熊猫:按时间间隔表示滚动

I'm new to Pandas.... I've got a bunch of polling data; I want to compute a rolling mean to get an estimate for each day based on a three-day window. As I understand from this question, the rolling_* functions compute the window based on a specified number of values, and not a specific datetime range.

我新的熊猫....我有一堆调查数据;我想计算一个滚动的平均值，以得到基于三天窗口的每天的估计值。正如我从这个问题中了解到的，rolling_*函数基于指定数量的值计算窗口，而不是特定的datetime范围。

Is there a different function that implements this functionality? Or am I stuck writing my own?

是否有一个不同的函数来实现这个功能?还是我自己写的?

EDIT:

编辑:

Sample input data:

样本输入数据:

polls_subset.tail(20)
Out[185]: 
            favorable  unfavorable  other

enddate                                  
2012-10-25       0.48         0.49   0.03
2012-10-25       0.51         0.48   0.02
2012-10-27       0.51         0.47   0.02
2012-10-26       0.56         0.40   0.04
2012-10-28       0.48         0.49   0.04
2012-10-28       0.46         0.46   0.09
2012-10-28       0.48         0.49   0.03
2012-10-28       0.49         0.48   0.03
2012-10-30       0.53         0.45   0.02
2012-11-01       0.49         0.49   0.03
2012-11-01       0.47         0.47   0.05
2012-11-01       0.51         0.45   0.04
2012-11-03       0.49         0.45   0.06
2012-11-04       0.53         0.39   0.00
2012-11-04       0.47         0.44   0.08
2012-11-04       0.49         0.48   0.03
2012-11-04       0.52         0.46   0.01
2012-11-04       0.50         0.47   0.03
2012-11-05       0.51         0.46   0.02
2012-11-07       0.51         0.41   0.00

Output would have only one row for each date.

每个日期的输出只有一行。

EDIT x2: fixed typo

编辑x2:固定的错误

7 个解决方案

#1

What about something like this:

像这样的东西:

First resample the data frame into 1D intervals. This takes the mean of the values for all duplicate days. Use the fill_method option to fill in missing date values. Next, pass the resampled frame into pd.rolling_mean with a window of 3 and min_periods=1 :

首先将数据帧重新采样到一维间隔中。这就取了所有重复天数的值的平均值。使用fill_method选项来填充缺失的日期值。接下来，将重新采样的框架传递到pd。rolling_means窗口为3,min_period =1:

pd.rolling_mean(df.resample("1D", fill_method="ffill"), window=3, min_periods=1)

            favorable  unfavorable     other
enddate
2012-10-25   0.495000     0.485000  0.025000
2012-10-26   0.527500     0.442500  0.032500
2012-10-27   0.521667     0.451667  0.028333
2012-10-28   0.515833     0.450000  0.035833
2012-10-29   0.488333     0.476667  0.038333
2012-10-30   0.495000     0.470000  0.038333
2012-10-31   0.512500     0.460000  0.029167
2012-11-01   0.516667     0.456667  0.026667
2012-11-02   0.503333     0.463333  0.033333
2012-11-03   0.490000     0.463333  0.046667
2012-11-04   0.494000     0.456000  0.043333
2012-11-05   0.500667     0.452667  0.036667
2012-11-06   0.507333     0.456000  0.023333
2012-11-07   0.510000     0.443333  0.013333

UPDATE: As Ben points out in the comments, with pandas 0.18.0 the syntax has changed. With the new syntax this would be:

更新:Ben在评论中指出，熊猫0.18.0的语法已经改变。新的语法将是:

df.resample("1d").sum().fillna(0).rolling(window=3, min_periods=1).mean()

#2

I just had the same question but with irregularly spaced datapoints. Resample is not really an option here. So I created my own function. Maybe it will be useful for others too:

我也有同样的问题，但有不规则间隔的数据点。在这里，重新采样并不是一个选项。所以我创建了我自己的函数。也许它对别人也有用:

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

def rolling_mean(data, window, min_periods=1, center=False):
    ''' Function that computes a rolling mean

    Parameters
    ----------
    data : DataFrame or Series
           If a DataFrame is passed, the rolling_mean is computed for all columns.
    window : int or string
             If int is passed, window is the number of observations used for calculating 
             the statistic, as defined by the function pd.rolling_mean()
             If a string is passed, it must be a frequency string, e.g. '90S'. This is
             internally converted into a DateOffset object, representing the window size.
    min_periods : int
                  Minimum number of observations in window required to have a value.

