我有一个有序(即排序)列表,其中包含按升序排序的日期(作为日期时间对象)。
我想写一个迭代这个列表的函数,并生成每个月的第一个可用日期的另一个列表。
例如,假设我的排序列表包含以下数据:
A = [
'2001/01/01',
'2001/01/03',
'2001/01/05',
'2001/02/04',
'2001/02/05',
'2001/03/01',
'2001/03/02',
'2001/04/10',
'2001/04/11',
'2001/04/15',
'2001/05/07',
'2001/05/12',
'2001/07/01',
'2001/07/10',
'2002/03/01',
'2002/04/01',
]
返回的列表将是
B = [
'2001/01/01',
'2001/02/04',
'2001/03/01',
'2001/04/10',
'2001/05/07',
'2001/07/01',
'2002/03/01',
'2002/04/01',
]
我建议的逻辑是这样的:
def extract_month_first_dates(input_list, start_date, end_date):
#note: start_date and end_date DEFINITELY exist in the passed in list
prev_dates, output = [],[] # <- is this even legal?
for (curr_date in input_list):
if ((curr_date < start_date) or (curr_date > end_date)):
continue
curr_month = curr_date.date.month
curr_year = curr_date.date.year
date_key = "{0}-{1}".format(curr_year, curr_month)
if (date_key in prev_dates):
continue
else:
output.append(curr_date)
prev_dates.append(date_key)
return output
有任何意见,建议吗? - 这可以改进为更多'Pythonic'吗?
答案 0 :(得分:7)
>>> import itertools
>>> [min(j) for i, j in itertools.groupby(A, key=lambda x: x[:7])]
['2001/01/01', '2001/02/04', '2001/03/01', '2001/04/10', '2001/05/07', '2001/07/01', '2002/03/01', '2002/04/01']
答案 1 :(得分:1)
搜索列表是O( n )操作。我想你可以简单地检查密钥是否是新的:
def extract_month_first_dates(input_list):
output = []
last_key = None
for curr_date in input_list:
date_key = curr_date.date.month, curr_date.date.year # no string key required
if date_key != last_key:
output.append(curr_date)
last_key = date_key
return output
答案 2 :(得分:0)
这是classic python
中的简单解决方案,即没有itertools;)和自解释
visited = {}
B = []
for a in A:
month = a[:7]
if month not in visited:
B.append(a)
visited[month] = 1
print B
输出继电器:
['2001/01/01', '2001/02/04', '2001/03/01', '2001/04/10', '2001/05/07', '2001/07/01', '2002/03/01', '2002/04/01']