有没有办法在python字典和列表中对值进行分组?

自我实现的方式是使用for循环,并添加到所需的结构中。

您还可以使用标准库中的模块 itertools.groupby , 当输入被排序时 (之前)。

分组

from itertools import groupby

lst = [
   {'Amount': 13.5, 'Name': 'Amy', 'date': '2022-01-30'}, 
   {'Amount': 15, 'Name': 'Bob', 'date': '2022-01-30'}, 
   {'Amount': 13.5, 'Name': 'Cara', 'date': '2022-01-31'}
]
keyfunc = lambda x: x['date']  # extracting the key for sort and group

groups = []
uniquekeys = []

data = sorted(lst, key=keyfunc)  # required for groupby
for k, g in groupby(data, keyfunc):
    groups.append({'date': k, 'details': list(g)})  # Store group iterator as a list
    uniquekeys.append(k)

print(groups)

印刷品:

[{'date': '2022-01-30', 'details': [{'Amount': 13.5, 'Name': 'Amy', 'date': '2022-01-30'}, {'Amount': 15, 'Name': 'Bob', 'date': '2022-01-30'}]}, {'date': '2022-01-31', 'details': [{'Amount': 13.5, 'Name': 'Cara', 'date': '2022-01-31'}]}]

取下钥匙

自从你的输出在 details 不只是一个分组列表,而是删除了键(用于分组方式),我们需要进行一些调整:

from itertools import groupby

lst = [
   {'Amount': 13.5, 'Name': 'Amy', 'date': '2022-01-30'}, 
   {'Amount': 15, 'Name': 'Bob', 'date': '2022-01-30'}, 
   {'Amount': 13.5, 'Name': 'Cara', 'date': '2022-01-31'}
]
keyfunc = lambda x: x['date']  # extracting the key for sort and group

groups = []

data = sorted(lst, key=keyfunc)  # required for groupby
for k, g in groupby(data, keyfunc):
    # remove the key from grouped list
    removed_key = []
    for v in list(g):
        del v['date']
        removed_key.append(v)
    # add to groups  
    groups.append({'date': k, 'details': removed_key})

print(groups)

所需打印:

[{'date': '2022-01-30', 'details': [{'Amount': 13.5, 'Name': 'Amy'}, {'Amount': 15, 'Name': 'Bob'}]}, {'date': '2022-01-31', 'details': [{'Amount': 13.5, 'Name': 'Cara'}]}]

我建议使用字典作为输出:

from collections import defaultdict
out = defaultdict(list)

for d in lst:
    d = d.copy() # avoids mutating the original dictionaries.
    out[d.pop('date')].append(d)

dict(out)

输出:

{'2022-01-30': [{'Amount': 13.5, 'Name': 'Amy'}, {'Amount': 15, 'Name': 'Bob'}],
'2022-01-31': [{'Amount': 13.5, 'Name': 'Cara'}]}

如果你真的想要你的格式:

out2 = [{'date': k, 'details': v} for k,v in out.items()]

输出:

[{'date': '2022-01-30', 'details': [{'Amount': 13.5, 'Name': 'Amy'}, {'Amount': 15, 'Name': 'Bob'}]},
 {'date': '2022-01-31', 'details': [{'Amount': 13.5, 'Name': 'Cara'}]}]

第一部分无违约条款:

out = {}

for d in lst:
    d = d.copy()
    date = d.pop('date')
    if date not in out:
        out[date] = []
    out[date].append(d)

一次完成两部分:

out = {}

for d in lst:
    d = d.copy()
    date = d.pop('date')
    if date not in out:
        out[date] = {'date': date, 'details': []}
    out[date]['details'].append(d)

out = list(out.values())