    Returns
    -------
    Series or DataFrame, if more than one column    
    '''
    def f(x):
        '''Function to apply that actually computes the rolling mean'''
        if center == False:
            dslice = col[x-pd.datetools.to_offset(window).delta+timedelta(0,0,1):x]
                # adding a microsecond because when slicing with labels start and endpoint
                # are inclusive
        else:
            dslice = col[x-pd.datetools.to_offset(window).delta/2+timedelta(0,0,1):
                         x+pd.datetools.to_offset(window).delta/2]
        if dslice.size < min_periods:
            return np.nan
        else:
            return dslice.mean()

    data = DataFrame(data.copy())
    dfout = DataFrame()
    if isinstance(window, int):
        dfout = pd.rolling_mean(data, window, min_periods=min_periods, center=center)
    elif isinstance(window, basestring):
        idx = Series(data.index.to_pydatetime(), index=data.index)
        for colname, col in data.iterkv():
            result = idx.apply(f)
            result.name = colname
            dfout = dfout.join(result, how='outer')
    if dfout.columns.size == 1:
        dfout = dfout.ix[:,0]
    return dfout


# Example
idx = [datetime(2011, 2, 7, 0, 0),
       datetime(2011, 2, 7, 0, 1),
       datetime(2011, 2, 7, 0, 1, 30),
       datetime(2011, 2, 7, 0, 2),
       datetime(2011, 2, 7, 0, 4),
       datetime(2011, 2, 7, 0, 5),
       datetime(2011, 2, 7, 0, 5, 10),
       datetime(2011, 2, 7, 0, 6),
       datetime(2011, 2, 7, 0, 8),
       datetime(2011, 2, 7, 0, 9)]
idx = pd.Index(idx)
vals = np.arange(len(idx)).astype(float)
s = Series(vals, index=idx)
rm = rolling_mean(s, window='2min')

#3

In the meantime, a time-window capability was added. See the link below:

同时，还添加了一个时间窗口功能。请参阅下面的链接:

https://github.com/pydata/pandas/pull/13513

In [1]: df = DataFrame({'B': range(5)})

In [2]: df.index = [Timestamp('20130101 09:00:00'),
   ...:             Timestamp('20130101 09:00:02'),
   ...:             Timestamp('20130101 09:00:03'),
   ...:             Timestamp('20130101 09:00:05'),
   ...:             Timestamp('20130101 09:00:06')]

In [3]: df
Out[3]: 
                     B
2013-01-01 09:00:00  0
2013-01-01 09:00:02  1
2013-01-01 09:00:03  2
2013-01-01 09:00:05  3
2013-01-01 09:00:06  4

In [4]: df.rolling(2, min_periods=1).sum()
Out[4]: 
                       B
2013-01-01 09:00:00  0.0
2013-01-01 09:00:02  1.0
2013-01-01 09:00:03  3.0
2013-01-01 09:00:05  5.0
2013-01-01 09:00:06  7.0

In [5]: df.rolling('2s', min_periods=1).sum()
Out[5]: 
                       B
2013-01-01 09:00:00  0.0
2013-01-01 09:00:02  1.0
2013-01-01 09:00:03  3.0
2013-01-01 09:00:05  3.0
2013-01-01 09:00:06  7.0

#4

user2689410's code was exactly what I needed. Providing my version (credits to user2689410), which is faster due to calculating mean at once for whole rows in the DataFrame.

user2689410的代码正是我所需要的。提供我的版本(归功于user2689410)，这更快了，因为它一次计算DataFrame中所有行的平均值。

Hope my suffix conventions are readable: _s: string, _i: int, _b: bool, _ser: Series and _df: DataFrame. Where you find multiple suffixes, type can be both.

