pandas 获取熊猫日期时间索引的先前值
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get previous value of pandas datetime index
提问by trbck
I have a pandas dataframe with datetime index
我有一个带有日期时间索引的Pandas数据框
Date
2013-02-22 00:00:00+00:00 0.280001
2013-02-25 00:00:00+00:00 0.109999
2013-02-26 00:00:00+00:00 -0.150000
2013-02-27 00:00:00+00:00 0.130001
2013-02-28 00:00:00+00:00 0.139999
Name: MOM12
and want to evaluate the previous three values of the given datetime index.
并且想要评估给定日期时间索引的前三个值。
date = "2013-02-27 00:00:00+00:00"
df.ix[date]
I searched for this but since my index is a date I can't do
我搜索了这个，但因为我的索引是一个我不能做的日期
df.ix[int-1]
采纳答案by Andy Hayden
Here's one way to do it, first grab the integer location of the index key via get_loc:
这是一种方法，首先通过以下方式获取索引键的整数位置get_loc：
In [15]: t = pd.Timestamp("2013-02-27 00:00:00+00:00")
In [16]: df1.index.get_loc(t)
Out[16]: 3
And then you can use iloc(to get the integer location, or slice by integer location):
然后您可以使用iloc（获取整数位置，或按整数位置切片）：
In [17]: loc = df1.index.get_loc(t)
In [18]: df.iloc[loc - 1]
Out[18]:
Date 2013-02-26 00:00:00
-0.15
Name: 2, Dtype: object
In [19]: df1.iloc[slice(max(0, loc-3), min(loc, len(df)))]
# the min and max feel slightly hacky (!) but needed incase it's within top or bottom 3
Out[19]:
Date
2013-02-22 0.280001
2013-02-25 0.109999
2013-02-26 -0.150000
See the indexing section of the docs.
I'm not quite sure how you set up your DataFrame, but that doesn't look like a Datetime Index to me.Here's how I got the DataFrame (with Timestamp index):
我不太确定您是如何设置 DataFrame 的，但这对我来说看起来不像是日期时间索引。这是我获得 DataFrame 的方式（带有时间戳索引）：
In [11]: df = pd.read_clipboard(sep='\s\s+', header=None, parse_dates=[0], names=['Date', None])
In [12]: df
Out[12]:
Date
0 2013-02-22 00:00:00 0.280001
1 2013-02-25 00:00:00 0.109999
2 2013-02-26 00:00:00 -0.150000
3 2013-02-27 00:00:00 0.130001
4 2013-02-28 00:00:00 0.139999
In [13]: df1 = df.set_index('Date')
In [14]: df1
Out[14]:
Date
2013-02-22 0.280001
2013-02-25 0.109999
2013-02-26 -0.150000
2013-02-27 0.130001
2013-02-28 0.139999
回答by dionysos137
Could you just do df.shift().loc[date]?
你能做df.shift().loc[date]吗？
