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，另一个是字典的列表，我的数据看起来像下面。

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

我需要检查是否 need_placeholders 项目等于 ID 值从 ad_dict. 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}')

输出是：

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

我怎样才能在不循环浏览字典或列表的情况下比较这些值呢？

Answer 1

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']

你可以使用 itertools.filterfalse

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

如果你不在乎秩序

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 2

你有你的 else 在循环中的错误位置。那个内循环每出一个都要运行几次。你可以完全避免那个循环，如果你拉出你的 ID 先将值放入一个集合中。

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}')

Prints:

no key exits for 1200
no key exits for 1300

如果顺序并不重要，你可以将整个过程作为一个单一的集合操作来完成。

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

Answer 3

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 4

一个快速的方法是使用嵌套的列表理解法

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

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