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[英]Remove duplicates from list of dictionaries created using groupby itertools in Python
[英]Itertools groupby to organize list of dictionaries by two values
我正在尝试按出生时的 state 以及他们是否有 0 钱来组织价值观。 Itertools groupby function 看起来是最简单的方法,但我正在努力实现它。 也对其他选项开放。
如果我有一个看起来像这样的字典列表
users = [
{"name": "John", "state_of_birth": "CA", "money": 0},
{"name": "Andrew", "state_of_birth": "CA", "money": 300},
{"name": "Scott", "state_of_birth": "OR", "money": 20},
{"name": "Travis", "state_of_birth": "NY", "money": 0},
{"name": "Bill", "state_of_birth": "CA", "money": 0},
{"name": "Mike", "state_of_birth": "NY", "money": 0}
]
我正在尝试获取此 output
desired_output = [
[{"name": "John", "state_of_birth": "CA", "money": 0}, {"name": "Bill", "state_of_birth": "CA", "money": 0}],
[{"name": "Andrew", "state_of_birth": "CA", "money": 300}],
[{"name": "Scott", "state_of_birth": "OR", "money": 20}],
[{"name": "Travis", "state_of_birth": "NY", "money": 0},{"name": "Mike", "state_of_birth": "NY", "money": 0}]
]
您可以像这样使用itertools
:
import itertools
def func(x):
return tuple([x['state_of_birth'], x['money'] != 0])
desired_output = list(list(v) for _,v in itertools.groupby(sorted(users, key=func), func))
group_by
function 是生成key
和value
的生成器。 密钥是从我们传递给itertools.groupb_by()
的key_function
派生的。 在您的情况下, keys
不重要,这就是为什么在for _, v
中忽略它的原因。
Output:
[{'name': 'John', 'state_of_birth': 'CA', 'money': 0}, {'name': 'Bill', 'state_of_birth': 'CA', 'money': 0}]
[{'name': 'Andrew', 'state_of_birth': 'CA', 'money': 300}]
[{'name': 'Travis', 'state_of_birth': 'NY', 'money': 0}, {'name': 'Mike', 'state_of_birth': 'NY', 'money': 0}]
[{'name': 'Scott', 'state_of_birth': 'OR', 'money': 20}]
代码:
users = [
{"name": "John", "state_of_birth": "CA", "money": 0},
{"name": "Andrew", "state_of_birth": "CA", "money": 300},
{"name": "Scott", "state_of_birth": "OR", "money": 20},
{"name": "Travis", "state_of_birth": "NY", "money": 0},
{"name": "Bill", "state_of_birth": "CA", "money": 0},
{"name": "Mike", "state_of_birth": "NY", "money": 0}
]
result = {}
for user in users:
key = (user["state_of_birth"],user["money"])
if key in result:
result[key].extend([user])
else:
result[key] = [user]
for _,v in result.items():
print(v)
结果:
[{'name': 'John', 'state_of_birth': 'CA', 'money': 0}, {'name': 'Bill', 'state_of_birth': 'CA', 'money': 0}]
[{'name': 'Andrew', 'state_of_birth': 'CA', 'money': 300}]
[{'name': 'Scott', 'state_of_birth': 'OR', 'money': 20}]
[{'name': 'Travis', 'state_of_birth': 'NY', 'money': 0}, {'name': 'Mike', 'state_of_birth': 'NY', 'money': 0}]
如果我理解这个问题是正确的,你有一个结构是List[Dict]
并且你想要一个List[List[Dict]]
,其中内部列表包含具有相同state_of_birth
和money > 0
boolean 的字典。
我想说最简单的解决方案实际上是使用pandas
import pandas as pd
users = [
{"name": "John", "state_of_birth": "CA", "money": 0},
{"name": "Andrew", "state_of_birth": "CA", "money": 300},
{"name": "Scott", "state_of_birth": "OR", "money": 20},
