Python:对列表中具有相同score1和teamB的所有dicts的teamA和teamB字段进行汇总？

Question

我对把字典算在单子上一点想法都没有。就像实例列表一样

我们有这样的名单

list_of_dict = [{'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '0', 'score2': '1'},
    {'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '0', 'score2': '1'},
    {'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'}]

我希望在像下面的列表那样计算输出之后

list_of_dict = [
{'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '1', 'score2': '2'}
{'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '3', 'score2': '0'}
]

Answer 1

@umn 's answer更优雅，而且可能更优化，因此如果您可以在脚本中使用numpy，则更倾向于这样做。下面是一种不需要额外库就可以这样做的简单方法：

list_of_dict = [{'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '0', 'score2': '1'},
    {'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '0', 'score2': '1'},
    {'teamA': 'Ocean Gaming ', 'teamB': 'PSISTORM Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'},
    {'teamA': 'Jin Air Green Wings ', 'teamB': 'Invictus Gaming', 'score1': '1', 'score2': '0'}]

intermediate_dict = {}

for d in list_of_dict:
    key = (d['teamA'], d['teamB'])
    if key in intermediate_dict.keys():
        intermediate_dict[key][0] += int(d['score1'])
        intermediate_dict[key][1] += int(d['score2'])
    else:
        intermediate_dict[key] = [int(d['score1']), int(d['score2'])]

final_list = []
for k,v in intermediate_dict.items():
    temp_dict = {}
    temp_dict['teamA'] = k[0]
    temp_dict['teamB'] = k[1]
    temp_dict['score1'] = v[0]
    temp_dict['score2'] = v[1]
    final_list.append(temp_dict)

print(final_list)