本文介绍了为什么大 pandas 使用单维ndarray的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我被动用pandas 滚动功能来执行滚动多因子回归(这个问题是 NOT 关于滚动多因子回归)。我希望能够在 df.rolling(2)之后使用 apply ,并将生成的 pd.DataFrame 用 .values 提取ndarray并执行必要的矩阵乘法。它没有这样做。



这是我发现的:

  import pandas as pd 
import numpy as np

np.random.seed([3,1415])
df = pd.DataFrame(np.random。 rand(5,2).round(2),columns = ['A','B'])
X = np.random.rand(2,1).round(2)

对象是什么样的:

 打印\\\
df = \ n,df
打印\\\
X = \ n,X
打印\\\
df.shape =,df.shape, ,X.shape =,X.shape

df =
AB
0 0.44 0.41
1 0.46 0.47
2 0.46 0.02
3 0.85 0.82
4 0.78 0.76

X =
[[0.93]
[0.83]]

df.shape =( 5,2),X.shape =(2L,1L)

矩阵乘法表现正常: p>

  df.values.dot(X)

数组([[0.7495],
[ 0.8179],
[0.4444],
[1.4711],
[1.3562]])

使用apply逐行执行产品的行为与预期相同:

  df.apply(lambda x:x.values.dot(X)[0],axis = 1)

0 0.7495
1 0.8179
2 0.4444
3 1.4711
4 1.3562
dtype:float64

Groupby - >应用行为正如我所料:

  df.groupby(level = 0).apply(lambda x:x.values.dot X)[0,0])

0 0.7495
1 0.8179
2 0.4444
3 1.4711
4 1.3562
dtype:float64

但是当我运行时:

  df.rolling(1).apply(lambda x:x.values.dot(X))

我得到:

好吧,熊猫在其<滚动执行。我可以处理。而不是使用 .values 来获取 ndarray ,我们来试试:

  df.rolling(1).apply(lambda x:x.dot(X))



lockquote

等等!什么?!



因此,我创建了一个自定义函数来查看滚动正在做什么。

  def print_type_sum(x):
print type(x),x.shape
return x.sum()

然后运行:

  print df.rolling(1)。 apply(print_type_sum)

< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L,)
< type'numpy.ndarray'> (1L)
AB
0 0.44 0.41
1 0.46 0.47
2 0.46 0.02
3 0.85 0.82
4 0.78 0.76

我的结果 pd.DataFrame 是一样的,这很好。但它打印出10个单维 ndarray 对象。关于 rolling(2)

  print df.rolling(2) .apply(print_type_sum)

< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L,)
< type'numpy.ndarray'> (2L)
AB
0 NaN NaN
1 0.90 0.88
2 0.92 0.49
3 1.31 0.84
4 1.63 1.58

同样的事情,期望输出,但它打印8 ndarray 对象。 rolling 产生长度窗口的单维 ndarray ,用于每一列与我期望的相反,这是一个 ndarray 形状(window,len(df.columns)) 。

问题是为什么?



我现在没有办法轻松地运行滚动多线程,因子回归。

解决方案

使用,这是一个向量化的方法 -

  get_sliding_window(df,2).dot(X)#window size = 2 

运行时测试 -

 在[101]中:df = pd.DataFrame(np.random.rand(5,2).round(2)在[102]中:X = np.array([2,3])

在[103]中,列= ['A','B'])

:roll_df = roll(df,2)

在[104]中:%timeit rolled_df.apply(lambda df:pd.Series(df.values.dot(X)))
100循环,最好每个循环3:5.51 ms

在[105]中:%timeit get_sliding_window(df,2).dot(X)
10000循环,最好是3:每循环43.7μs

验证结果 -

  In [106]:rolled_df.apply(lambda df:pd.Series(df.values.dot(X)))
Out [106]:
0 1
1 2.70 4.09
2 4.09 2.52
3 2.52 1.78
4 1.78 3.50

在[107]中:get_sliding_window(df,2).dot(X)
[107]:
数组([[2.7,4.09],
[4.09,2.52],
[2.52,1.78],
[1.78,3.5]])

巨大的改进,我希望可以在更大的数组上保持显着!


