pandas.DataFrame.rolling中没有步长选项吗?还有另一个函数可以帮我吗?

问题描述 投票:1回答:3

在R中,您可以计算具有指定窗口的滚动平均值,该窗口可以每次移动指定量。

但是也许我只是在任何地方都没有找到它,但似乎无法在熊猫或其他Python库中做到这一点?

有人知道解决此问题的方法吗?我将举例说明我的意思:

example

这里有半个月的数据,我正在计算两个月的移动平均值,每个月都在变化。

所以在R中,我将执行以下操作:two_month__movavg=rollapply(mydata,4,mean,by = 2,na.pad = FALSE) Python中没有等效项吗?

EDIT1:

DATE  A DEMAND   ...     AA DEMAND  A Price
    0  2006/01/01 00:30:00  8013.27833   ...     5657.67500    20.03
    1  2006/01/01 01:00:00  7726.89167   ...     5460.39500    18.66
    2  2006/01/01 01:30:00  7372.85833   ...     5766.02500    20.38
    3  2006/01/01 02:00:00  7071.83333   ...     5503.25167    18.59
    4  2006/01/01 02:30:00  6865.44000   ...     5214.01500    17.53
python r pandas numpy zoo
3个回答
1
投票

您可以再次使用滚动,只需要一点点分配索引就可以了

这里by = 2

by = 2

df.loc[df.index[np.arange(len(df))%by==1],'New']=df.Price.rolling(window=4).mean()
df
    Price    New
0      63    NaN
1      92    NaN
2      92    NaN
3       5  63.00
4      90    NaN
5       3  47.50
6      81    NaN
7      98  68.00
8     100    NaN
9      58  84.25
10     38    NaN
11     15  52.75
12     75    NaN
13     19  36.75

1
投票

现在,对于一维数据数组来说,这有点过头了,但是您可以简化它并提取所需的内容。由于熊猫可以依赖numpy,因此您可能需要检查一下熊猫的滚动/跨步功能(如果实现了)。结果为20个连续数字。 7天的窗口,大步/滑动2

    z = np.arange(20)
    z   #array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
    s = stride(z, (7,), (2,))

np.mean(s, axis=1)  # array([ 3.,  5.,  7.,  9., 11., 13., 15.])

这是我使用的代码,不包含文档的主要部分。它源自numpy中stridd函数的许多实现,可以在此站点上找到。有变体和化身,这只是另一个。

def stride(a, win=(3, 3), stepby=(1, 1)):
    """Provide a 2D sliding/moving view of an array.
    There is no edge correction for outputs. Use the `pad_` function first."""
    err = """Array shape, window and/or step size error.
    Use win=(3,) with stepby=(1,) for 1D array
    or win=(3,3) with stepby=(1,1) for 2D array
    or win=(1,3,3) with stepby=(1,1,1) for 3D
    ----    a.ndim != len(win) != len(stepby) ----
    """
    from numpy.lib.stride_tricks import as_strided
    a_ndim = a.ndim
    if isinstance(win, int):
        win = (win,) * a_ndim
    if isinstance(stepby, int):
        stepby = (stepby,) * a_ndim
    assert (a_ndim == len(win)) and (len(win) == len(stepby)), err
    shp = np.array(a.shape)    # array shape (r, c) or (d, r, c)
    win_shp = np.array(win)    # window      (3, 3) or (1, 3, 3)
    ss = np.array(stepby)      # step by     (1, 1) or (1, 1, 1)
    newshape = tuple(((shp - win_shp) // ss) + 1) + tuple(win_shp)
    newstrides = tuple(np.array(a.strides) * ss) + a.strides
    a_s = as_strided(a, shape=newshape, strides=newstrides, subok=True).squeeze()
    return a_s

我未能指出您可以创建输出,并可以将其作为列附加到熊猫中。回到上面使用的原始定义

nans = np.full_like(z, np.nan, dtype='float')  # z is the 20 number sequence
means = np.mean(s, axis=1)   # results from the strided mean
# assign the means to the output array skipping the first and last 3 and striding by 2

nans[3:-3:2] = means        

nans # array([nan, nan, nan,  3., nan,  5., nan,  7., nan,  9., nan, 11., nan, 13., nan, 15., nan, nan, nan, nan])

0
投票

如果数据大小不是太大,这是一个简单的方法:

by = 2
win = 4
start = 3 ## it's the index of your 1st valid value.
df.rolling(win).mean()[start::by] ## caculate all, choos what you need.
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