Question

我有一个csv.DictReader dicts实例和一个dicts列表。

实例：

{'Salary': '3000', 'Name': 'James Jones', 'GameInfo': 'Den@Cle 07:30PM ET', 
 'AvgPointsPerGame': '4.883', 'teamAbbrev': 'Cle', 'Position': 'SG'}
{'Salary': '3000', 'Name': 'Justin Anderson', 'GameInfo': 'Orl@Dal 09:00PM ET',
 'AvgPointsPerGame': '13.161', 'teamAbbrev': 'Dal', 'Position': 'SF'}

列表：

[

    {'playername': 'Justin Anderson', 'points': '6.94'},

    {'playername': 'DeAndre Liggins', 'points': '11.4'},

]

我无法弄清楚如何迭代这些字典列表，匹配Name和playername键，然后从一个字典中吐出['Name']，从匹配的字典中吐出['points']。在上面的例子中，我将从两组词中匹配Justin Anderson然后打印出Justin Anderson，6.94

该应用程序的核心需要2个CSV，并使其成为dicts列表。

Answer 1

这种方式效率不高，但不需要任何预处理：

# Instead of your CSVReader:
dicts = [{'Salary': '3000', 'Name': 'James Jones', 'GameInfo': 'Den@Cle 07:30PM ET', 'AvgPointsPerGame': '4.883', 'teamAbbrev': 'Cle', 'Position': 'SG'},
         {'Salary': '3000', 'Name': 'Justin Anderson', 'GameInfo': 'Orl@Dal 09:00PM ET', 'AvgPointsPerGame': '13.161', 'teamAbbrev': 'Dal', 'Position': 'SF'}]

list_of_dicts = [
    {'playername': 'Justin Anderson', 'points': '6.94'},
    {'playername': 'DeAndre Liggins', 'points': '11.4'},
]

# For each dictionary in the CSVReader
for dct in dicts:
    # For each dictionary in your list of dictionaries
    for subdict in list_of_dicts:
        # Check if the name and playername matches
        if dct['Name'] == subdict['playername']:
            # I just print out the results, you need to do your logic here
            print(dct['Name'])
            print(dct)
            print('matching')
            print(subdict)

并打印出来：

Justin Anderson
{'Salary': '3000', 'Name': 'Justin Anderson', 'GameInfo': 'Orl@Dal 09:00PM ET', 'AvgPointsPerGame': '13.161', 'Position': 'SF', 'teamAbbrev': 'Dal'}
matching
{'playername': 'Justin Anderson', 'points': '6.94'}

如果您希望它比预处理词典列表更快，那么您只需查找playername：

>>> dict_of_dicts = {dct['playername']: dct for dct in list_of_dicts}
>>> dict_of_dicts
{'DeAndre Liggins': {'playername': 'DeAndre Liggins', 'points': '11.4'},
 'Justin Anderson': {'playername': 'Justin Anderson', 'points': '6.94'}}

然后循环简化为：

for dct in dicts:
    if dct['Name'] in dict_of_dicts:
        print(dct['Name'])
        print(dct)
        print('matching')
        print(dict_of_dicts[dct['Name']])

给出相同的结果。

在不同的dicts列表中匹配键

1 个答案: