Pandas滑动/滚动窗口在不规则时间序列上的应用

Question

请原谅我的风格不太好，解决方案也不够高效。非常感谢您的所有帮助。 背景： 尝试在一年的时间内，隔离出6周内最佳的骑车表现提升率。性能是通过测量任何给定时间段内产生的最大功率来衡量的，例如1、5、20分钟的努力等等... 任务：

1. 创建滚动窗口
2. 为每个窗口创建最佳趋势线
3. 保留对应于最大正斜率的窗口

数据：

ap1 = np.array([[datetime(2015, 10, 17, 12, 45, 13),
   datetime(2015, 10, 18, 11, 56, 35),
   datetime(2015, 10, 20, 9, 24, 52),
   datetime(2015, 10, 23, 9, 27, 12),
   datetime(2015, 10, 24, 12, 26, 33)], 
[281.0, 343.0, 270.0, 312.0, 320.0], 
[246.0, 305.0, 260.0, 283.0, 289.0], 
[236.0, 250.0, 239.0, 257.0, 245.0]], dtype=object)

问题：我目前卡在任务1上。我一直在尝试遵循user2689410对于计算不规则时间序列的rolling_mean的响应。我希望能够使用他的数据切片方法。

我只想将数据集切成45天的滚动间隔。以下是进展：

from pandas import Series, DataFrame
import pandas as pd
from datetime import datetime, timedelta
import numpy as np

idx = ap1[0]
idx = pd.Index(idx)

ap1=np.transpose(ap1)
ap1=pd.DataFrame(ap1, index = idx, columns = ['date', 'cp1', 'cp2', 'cp3'])
ap2=ap1.drop('date', 1)

ap2 = DataFrame(ap2.copy())
idx = Series(ap2.index.to_pydatetime(), index=ap2.index)

for colname, col in ap2.iteritems():
    dslice = col[idx-pd.tseries.frequencies.to_offset('42D').delta:idx]

for循环给我报错：

Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "/usr/local/lib64/python2.7/site-packages/pandas/core/series.py", line 642, in __getitem__
return self._get_with(key)
File "/usr/local/lib64/python2.7/site-packages/pandas/core/series.py", line 647, in _get_with
indexer = self.index._convert_slice_indexer(key, kind='getitem')
File "/usr/local/lib64/python2.7/site-packages/pandas/indexes/base.py", line 1208, in _convert_slice_indexer
indexer = self.slice_indexer(start, stop, step, kind=kind)
File "/usr/local/lib64/python2.7/site-packages/pandas/tseries/index.py", line 1497, in slice_indexer
return Index.slice_indexer(self, start, end, step, kind=kind)
File "/usr/local/lib64/python2.7/site-packages/pandas/indexes/base.py", line 2962, in slice_indexer
kind=kind)
File "/usr/local/lib64/python2.7/site-packages/pandas/indexes/base.py", line 3141, in slice_locs
start_slice = self.get_slice_bound(start, 'left', kind)
File "/usr/local/lib64/python2.7/site-packages/pandas/indexes/base.py", line 3084, in get_slice_bound
slc = self.get_loc(label)
File "/usr/local/lib64/python2.7/site-packages/pandas/tseries/index.py", line 1419, in get_loc
stamp = Timestamp(key, tz=self.tz)
File "pandas/tslib.pyx", line 405, in pandas.tslib.Timestamp.__new__ (pandas/tslib.c:9932)
File "pandas/tslib.pyx", line 1475, in pandas.tslib.convert_to_tsobject (pandas/tslib.c:26432)
TypeError: Cannot convert input to Timestamp

我接下来该怎么办？

- Rudabagle

Answer 1

现如今，pandas.DataFrame.rolling 可以处理不规则时间序列。

- Robert Pollak

Answer 2

我找到了一个解决方案，虽然不太美观但是可行。请提供反馈以提高效率。这个解决方案为我提供了与特定列的移动窗口对应的子数组数组。

 idx = ap2[1]
 idx = pd.Index(idx)

 ap2 = np.transpose(ap2)
 ap2 = pd.DataFrame(ap2, index = idx, columns = ['date', 'cp1', 'cp2', 'cp3'])
 ap2=ap2.drop('date', 1)
 ap2=ap2.astype(float)
 ap2 = DataFrame(ap2.copy())
 dfout = DataFrame()

 idx = Series(ap2.index.to_pydatetime(), index=ap2.index)

 window = '42D'

 idxwindow = idx[idx[0]:idx[len(idx)-1]-pd.tseries.frequencies.to_offset(window).delta]

for i in ap2:
    exec(i +"= []")
for colname, col in ap2.iteritems():
    for i in idxwindow:
        result=col[i:i+pd.tseries.frequencies.to_offset(window).delta]
        result=np.stack((result.index.date, result.values), axis=-1)
        if colname == 'cp1':
            cp1.append(result)
        elif colname == 'cp2':
            cp2.append(result)
        else:
            cp3.append(result)

- Rudabagle