Question

我有一个结果行块，并且我试图将它们分组为嵌套[{key: value[{key:value[]}]}]的两个级别。顶级键上的值不唯一。

我一直在尝试使用defaultdict，但由于存在非唯一性，所以在两个级别上都没有成功分组。遍历数据可能会更好，但我也没有成功。

开始数据：

data = 
[{'Name': 'Bob', 'Time': 12, 'Place': 'Home'}, 
{'Name': 'Bob', 'Time': 11, 'Place': 'Home'}, 
{'Name': 'Jerry', 'Time': 5, 'Place': 'Home'}, 
{'Name': 'Jerry', 'Time': 11, 'Place': '-----'}, 
{'Name': 'Jerry', 'Time': 11, 'Place': 'Work'}]

最终所需数据：

[{"Name": "Bob", "Details":[{"Place":"Home", "Time":[12, 11]}]}, 
{"Name": "Jerry", "Details":[{"Place":"Home", "Time":[5]}, 
                             {"Place":"-----", "Time":[11]}, 
                             {"Place":"Work", "Time":[11]}]}]

Answer 1

您可以使用Name按Place和itertools.groupby分组，

>>> import itertools
>>> from collections import defaultdict
>>> data
[{'Name': 'Bob', 'Time': 12, 'Place': 'Home'}, {'Name': 'Bob', 'Time': 11, 'Place': 'Home'}, {'Name': 'Jerry', 'Time': 5, 'Place': 'Home'}, {'Name': 'Jerry', 'Time': 11, 'Place': '-----'}, {'Name': 'Jerry', 'Time': 11, 'Place': 'Work'}]
>>> sorted_data = sorted(data, key=lambda x: (x['Name'], x['Place'])) # sorting before grouping as suggested by @wwii, because The returned group is itself an iterator that shares the underlying iterable with groupby(). Please see (https://docs.python.org/3/library/itertools.html#itertools.groupby)
>>> d = defaultdict(list)
>>> y = itertools.groupby(sorted_data, lambda x: (x['Name'], x['Place']))
>>> for group, grouper in y:
...   time_ = [item['Time'] for item in grouper]
...   name, place = group
...   d[name].append({'Place': place, 'Time': time_})
... 
>>> d
defaultdict(<class 'list'>, {'Bob': [{'Place': 'Home', 'Time': [12, 11]}], 'Jerry': [{'Place': 'Home', 'Time': [5]}, {'Place': '-----', 'Time': [11]}, {'Place': 'Work', 'Time': [11]}]})
>>> pprint.pprint(dict(d))
{'Bob': [{'Place': 'Home', 'Time': [12, 11]}],
 'Jerry': [{'Place': 'Home', 'Time': [5]},
           {'Place': '-----', 'Time': [11]},
           {'Place': 'Work', 'Time': [11]}]}

如果您需要显示的确切结构，

>>> f_data = []
>>> for key, value in d.items():
...   f_data.append({'Name': key, 'Details': value})
... 
>>> pprint.pprint(f_data)
[{'Details': [{'Place': 'Home', 'Time': [12, 11]}], 'Name': 'Bob'},
 {'Details': [{'Place': '-----', 'Time': [11]},
              {'Place': 'Home', 'Time': [5]},
              {'Place': 'Work', 'Time': [11]}],
  'Name': 'Jerry'}]

Answer 2

排序数据；按'Name'分组，按'Place'分组结果；提取时间。

import operator
name = operator.itemgetter('Name')
where = operator.itemgetter('Place')
time = operator.itemgetter('Time')

data.sort(key=lambda x: (name(x),where(x)))
result = []
for name, group in itertools.groupby(data,key=name):
    d = {'Name':name, 'Details':[]}
    for place, times in itertools.groupby(group,key=where):
        times = map(time, times)
        d['Details'].append({'Place':place, 'Time':list(times)})
    result.append(d)

我喜欢使用operator.itemgetter而不是lambda函数，如果它将多次使用。只是我个人的喜好。

Answer 3

我试图在熊猫的帮助下解决它。看看：

import pandas as pd

data = [{'Name': 'Bob', 'Time': 12, 'Place': 'Home'}, 
{'Name': 'Bob', 'Time': 11, 'Place': 'Home'}, 
{'Name': 'Jerry', 'Time': 5, 'Place': 'Home'}, 
{'Name': 'Jerry', 'Time': 11, 'Place': '-----'}, 
{'Name': 'Jerry', 'Time': 11, 'Place': 'Work'}]

df = pd.DataFrame.from_dict(data)

#Take the unique names only
names = df["Name"].unique()

#This list will hold the desired values
new_list = []

# Iterate over names
for n in names:
    # Make subset off the data set where name is n
    subset = df[df["Name"]==n]
    # Get Unique Places in the subset
    places = subset["Place"].unique()
    # This will hold the details
    details = []
    # Iterate over unique places
    for p in places:
        # Get times from subset where place is  and convert to list
        times = subset[subset["Place"]==p]["Time"].tolist()
        # Append to details list
        details.append({"Place":p,"Time":times})
    # Add the details in new_list as the format you preferred
    new_list.append({"Name":n, "Details":details})

print(new_list)

Answer 4

通过defaultdict加上迭代，您已经有了正确的想法。唯一有点棘手的地方是嵌套defaultdict。

from collections import defaultdict

def timegroup(data):
    grouped = defaultdict(lambda:defaultdict(list))
    for d in data:
        grouped[d['Name']][d['Place']].append(d['Time'])
    for name, details in grouped.items():
        yield {'Name': name,
               'Details': [{'Place': p, 'Time': t} for p, t in details.items()]}

（我喜欢将生成器用于此类操作，因为有时您只想迭代结果，在这种情况下，您不需要列表，如果需要列表，则很容易创建一个列表。）

嵌套的defaultdict（list）分组

4 个答案: