从元组列表中获取最小唯一元组对

解决方案1:

min_dates = {}
for item in transactions:
    key, date = item
    if key not in min_dates or date < min_dates[key]:
        min_dates[key] = date
result = [(key, min_dates[key]) for key in min_dates]
result.sort(key=lambda x: x[1])

解决方案2:

from itertools import groupby
# Sort transactions by key and then by date
sorted_transactions = sorted(transactions)

# Group transactions by key
grouped_transactions = groupby(sorted_transactions, key=lambda x: x[0])

# Construct final list of tuples with the minimum date for each key
result = [(key, min(group, key=lambda x: x[1])[1]) for key, group in grouped_transactions]

#print(result)
# Prints : [('GBP.USD', '2022-04-27'), ('MSFT', '2022-04-11'), ('SHOP', '2022-04-21'), ('SNOW', '2022-04-26')]

你应该能够做到:

output = dict(reversed(sorted(transactions)))
output = list(output.items())

sorted 可以自动处理元组,并且由于您需要最小值,因此可以使用 reversed (而不是指定 key=lambda ... ).

因为你的约会是 yyyy-mm-dd ,字符串排序应该正确。

这是创建一个dict,并在每个键到达时用排序较少的值覆盖它。

我的计划是制作一本词典 {name: date} 其中日期是对应名称的最小(或最早)。

transactions = [
    ('GBP.USD', '2022-04-29'),
    ('SNOW', '2022-04-26'),
    ('SHOP', '2022-04-21'),
    ('GBP.USD', '2022-04-27'),
    ('MSFT', '2022-04-11'),
    ('MSFT', '2022-04-21'),
    ('SHOP', '2022-04-25')
]

earliest = {}
for name, date in transactions:
    earliest.setdefault(name, date)
    if earliest[name] > date:
        earliest[name] = date
out = list(earliest.items())

out 则为:

[('GBP.USD', '2022-04-27'),
 ('SNOW', '2022-04-26'),
 ('SHOP', '2022-04-21'),
 ('MSFT', '2022-04-11')]

笔记

• 这个 setdefault 方法很神奇:如果字典中不存在键,它会将键/值插入字典中。如果密钥确实存在,则不会插入(不执行任何操作)。
• 设置默认值后,我将仅在值较小时更新该值
• 有一次我造了字典 list(...items()) 会得到我们的结果。
• 此功能不会改变 transactions 无论如何。

由于考虑太多,我的直觉反应导致我找到了下面的解决方案。注意,我并不认为这是最佳的。

# Import pandas to leverage built-in query methods
import pandas as pd

# Define the test data, list of tuples
transactions = [ ('GBP.USD', '2022-04-29'),
                 ('SNOW', '2022-04-26'),
                 ('SHOP', '2022-04-21'),
                 ('GBP.USD', '2022-04-27'),
                 ('MSFT', '2022-04-11'),
                 ('MSFT', '2022-04-21'),
                 ('SHOP', '2022-04-25') ]

# Convert the test data into a pandas dataframe, columns can be named anything
df = pd.DataFrame( transactions, columns=[ 'Symbol', 'Date' ] )

# Convert the Date column into proper dates, very powerful option going forward
df[ 'Date' ] = pd.to_datetime( df[ 'Date' ] )

# Group the dataframe by the unique values in the Symbol column and reduce to instances where the minimum value appears in the Date column for a given unique value of the Symbol column
df = df.groupby( 'Symbol' ).apply( lambda x: x.loc[ x.Date.idxmin() ] ).reset_index( drop=True )

# Optional, convert the dataframe back into a list of tuples, maintains dates as Timestamps for future leveraging of pandas built in methods for date/time data
transactions = list( df.itertuples( index=False, name=None ) )

如果打印变量 df 在最后一次分配后,您将获得:

print(df)
    Symbol       Date
0  GBP.USD 2022-04-27
1     MSFT 2022-04-11
2     SHOP 2022-04-21
3     SNOW 2022-04-26

如果打印变量 transactions 在最后一次分配后,您将获得:

print(transactions)
[('GBP.USD', Timestamp('2022-04-27 00:00:00')), ('MSFT', Timestamp('2022-04-11 00:00:00')), ('SHOP', Timestamp('2022-04-21 00:00:00')), ('SNOW', Timestamp('2022-04-26 00:00:00'))]

您可以使用 pandas 喜欢下面

import pandas as pd
list(pd.DataFrame(transactions).groupby(0).min().itertuples(name=None))

应该给

[('GBP.USD', '2022-04-27'),
 ('MSFT', '2022-04-11'),
 ('SHOP', '2022-04-21'),
 ('SNOW', '2022-04-26')]