[英]Convert integer index from Fama-French factors to datetime index in pandas
I get the Fama-French factors from Ken French's data library using pandas.io.data
, but I can't figure out how to convert the integer year-month date index (eg, 200105
) to a datetime
index so that I can take advantage of more pandas
features. 我使用
pandas.io.data
从Ken French的数据库中获得了Fama-French因子,但是我不知道如何将整数年月日期索引(例如200105
)转换为datetime
索引,以便我可以更多pandas
功能的优势。
The following code runs, but my index attempt in the last un-commented line drops all data in DataFrame ff
. 下面的代码运行,但是我在未注释的最后一行中的索引尝试将删除DataFrame
ff
所有数据。 I also tried .reindex()
, but this doesn't change the index to range
. 我也尝试过
.reindex()
,但这不会将索引更改为range
。 What is the pandas
way? pandas
是什么方式? Thanks! 谢谢!
import pandas as pd
from pandas.io.data import DataReader
import datetime as dt
ff = pd.DataFrame(DataReader("F-F_Research_Data_Factors", "famafrench")[0])
ff.columns = ['Mkt_rf', 'SMB', 'HML', 'rf']
start = ff.index[0]
start = dt.datetime(year=start//100, month=start%100, day=1)
end = ff.index[-1]
end = dt.datetime(year=end//100, month=end%100, day=1)
range = pd.DateRange(start, end, offset=pd.datetools.MonthEnd())
ff = pd.DataFrame(ff, index=range)
#ff.reindex(range)
reindex
realigns the existing index to the given index rather than changing the index. reindex
将现有索引与给定reindex
重新对齐,而不是更改索引。 you can just do ff.index = range
if you've made sure the lengths and the alignment matches. 如果您确定长度和对齐方式匹配,则只需执行
ff.index = range
。
Parsing each original index value is much safer. 解析每个原始索引值要安全得多。 The easy approach is to do this by converting to a string:
一种简单的方法是通过转换为字符串来做到这一点:
In [132]: ints
Out[132]: Int64Index([201201, 201201, 201201, ..., 203905, 203905, 203905])
In [133]: conv = lambda x: datetime.strptime(str(x), '%Y%m')
In [134]: dates = [conv(x) for x in ints]
In [135]: %timeit [conv(x) for x in ints]
1 loops, best of 3: 222 ms per loop
This is kind of slow, so if you have a lot observations you might want to use an optimize cython function in pandas: 这有点慢,因此,如果您有很多观察,则可能要在熊猫中使用优化cython函数:
In [144]: years = (ints // 100).astype(object)
In [145]: months = (ints % 100).astype(object)
In [146]: days = np.ones(len(years), dtype=object)
In [147]: import pandas.lib as lib
In [148]: %timeit Index(lib.try_parse_year_month_day(years, months, days))
100 loops, best of 3: 5.47 ms per loop
Here ints
has 10000 entries. 在这里,
ints
具有10000个条目。
Try this list comprehensions, it works for me: 试试这个列表理解,它对我有用:
ff = pd.DataFrame(DataReader("F-F_Research_Data_Factors", "famafrench")[0])
ff.columns = ['Mkt_rf', 'SMB', 'HML', 'rf']
ff.index = [dt.datetime(d/100, d%100, 1) for d in ff.index]
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