
    adpoint_key = item
    duplicate_exist = False # flag that will be update if key exists
    for index, my_dict in enumerate(ad_dict):
        if my_dict["ID"] == adpoint_key:
            duplicate_exist = True    # flag is updated as key exists
    not duplicate_exist and print(f'No key exists for {adpoint_key}') # for all non duplicate print function is invoked at the end of inner loop

根据字典中的键值检查列表中的值 Python

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