Question

我想使用滚动窗口计算DataFrame的两列之间的Spearman和/或Pearson Correlation。

我试过df['corr'] = df['col1'].rolling(P).corr(df['col2'])
（P是窗口大小）

但我似乎无法定义该方法。（添加method='spearman'作为参数会产生错误：

File "main.py", line 29, in __init__
_df['corr'] = g['col1'].rolling(P).corr(g['col2'], method = corr_function)
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1287, in corr
**kwargs)
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1054, in corr
_get_corr, pairwise=bool(pairwise))
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1866, in _flex_binary_moment
return f(X, Y)
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1051, in _get_corr
return a.cov(b, **kwargs) / (a.std(**kwargs) * b.std(**kwargs))
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1280, in cov
ddof=ddof, **kwargs)
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1020, in cov
_get_cov, pairwise=bool(pairwise))
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1866, in _flex_binary_moment
return f(X, Y)
File "~\Python36\lib\site-packages\pandas\core\window.py", line 1015, in _get_cov
center=self.center).count(**kwargs)
TypeError: count() got an unexpected keyword argument 'method'

公平地说，我不希望这个工作，因为阅读文档，没有提到rolling.corr支持方法......

考虑到数据帧非常大（> 10M行），有关如何执行此操作的任何建议吗？

Answer 1

rolling.corr做Pearson，所以你可以用它。对于斯皮尔曼，请使用以下内容：

import pandas as pd
from numpy.lib.stride_tricks import as_strided
from numpy.lib import pad
import numpy as np
def rolling_spearman(seqa, seqb, window):
    stridea = seqa.strides[0]
    ssa = as_strided(seqa, shape=[len(seqa) - window + 1, window], strides=[stridea, stridea])
    strideb = seqa.strides[0]
    ssb = as_strided(seqb, shape=[len(seqb) - window + 1, window], strides =[strideb, strideb])
    ar = pd.DataFrame(ssa)
    br = pd.DataFrame(ssb)
    ar = ar.rank(1)
    br = br.rank(1)
    corrs = ar.corrwith(br, 1)
    return pad(corrs, (window - 1, 0), 'constant', constant_values=np.nan)

E.g：

In [144]: df = pd.DataFrame(np.random.randint(0,1000,size=(10,2)), columns = list('ab'))
In [145]: df['corr'] = rolling_spearman(df.a, df.b, 4)
In [146]: df
Out[146]: 
     a    b  corr
0  429  922   NaN
1  618  382   NaN
2  476  517   NaN
3  267  536  -0.8
4  582  844  -0.4
5  254  895  -0.4
6  583  974   0.4
7  687  298  -0.4
8  697  447  -0.6
9  383   35   0.4

说明：numpy.lib.stride_tricks.as_strided是一种hacky方法，在这种情况下，我们可以看到一个看起来像二维数组的序列，其中包含我们正在查看的序列的滚动窗口部分。

从那时起，它很简单。 Spearman相关等效于将序列转换为等级，并采用Pearson相关系数。在DataFrame s上，Pandas有助于快速实现这一行。然后在最后我们使用NaN值填充结果Series的开头（这样您就可以将它作为列添加到数据框或其他任何位置）。

（个人提示：在我意识到你需要的所有东西已经在熊猫中之前......我花了很长时间才试图弄清楚如何有效地使用numpy和scipy来做这件事。）。

为了显示这种方法的速度优势而不仅仅是在滑动窗口上循环，我创建了一个名为srsmall.py的小文件，其中包含：

import pandas as pd
from numpy.lib.stride_tricks import as_strided
import scipy.stats
from numpy.lib import pad
import numpy as np

def rolling_spearman_slow(seqa, seqb, window):
    stridea = seqa.strides[0]
    ssa = as_strided(seqa, shape=[len(seqa) - window + 1, window], strides=[stridea, stridea])
    strideb = seqa.strides[0]
    ssb = as_strided(seqb, shape=[len(seqb) - window + 1, window], strides =[strideb, strideb])
    corrs = [scipy.stats.spearmanr(a, b)[0] for (a,b) in zip(ssa, ssb)]
    return pad(corrs, (window - 1, 0), 'constant', constant_values=np.nan)

def rolling_spearman_quick(seqa, seqb, window):
    stridea = seqa.strides[0]
    ssa = as_strided(seqa, shape=[len(seqa) - window + 1, window], strides=[stridea, stridea])
    strideb = seqa.strides[0]
    ssb = as_strided(seqb, shape=[len(seqb) - window + 1, window], strides =[strideb, strideb])
    ar = pd.DataFrame(ssa)
    br = pd.DataFrame(ssb)
    ar = ar.rank(1)
    br = br.rank(1)
    corrs = ar.corrwith(br, 1)
    return pad(corrs, (window - 1, 0), 'constant', constant_values=np.nan)

并比较表现：

In [1]: import pandas as pd
In [2]: import numpy as np
In [3]: from srsmall import rolling_spearman_slow as slow
In [4]: from srsmall import rolling_spearman_quick as quick
In [5]: for i in range(6):
   ...:     df = pd.DataFrame(np.random.randint(0,1000,size=(10*(10**i),2)), columns=list('ab'))
   ...:     print len(df), " rows"
   ...:     print "quick: ",
   ...:     %timeit quick(df.a, df.b, 4)
   ...:     print "slow: ",
   ...:     %timeit slow(df.a, df.b, 4)
   ...:     
10  rows
quick: 100 loops, best of 3: 3.52 ms per loop
slow: 100 loops, best of 3: 3.2 ms per loop
100  rows
quick: 100 loops, best of 3: 3.53 ms per loop
slow: 10 loops, best of 3: 42 ms per loop
1000  rows
quick: 100 loops, best of 3: 3.82 ms per loop
slow: 1 loop, best of 3: 430 ms per loop
10000  rows
quick: 100 loops, best of 3: 7.47 ms per loop
slow: 1 loop, best of 3: 4.33 s per loop
100000  rows
quick: 10 loops, best of 3: 50.2 ms per loop
slow: 1 loop, best of 3: 43.4 s per loop
1000000  rows
quick: 1 loop, best of 3: 404 ms per loop
slow:

在一百万行（在我的机器上），快速（熊猫）版本在不到半秒的时间内运行。上面没有显示但是1000万花了8.43秒。缓慢的仍然在运行，但假设线性增长继续下去，1M应该需要7分钟，10M需要1个多小时。

熊猫滚动窗口Spearman相关

1 个答案: