python pandas从时间序列中提取唯一的日期
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14673394/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
python pandas extract unique dates from time series
提问by tesla1060
I have a DataFrame which contains a lot of intraday data, the DataFrame has several days of data, dates are not continuous.
我有一个包含大量日内数据的 DataFrame，DataFrame 有几天的数据，日期不连续。
2012-10-08 07:12:22 0.0 0 0 2315.6 0 0.0 0
2012-10-08 09:14:00 2306.4 20 326586240 2306.4 472 2306.8 4
2012-10-08 09:15:00 2306.8 34 249805440 2306.8 361 2308.0 26
2012-10-08 09:15:01 2308.0 1 53309040 2307.4 77 2308.6 9
2012-10-08 09:15:01.500000 2308.2 1 124630140 2307.0 180 2308.4 1
2012-10-08 09:15:02 2307.0 5 85846260 2308.2 124 2308.0 9
2012-10-08 09:15:02.500000 2307.0 3 128073540 2307.0 185 2307.6 11
......
2012-10-10 07:19:30 0.0 0 0 2276.6 0 0.0 0
2012-10-10 09:14:00 2283.2 80 98634240 2283.2 144 2283.4 1
2012-10-10 09:15:00 2285.2 18 126814260 2285.2 185 2285.6 3
2012-10-10 09:15:01 2285.8 6 98719560 2286.8 144 2287.0 25
2012-10-10 09:15:01.500000 2287.0 36 144759420 2288.8 211 2289.0 4
2012-10-10 09:15:02 2287.4 6 109829280 2287.4 160 2288.6 5
......
How can I extract the unique date in the datetime format from the above DataFrame? To have result like [2012-10-08, 2012-10-10]
如何从上述 DataFrame 中提取日期时间格式的唯一日期？有这样的结果[2012-10-08, 2012-10-10]
采纳答案by DSM
If you have a Serieslike:
如果你有一个Series喜欢：
In [116]: df["Date"]
Out[116]:
0 2012-10-08 07:12:22
1 2012-10-08 09:14:00
2 2012-10-08 09:15:00
3 2012-10-08 09:15:01
4 2012-10-08 09:15:01.500000
5 2012-10-08 09:15:02
6 2012-10-08 09:15:02.500000
7 2012-10-10 07:19:30
8 2012-10-10 09:14:00
9 2012-10-10 09:15:00
10 2012-10-10 09:15:01
11 2012-10-10 09:15:01.500000
12 2012-10-10 09:15:02
Name: Date
where each object is a Timestamp:
其中每个对象是一个Timestamp：
In [117]: df["Date"][0]
Out[117]: <Timestamp: 2012-10-08 07:12:22>
you can get only the date by calling .date():
您只能通过调用获取日期.date()：
In [118]: df["Date"][0].date()
Out[118]: datetime.date(2012, 10, 8)
and Series have a .unique()method. So you can use mapand a lambda:
和系列有一个.unique()方法。所以你可以使用map和一个lambda：
In [126]: df["Date"].map(lambda t: t.date()).unique()
Out[126]: array([2012-10-08, 2012-10-10], dtype=object)
or use the Timestamp.datemethod:
或使用以下Timestamp.date方法：
In [127]: df["Date"].map(pd.Timestamp.date).unique()
Out[127]: array([2012-10-08, 2012-10-10], dtype=object)
回答by iTayb
Using regex:
使用正则表达式：
(\d{4}-\d{2}-\d{2})
Run it with re.findallfunction to get all matches:
使用re.findall函数运行它以获取所有匹配项：
result = re.findall(r"(\d{4}-\d{2}-\d{2})", subject)
回答by Nicolás Trejo
Just to give an alternative answer to @DSM, look at this other answerfrom @Psidom
只是为了给@DSM 提供一个替代答案，请查看@Psidom 的另一个答案
It would be something like:
它会是这样的：
pd.to_datetime(df['DateTime']).dt.date.unique()
It seems to me that it performs slightly better
在我看来它的表现要好一些