希望我的后缀约定是可读的:_s: string， _i: int， _b: bool， _ser: Series和_df: DataFrame。在你找到多个后缀的地方，类型可以都是。

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

def time_offset_rolling_mean_df_ser(data_df_ser, window_i_s, min_periods_i=1, center_b=False):
    """ Function that computes a rolling mean

    Credit goes to user2689410 at http://*.com/questions/15771472/pandas-rolling-mean-by-time-interval

    Parameters
    ----------
    data_df_ser : DataFrame or Series
         If a DataFrame is passed, the time_offset_rolling_mean_df_ser is computed for all columns.
    window_i_s : int or string
         If int is passed, window_i_s is the number of observations used for calculating
         the statistic, as defined by the function pd.time_offset_rolling_mean_df_ser()
         If a string is passed, it must be a frequency string, e.g. '90S'. This is
         internally converted into a DateOffset object, representing the window_i_s size.
    min_periods_i : int
         Minimum number of observations in window_i_s required to have a value.

    Returns
    -------
    Series or DataFrame, if more than one column

    >>> idx = [
    ...     datetime(2011, 2, 7, 0, 0),
    ...     datetime(2011, 2, 7, 0, 1),
    ...     datetime(2011, 2, 7, 0, 1, 30),
    ...     datetime(2011, 2, 7, 0, 2),
    ...     datetime(2011, 2, 7, 0, 4),
    ...     datetime(2011, 2, 7, 0, 5),
    ...     datetime(2011, 2, 7, 0, 5, 10),
    ...     datetime(2011, 2, 7, 0, 6),
    ...     datetime(2011, 2, 7, 0, 8),
    ...     datetime(2011, 2, 7, 0, 9)]
    >>> idx = pd.Index(idx)
    >>> vals = np.arange(len(idx)).astype(float)
    >>> ser = pd.Series(vals, index=idx)
    >>> df = pd.DataFrame({'s1':ser, 's2':ser+1})
    >>> time_offset_rolling_mean_df_ser(df, window_i_s='2min')
                          s1   s2
    2011-02-07 00:00:00  0.0  1.0
    2011-02-07 00:01:00  0.5  1.5
    2011-02-07 00:01:30  1.0  2.0
    2011-02-07 00:02:00  2.0  3.0
    2011-02-07 00:04:00  4.0  5.0
    2011-02-07 00:05:00  4.5  5.5
    2011-02-07 00:05:10  5.0  6.0
    2011-02-07 00:06:00  6.0  7.0
    2011-02-07 00:08:00  8.0  9.0
    2011-02-07 00:09:00  8.5  9.5
    """

    def calculate_mean_at_ts(ts):
        """Function (closure) to apply that actually computes the rolling mean"""
        if center_b == False:
            dslice_df_ser = data_df_ser[
                ts-pd.datetools.to_offset(window_i_s).delta+timedelta(0,0,1):
                ts
            ]
            # adding a microsecond because when slicing with labels start and endpoint
            # are inclusive
        else:
            dslice_df_ser = data_df_ser[
                ts-pd.datetools.to_offset(window_i_s).delta/2+timedelta(0,0,1):
                ts+pd.datetools.to_offset(window_i_s).delta/2
            ]
        if  (isinstance(dslice_df_ser, pd.DataFrame) and dslice_df_ser.shape[0] < min_periods_i) or \
            (isinstance(dslice_df_ser, pd.Series) and dslice_df_ser.size < min_periods_i):
            return dslice_df_ser.mean()*np.nan   # keeps number format and whether Series or DataFrame
        else:
            return dslice_df_ser.mean()

    if isinstance(window_i_s, int):
        mean_df_ser = pd.rolling_mean(data_df_ser, window=window_i_s, min_periods=min_periods_i, center=center_b)
    elif isinstance(window_i_s, basestring):
        idx_ser = pd.Series(data_df_ser.index.to_pydatetime(), index=data_df_ser.index)
        mean_df_ser = idx_ser.apply(calculate_mean_at_ts)

    return mean_df_ser

#5

This example seems to call for a weighted mean as suggested in @andyhayden's comment. For example, there are two polls on 10/25 and one each on 10/26 and 10/27. If you just resample and then take the mean, this effectively gives twice as much weighting to the polls on 10/26 and 10/27 compared to the ones on 10/25.

