python pandas数据帧按日期条件切片
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/16175874/
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 dataframe slicing by date conditions
提问by Rishabh Sagar
I am able to read and slice pandas dataframe using python datetime objects, however I am forced to use only existing datesin index. For example, this works:
我能够使用 python datetime 对象读取和切片 pandas 数据帧，但是我被迫仅使用索引中的现有日期。例如，这有效：
>>> data
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 252 entries, 2010-12-31 00:00:00 to 2010-04-01 00:00:00
Data columns:
Adj Close 252 non-null values
dtypes: float64(1)
>>> st = datetime.datetime(2010, 12, 31, 0, 0)
>>> en = datetime.datetime(2010, 12, 28, 0, 0)
>>> data[st:en]
Adj Close
Date
2010-12-31 593.97
2010-12-30 598.86
2010-12-29 601.00
2010-12-28 598.92
However if I use a start or end date that is not present in the DF, I get python KeyError.
但是，如果我使用 DF 中不存在的开始或结束日期，则会出现 python KeyError。
My Question : How do I query the dataframe object for a date range; even when the start and end dates are not present in the DataFrame. Does pandas allow for range based slicing?
我的问题：如何查询数据框对象的日期范围；即使开始和结束日期不存在于 DataFrame 中。熊猫是否允许基于范围的切片？
I am using pandas version 0.10.1
我正在使用熊猫版本 0.10.1
采纳答案by waitingkuo
Use searchsortedto find the nearest times first, and then use it to slice.
先用searchsorted找最近的时间，再用它来切片。
In [15]: df = pd.DataFrame([1, 2, 3], index=[dt.datetime(2013, 1, 1), dt.datetime(2013, 1, 3), dt.datetime(2013, 1, 5)])
In [16]: df
Out[16]:
0
2013-01-01 1
2013-01-03 2
2013-01-05 3
In [22]: start = df.index.searchsorted(dt.datetime(2013, 1, 2))
In [23]: end = df.index.searchsorted(dt.datetime(2013, 1, 4))
In [24]: df.iloc[start:end]
Out[24]:
0
2013-01-03 2
回答by Dan Allan
Short answer: Sort your data (data.sort()) and then I think everything will work the way you are expecting.
简短回答：对您的数据进行排序 ( data.sort())，然后我认为一切都会按照您期望的方式进行。
Yes, you can slice using datetimes not present in the DataFrame. For example:
是的，您可以使用 DataFrame 中不存在的日期时间进行切片。例如：
In [12]: df
Out[12]:
0
2013-04-20 1.120024
2013-04-21 -0.721101
2013-04-22 0.379392
2013-04-23 0.924535
2013-04-24 0.531902
2013-04-25 -0.957936
In [13]: df['20130419':'20130422']
Out[13]:
0
2013-04-20 1.120024
2013-04-21 -0.721101
2013-04-22 0.379392
As you can see, you don't even have to build datetime objects; strings work.
如您所见，您甚至不必构建日期时间对象；字符串工作。
Because the datetimes in your index are not sequential, the behavior is weird. If we shuffle the index of my example here...
因为索引中的日期时间不是连续的，所以行为很奇怪。如果我们在这里洗牌我的例子的索引......
In [17]: df
Out[17]:
0
2013-04-22 1.120024
2013-04-20 -0.721101
2013-04-24 0.379392
2013-04-23 0.924535
2013-04-21 0.531902
2013-04-25 -0.957936
...and take the same slice, we get a different result. It returns the first element inside the range and stops at the first element outside the range.
...并取相同的切片，我们得到不同的结果。它返回范围内的第一个元素并在范围外的第一个元素处停止。
In [18]: df['20130419':'20130422']
Out[18]:
0
2013-04-22 1.120024
2013-04-20 -0.721101
2013-04-24 0.379392
This is probably not useful behavior. If you want to select ranges of dates, would it make sense to sort it by date first?
这可能不是有用的行为。如果您想选择日期范围，先按日期排序是否有意义？
df.sort_index()
回答by watsonic
You can use a simple mask to accomplish this:
您可以使用一个简单的掩码来完成此操作：
date_mask = (data.index > start) & (data.index < end)
dates = data.index[date_mask]
data.ix[dates]
By the way, this works for hierarchical indexing as well. In that case data.indexwould be replaced with data.index.levels[0]or similar.
顺便说一下，这也适用于分层索引。在这种情况下，data.index将替换为data.index.levels[0]或类似的。
回答by R. Cox
I had difficulty with other approaches but I found that the following approach worked for me:
我在使用其他方法时遇到了困难，但我发现以下方法对我有用：
# Set the Index to be the Date
df['Date'] = pd.to_datetime(df['Date_1'], format='%d/%m/%Y')
df.set_index('Date', inplace=True)
# Sort the Data
df = df.sort_values('Date_1')
# Slice the Data
From = '2017-05-07'
To = '2017-06-07'
df_Z = df.loc[From:To,:]