I was motivated to use pandas rolling feature to perform a rolling multi-factor regression (This question is NOT about rolling multi-factor regression). I expected that I'd be able to use apply after a df.rolling(2) and take the resulting pd.DataFrame extract the ndarray with .values and perform the requisite matrix multiplication. It didn't work out that way.

Here is what I found:

import pandas as pd
import numpy as np

np.random.seed([3,1415])
df = pd.DataFrame(np.random.rand(5, 2).round(2), columns=['A', 'B'])
X = np.random.rand(2, 1).round(2)

What do objects look like:

print "\ndf = \n", df
print "\nX = \n", X
print "\ndf.shape =", df.shape, ", X.shape =", X.shape

df = 
      A     B
0  0.44  0.41
1  0.46  0.47
2  0.46  0.02
3  0.85  0.82
4  0.78  0.76

X = 
[[ 0.93]
 [ 0.83]]

df.shape = (5, 2) , X.shape = (2L, 1L)

Matrix multiplication behaves normally:

df.values.dot(X)

array([[ 0.7495],
       [ 0.8179],
       [ 0.4444],
       [ 1.4711],
       [ 1.3562]])

Using apply to perform row by row dot product behaves as expected:

df.apply(lambda x: x.values.dot(X)[0], axis=1)

0    0.7495
1    0.8179
2    0.4444
3    1.4711
4    1.3562
dtype: float64

Groupby -> Apply behaves as I'd expect:

df.groupby(level=0).apply(lambda x: x.values.dot(X)[0, 0])

0    0.7495
1    0.8179
2    0.4444
3    1.4711
4    1.3562
dtype: float64

But when I run:

df.rolling(1).apply(lambda x: x.values.dot(X))

I get:

Ok, so pandas is using straight ndarray within its rolling implementation. I can handle that. Instead of using .values to get the ndarray, let's try:

df.rolling(1).apply(lambda x: x.dot(X))

Wait! What?!

So I created a custom function to look at the what rolling is doing.

def print_type_sum(x):
    print type(x), x.shape
    return x.sum()

Then ran:

print df.rolling(1).apply(print_type_sum)

<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
<type 'numpy.ndarray'> (1L,)
      A     B
0  0.44  0.41
1  0.46  0.47
2  0.46  0.02
3  0.85  0.82
4  0.78  0.76

My resulting pd.DataFrame is the same, that's good. But it printed out 10 single dimensional ndarray objects. What about rolling(2)

print df.rolling(2).apply(print_type_sum)

<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
<type 'numpy.ndarray'> (2L,)
      A     B
0   NaN   NaN
1  0.90  0.88
2  0.92  0.49
3  1.31  0.84
4  1.63  1.58

Same thing, expect output but it printed 8 ndarray objects. rolling is producing a single dimensional ndarray of length window for each column as opposed to what I expected which was an ndarray of shape (window, len(df.columns)).

Question is Why?

I now don't have a way to easily run a rolling multi-factor regression.

解决方案

Using the strides views concept on dataframe, here's a vectorized approach -

get_sliding_window(df, 2).dot(X) # window size = 2

Runtime test -

In [101]: df = pd.DataFrame(np.random.rand(5, 2).round(2), columns=['A', 'B'])

In [102]: X = np.array([2, 3])

In [103]: rolled_df = roll(df, 2)

In [104]: %timeit rolled_df.apply(lambda df: pd.Series(df.values.dot(X)))
100 loops, best of 3: 5.51 ms per loop

In [105]: %timeit get_sliding_window(df, 2).dot(X)
10000 loops, best of 3: 43.7 µs per loop

Verify results -

In [106]: rolled_df.apply(lambda df: pd.Series(df.values.dot(X)))
Out[106]: 
      0     1
1  2.70  4.09
2  4.09  2.52
3  2.52  1.78
4  1.78  3.50

In [107]: get_sliding_window(df, 2).dot(X)
Out[107]: 
array([[ 2.7 ,  4.09],
       [ 4.09,  2.52],
       [ 2.52,  1.78],
       [ 1.78,  3.5 ]])

Huge improvement there, which I am hoping would stay noticeable on larger arrays!

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10-28 07:59