这个例子似乎需要一个加权平均数，如@andyhayden的评论所建议的那样。例如，10/25有两个投票，10/26和10/27各有一个。如果你重新抽样，然后取平均值，这实际上是10/26和10/27的投票权重是10/25的两倍。

To give equal weight to each poll rather than equal weight to each day, you could do something like the following.

要给每个投票相同的权重而不是每天相同的权重，您可以做如下的事情。

>>> wt = df.resample('D',limit=5).count()

            favorable  unfavorable  other
enddate                                  
2012-10-25          2            2      2
2012-10-26          1            1      1
2012-10-27          1            1      1

>>> df2 = df.resample('D').mean()

            favorable  unfavorable  other
enddate                                  
2012-10-25      0.495        0.485  0.025
2012-10-26      0.560        0.400  0.040
2012-10-27      0.510        0.470  0.020

That gives you the raw ingredients for doing a poll-based mean instead of a day-based mean. As before, the polls are averaged on 10/25, but the weight for 10/25 is also stored and is double the weight on 10/26 or 10/27 to reflect that two polls were taken on 10/25.

这就为你提供了做基于民意的平均数而不是基于日的平均数的原材料。和以前一样，投票的平均值是10/25，但是10/25的权重也被存储，并且是10/26或10/27的两倍，以反映在10/25上进行了两次投票。

>>> df3 = df2 * wt
>>> df3 = df3.rolling(3,min_periods=1).sum()
>>> wt3 = wt.rolling(3,min_periods=1).sum()

>>> df3 = df3 / wt3  

            favorable  unfavorable     other
enddate                                     
2012-10-25   0.495000     0.485000  0.025000
2012-10-26   0.516667     0.456667  0.030000
2012-10-27   0.515000     0.460000  0.027500
2012-10-28   0.496667     0.465000  0.041667
2012-10-29   0.484000     0.478000  0.042000
2012-10-30   0.488000     0.474000  0.042000
2012-10-31   0.530000     0.450000  0.020000
2012-11-01   0.500000     0.465000  0.035000
2012-11-02   0.490000     0.470000  0.040000
2012-11-03   0.490000     0.465000  0.045000
2012-11-04   0.500000     0.448333  0.035000
2012-11-05   0.501429     0.450000  0.032857
2012-11-06   0.503333     0.450000  0.028333
2012-11-07   0.510000     0.435000  0.010000

Note that the rolling mean for 10/27 is now 0.51500 (poll-weighted) rather than 52.1667 (day-weighted).

注意，10/27的滚动平均值现在是0.51500(授粉加权)，而不是52.1667(日加权)。

Also note that there have been changes to the APIs for resample and rolling as of version 0.18.0.

还需要注意的是，在0.18.0版本中，对重新采样和滚动的api进行了修改。

rolling (what's new in pandas 0.18.0)

滚动(熊猫有什么新鲜事吗? 0.18.0)

resample (what's new in pandas 0.18.0)

重新采样(熊猫有什么新特性? 0.18.0)

#6

I found that user2689410 code broke when I tried with window='1M' as the delta on business month threw this error:

我发现user2689410代码在我尝试使用window='1M'时崩溃了，因为业务月的delta抛出了这个错误:

AttributeError: 'MonthEnd' object has no attribute 'delta'

I added the option to pass directly a relative time delta, so you can do similar things for user defined periods.

我添加了直接传递相对时间增量的选项，因此您可以对用户定义的周期执行类似的操作。

Thanks for the pointers, here's my attempt - hope it's of use.

谢谢你的指点，这是我的尝试——希望有用。

def rolling_mean(data, window, min_periods=1, center=False):
""" Function that computes a rolling mean
Reference:
    http://*.com/questions/15771472/pandas-rolling-mean-by-time-interval

Parameters
----------
data : DataFrame or Series
       If a DataFrame is passed, the rolling_mean is computed for all columns.
window : int, string, Timedelta or Relativedelta
         int - number of observations used for calculating the statistic,
               as defined by the function pd.rolling_mean()
         string - must be a frequency string, e.g. '90S'. This is
                  internally converted into a DateOffset object, and then
                  Timedelta representing the window size.
         Timedelta / Relativedelta - Can directly pass a timedeltas.
min_periods : int
              Minimum number of observations in window required to have a value.
center : bool
         Point around which to 'center' the slicing.

Returns
-------
Series or DataFrame, if more than one column
"""
def f(x, time_increment):
    """Function to apply that actually computes the rolling mean
    :param x:
    :return:
    """
    if not center:
        # adding a microsecond because when slicing with labels start
        # and endpoint are inclusive
        start_date = x - time_increment + timedelta(0, 0, 1)
        end_date = x
    else:
        start_date = x - time_increment/2 + timedelta(0, 0, 1)
        end_date = x + time_increment/2
    # Select the date index from the
    dslice = col[start_date:end_date]

    if dslice.size < min_periods:
        return np.nan
    else:
        return dslice.mean()

data = DataFrame(data.copy())
dfout = DataFrame()
if isinstance(window, int):
    dfout = pd.rolling_mean(data, window, min_periods=min_periods, center=center)

elif isinstance(window, basestring):
    time_delta = pd.datetools.to_offset(window).delta
    idx = Series(data.index.to_pydatetime(), index=data.index)
    for colname, col in data.iteritems():
        result = idx.apply(lambda x: f(x, time_delta))
        result.name = colname
        dfout = dfout.join(result, how='outer')

elif isinstance(window, (timedelta, relativedelta)):
    time_delta = window
    idx = Series(data.index.to_pydatetime(), index=data.index)
    for colname, col in data.iteritems():
        result = idx.apply(lambda x: f(x, time_delta))
        result.name = colname
        dfout = dfout.join(result, how='outer')

if dfout.columns.size == 1:
    dfout = dfout.ix[:, 0]
return dfout

And the example with a 3 day time window to calculate the mean:

用3天时间窗口计算平均值的例子:

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

idx = [datetime(2011, 2, 7, 0, 0),
           datetime(2011, 2, 7, 0, 1),
           datetime(2011, 2, 8, 0, 1, 30),
           datetime(2011, 2, 9, 0, 2),
           datetime(2011, 2, 10, 0, 4),
           datetime(2011, 2, 11, 0, 5),
           datetime(2011, 2, 12, 0, 5, 10),
           datetime(2011, 2, 12, 0, 6),
           datetime(2011, 2, 13, 0, 8),
           datetime(2011, 2, 14, 0, 9)]
idx = pd.Index(idx)
vals = np.arange(len(idx)).astype(float)
s = Series(vals, index=idx)
# Now try by passing the 3 days as a relative time delta directly.
rm = rolling_mean(s, window=relativedelta(days=3))
>>> rm
Out[2]: 
2011-02-07 00:00:00    0.0
2011-02-07 00:01:00    0.5
2011-02-08 00:01:30    1.0
2011-02-09 00:02:00    1.5
2011-02-10 00:04:00    3.0
2011-02-11 00:05:00    4.0
2011-02-12 00:05:10    5.0
2011-02-12 00:06:00    5.5
2011-02-13 00:08:00    6.5
2011-02-14 00:09:00    7.5
Name: 0, dtype: float64

#7

To keep it basic, I used a loop and something like this to get you started (my index are datetimes):

为了保持基本，我使用了一个循环和类似这样的东西来启动(我的索引是datetimes):

import pandas as pd
import datetime as dt

#populate your dataframe: "df"
#...

df[df.index<(df.index[0]+dt.timedelta(hours=1))] #gives you a slice. you can then take .sum() .mean(), whatever

and then you can run functions on that slice. You can see how adding an iterator to make the start of the window something other than the first value in your dataframes index would then roll the window (you could use a > rule for the start as well for example).

然后你可以在那个切片上运行函数。您可以看到如何添加一个迭代器以使窗口的开始不是dataframes索引中的第一个值，然后滚动窗口(例如，您也可以使用>规则作为开始)。

Note, this may be less efficient for SUPER large data or very small increments as your slicing may become more strenuous (works for me well enough for hundreds of thousands of rows of data and several columns though for hourly windows across a few weeks)

注意，对于超大数据或非常小的增量，这可能会降低效率，因为您的切片可能会变得更费力(对我来说，这对于成千上万行数据和几列数据足够有效，尽管对于几周内的每小时窗口)

#1