{"name": "Travis", "state_of_birth": "NY", "money": 0},
{"name": "Bill", "state_of_birth": "CA", "money": 0},
{"name": "Mike", "state_of_birth": "NY", "money": 0}
]
df = pd.DataFrame.from_records(users)
# we need a column to indicate if money > 0
df["money_bool"] = df["money"] > 0
# groupby gives you an iterator of Tuple[key, sub-dataframe]
# dfs now holds a list of your grouped dataframes
dfs = [tup[1] for tup in df.groupby(["state_of_birth", "money_bool"])]
# you can now drop the money_bool column if you want
dfs = [df.drop("money_bool", axis=1) for df in dfs]
desired_output = [df.to_dict("records") for df in dfs]
根据问题的上下文,您最好保留数据框/表格格式
您需要确保对groupby
function 的输入进行排序。 您可以使用与分组相同的密钥 function :
users = [
{"name": "John", "state_of_birth": "CA", "money": 0},
{"name": "Andrew", "state_of_birth": "CA", "money": 300},
{"name": "Scott", "state_of_birth": "OR", "money": 20},
{"name": "Travis", "state_of_birth": "NY", "money": 0},
{"name": "Bill", "state_of_birth": "CA", "money": 0},
{"name": "Mike", "state_of_birth": "NY", "money": 0}
]
def selector(item): return (item.get('state_of_birth'), item.get('money') != 0)
sorted_users = sorted(users, key=selector)
result = [list(group) for _, group in groupby(sorted_users, selector) ]
Output:
[
[{'name': 'John', 'state_of_birth': 'CA', 'money': 0}, {'name': 'Bill', 'state_of_birth': 'CA', 'money': 0}],
[{'name': 'Andrew', 'state_of_birth': 'CA', 'money': 300}],
[{'name': 'Travis', 'state_of_birth': 'NY', 'money': 0}, {'name': 'Mike', 'state_of_birth': 'NY', 'money': 0}],
[{'name': 'Scott', 'state_of_birth': 'OR', 'money': 20}]
]
虽然它的名字看起来应该是 go 的方式,但itertools.groupby
不是正确的 function 使用,因为它需要对数据进行预排序。 对于应该为 O(n) 的算法,排序会将您的时间复杂度提高到 O(n log(n))。
换个角度来看,如果你有一百万条记录要排序,而不是一百万次迭代,如果你使用groupby
而不是循环和字典,你现在有 2000 万次迭代。 这是一个相当大的性能损失。
如果groupby
写起来更干净或者没有导入,它可能是合理的,但它比使用普通循环和字典的更简单方法可读性差。
Pandas 很好,但除非你已经这样做了,否则真的没有理由使用它。 这就像带上航天飞机烤西葫芦一样。
您可以使用defaultdict
和循环:
from collections import defaultdict
from pprint import pprint
users = [
{"name": "John", "state_of_birth": "CA", "money": 0},
{"name": "Andrew", "state_of_birth": "CA", "money": 300},
{"name": "Scott", "state_of_birth": "OR", "money": 20},
{"name": "Travis", "state_of_birth": "NY", "money": 0},
{"name": "Bill", "state_of_birth": "CA", "money": 0},
{"name": "Mike", "state_of_birth": "NY", "money": 0},
]
grouped = defaultdict(list)
groupby = "state_of_birth", "money"
for user in users:
grouped[tuple([user[k] for k in groupby])].append(user)
pprint([*grouped.values()])
如果您想要“钱不是零”而不仅仅是"money"
值本身,您可以使用自定义分组 function:
grouped = defaultdict(list)
def group_by(x):
return x["state_of_birth"], x["money"] != 0
for user in users:
grouped[group_by(user)].append(user)
result = [*grouped.values()]
或内联逻辑:
grouped = defaultdict(list)
for user in users:
grouped[user["state_of_birth"], user["money"] != 0].append(user)
result = [*grouped.values()]
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