Python：獲取每個公司的最新日期

Question

我有一個元組列表，其中包含日期和公司名稱。 公司可以列出多個日期的信息：

 [(Company A, datetime.date(1980,1,30)),
  (Company A, datetime.date(1990,1,30)),
  (Company B, datetime.date(1990,1,30)),
  (Company B, datetime.date(2000,1,30))]

我想要做的是有一個列表，其中僅包含每個公司的最新日期，即結果：

 [(Company A, datetime.date(1990,1,30)),
  (Company B, datetime.date(2000,1,30))]

有任何想法嗎？

Answer 1

如何從itertools使用groupby ，然后取最大：

import datetime
x = [('Company A', datetime.date(1980,1,30)),
  ('Company A', datetime.date(1990,1,30)),
  ('Company B', datetime.date(1990,1,30)),
  ('Company B', datetime.date(2000,1,30))]

import itertools
out = []
for k,g in itertools.groupby(sorted(x, key = lambda y: y[0]), lambda y: y[0]):
    out.append(max(g, key = lambda y:y[1]))

out
[('Company A', datetime.date(1990, 1, 30)),
 ('Company B', datetime.date(2000, 1, 30))]

Answer 2

您也可以使用字典...

data = [('Company A', '1980,1,30'),
  ('Company A', '1990,1,30'),
  ('Company B', '1990,1,30'),
  ('Company B', '2000,1,30')]

datadict = { a:b for a,b in data }

for a, b in data:
    datadict[a] = max(b, datadict[a])

print(datadict)

Answer 3

這是使用reduce()的示例：

import datetime

company_dates = [
  ('Company A', datetime.date(1980,1,30)),
  ('Company A', datetime.date(1990,1,30)),
  ('Company B', datetime.date(1990,1,30)),
  ('Company B', datetime.date(2000,1,30)),
]

def reducer(acc, company_date):
  try:
    acc[company_date[0]] = max(acc[company_date[0]], company_date[1])
  except KeyError:
    acc[company_date[0]] = company_date[1]

  return acc

sorted = reduce(reducer, company_dates, {})

print sorted.items()

這是另一個使用不同功能的替代解決方案：

import datetime
import operator

company_dates = [
  ('Company A', datetime.date(1980,1,30)),
  ('Company A', datetime.date(1990,1,30)),
  ('Company B', datetime.date(1990,1,30)),
  ('Company B', datetime.date(2000,1,30)),
]

sorted = sorted(company_dates, key=operator.itemgetter(0, 1), reverse=True)
unique = set([company_date[0] for company_date in sorted])
top = [next(c for c in sorted if c[0] == company) for company in unique]

print top