問題描述
假設(shè)我有一個(gè)如下所示的列表:
Say I have a list looks like this:
[(datetime.datetime(2013, 8, 8, 1, 20, 15), 2060), (datetime.datetime(2013, 8, 9, 1, 6, 14), 2055), (datetime.datetime(2013, 8, 9, 1, 21, 1), 2050), (datetime.datetime(2013, 8, 10, 1, 5, 49), 2050), (datetime.datetime(2013, 8, 10, 1, 19, 51), 2050), (datetime.datetime(2013, 8, 11, 2, 4, 53), 2050), (datetime.datetime(2013, 8, 12, 0, 29, 45), 2050), (datetime.datetime(2013, 8, 12, 0, 44, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 34, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 47, 29), 2050), (datetime.datetime(2013, 8, 14, 1, 30, 39), 2050), (datetime.datetime(2013, 8, 14, 1, 33, 51), 2050), (datetime.datetime(2013, 8, 15, 0, 41, 1), 2050), (datetime.datetime(2013, 8, 15, 0, 54, 45), 2050), (datetime.datetime(2013, 8, 16, 0, 29, 57), 1950), (datetime.datetime(2013, 8, 16, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 17, 0, 27, 4), 1950), (datetime.datetime(2013, 8, 17, 0, 42, 30), 1950), (datetime.datetime(2013, 8, 18, 0, 26, 26), 1950), (datetime.datetime(2013, 8, 18, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 19, 0, 41, 49), 1950), (datetime.datetime(2013, 8, 20, 1, 10, 23), 1950), (datetime.datetime(2013, 8, 20, 1, 23, 44), 1950), (datetime.datetime(2013, 8, 21, 0, 47, 25), 1950), (datetime.datetime(2013, 8, 21, 1, 0, 12), 1950), (datetime.datetime(2013, 8, 22, 0, 45, 21), 1950), (datetime.datetime(2013, 8, 22, 1, 4, 33), 1950), (datetime.datetime(2013, 8, 23, 0, 51, 27), 1950), (datetime.datetime(2013, 8, 23, 1, 6, 36), 1950), (datetime.datetime(2013, 8, 24, 0, 41, 3), 1950), (datetime.datetime(2013, 8, 24, 0, 53, 14), 1950), (datetime.datetime(2013, 8, 25, 0, 29, 24), 1950), (datetime.datetime(2013, 8, 25, 0, 42, 40), 1950), (datetime.datetime(2013, 8, 26, 0, 28, 13), 1950), (datetime.datetime(2013, 8, 26, 0, 43, 30), 1950), (datetime.datetime(2013, 8, 27, 0, 30, 1), 1950), (datetime.datetime(2013, 8, 27, 0, 43, 43), 1950), (datetime.datetime(2013, 8, 28, 0, 33, 19), 1950), (datetime.datetime(2013, 8, 28, 0, 49, 11), 1950), (datetime.datetime(2013, 8, 29, 0, 26, 49), 1950), (datetime.datetime(2013, 8, 29, 0, 41, 21), 1950), (datetime.datetime(2013, 8, 30, 0, 26, 13), 1950), (datetime.datetime(2013, 8, 30, 0, 42, 9), 1950), (datetime.datetime(2013, 8, 31, 0, 23, 40), 1950), (datetime.datetime(2013, 8, 31, 0, 39, 49), 1950), (datetime.datetime(2013, 9, 1, 0, 22, 2), 1950), (datetime.datetime(2013, 9, 1, 0, 38, 16), 1950), (datetime.datetime(2013, 9, 2, 0, 21, 2), 1950), (datetime.datetime(2013, 9, 2, 0, 36, 19), 1950), (datetime.datetime(2013, 9, 3, 0, 22, 16), 1950), (datetime.datetime(2013, 9, 3, 0, 39, 2), 1900)]
很明顯，您可以看到這是一個(gè)元組列表，每個(gè)元組中的第一個(gè)元素是一個(gè)時(shí)間戳.已采用良好格式，由以下人員生成:
clearly you could see that this is a list of tuple and the first element in each tuple is a timestamp. Already in good format, generated by:
datetime.strptime(record[0], timeFormat)
第二個(gè)元素是監(jiān)控值.但是，每天可能有多個(gè)記錄.例如，datetime.datetime(2013, 8, 9..) 上有兩條記錄，它們有兩個(gè)不同的值 2055 和 2050.我想要的是實(shí)際上每天的最大值.所以在這種情況下.2055 將是 (2013, 8, 9) 的唯一記錄.
And the second element is the monitoring value. However, there might be multiple records in each day. For example, there are two records on datetime.datetime(2013, 8, 9..), which have two different values 2055 and 2050. What I want is the actually the maximum in each day. So in this case. 2055 would be the only records for (2013, 8, 9).
我想知道 Python 中是否有一種方便的方法可以做到這一點(diǎn).類似mysql的東西:
I am wondering would there be a handy way in Python to do that. Some thing similar like mysql:
select
date(timestamp),
max(value)
from table
group by date(timestamp)
mysql 語句只是為了展示這個(gè)想法，我絕對想要一個(gè) python 解決方案.
The mysql statement is just to show the idea and I definitely want a python solution.
推薦答案
使用 itertools.groupby:
>>> records = [(datetime.datetime(2013, 8, 8, 1, 20, 15), 2060), ....]
>>> import itertools
>>> [(dt, max(v for d, v in grp)) for dt, grp in itertools.groupby(records, key=lambda x: x[0].date())]
[(datetime.date(2013, 8, 8), 2060),
(datetime.date(2013, 8, 9), 2055),
(datetime.date(2013, 8, 10), 2050),
...
]
注意:假設(shè)記錄已排序.如果沒有，您應(yīng)該先按日期對它們進(jìn)行排序.
NOTE: assumed that the records are sorted. If not, you should sort them first by dates.
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