根据字典 Python 的键值检查列表中的值答案

【问题标题】：check the value from the list against key value from dictionary Python根据字典 Python 的键值检查列表中的值
【发布时间】：2020-08-05 14:30:11
【问题描述】：

我有两个数据源，一个是列表，另一个是字典列表。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 项目是否等于来自ad_dict 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

如何在不遍历字典或列表的情况下比较这些值？

【问题讨论】：

我建议使用一个类而不是一个列表中的那么多字典
我必须保持当前格式。

标签： python list dictionary compare key-value

【解决方案1】：

您的else 在循环中的错误位置。该内部循环为每个输出运行多次。如果您首先将 ID 值提取到集合中，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，您可以将整个事情作为一个单独的集合操作来完成：

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

【讨论】：

【解决方案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']

如果你不关心订单

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

【讨论】：

【解决方案3】：

一种快速的方法是使用嵌套列表推导

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

【讨论】：

【解决方案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}')

【讨论】：

【解决方案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

【讨论】：