python - 在数据框上滚动函数

Question

我有以下数据框 C。

>>> C
              a    b   c
2011-01-01    0    0 NaN
2011-01-02   41   12 NaN
2011-01-03   82   24 NaN
2011-01-04  123   36 NaN
2011-01-05  164   48 NaN
2011-01-06  205   60   2
2011-01-07  246   72   4
2011-01-08  287   84   6
2011-01-09  328   96   8
2011-01-10  369  108  10

我想添加一个新列 d，我在固定窗口(此处为 6)上应用滚动函数，我以某种方式为每一行(或日期)< em>修复值c。这个滚动函数中的一个循环应该是(伪):

              a    b   c   d
2011-01-01    0    0 NaN   a + b*2 (a,b from this row, '2' is from 'c' on 2011-01-06)
2011-01-02   41   12 NaN   a + b*2 (a,b from this row, '2' is still from 2011-01-06)
2011-01-03   82   24 NaN   a + b*2
2011-01-04  123   36 NaN   a + b*2
2011-01-05  164   48 NaN   a + b*2
2011-01-06  205   60   2   a + b*2
2011-01-07  246   72   4   
2011-01-08  287   84   6   
2011-01-09  328   96   8   
2011-01-10  369  108  10

在这个“循环”之后，我想在 d 中获取所有这 6 个计算行并运行一个函数调用，该函数调用将返回 一个 值，这应该存储在另一列中，e 说:

              a    b   c   d                               e

2011-01-01    0    0 NaN   a + b*2 ---|                   NaN
2011-01-02   41   12 NaN   a + b*2    |                   NaN
2011-01-03   82   24 NaN   a + b*2    | These values      NaN
2011-01-04  123   36 NaN   a + b*2    | are input to      NaN
2011-01-05  164   48 NaN   a + b*2    | function          NaN
2011-01-06  205   60   2   a + b*2 ---| yielding          X
2011-01-07  246   72   4                value X in
2011-01-08  287   84   6                column 'e'
2011-01-09  328   96   8   
2011-01-10  369  108  10

然后，此过程将迭代到下一个 窗口(同样为 6 长)，例如:

              a    b   c   d             e
2011-01-01    0    0 NaN   
2011-01-02   41   12 NaN   a + b*4 (a,b from this row, '4' is from 'c' now from 2011-01-07)
2011-01-03   82   24 NaN   a + b*4 (a,b from this row, '4' is still from 2011-01-07)
2011-01-04  123   36 NaN   a + b*4
2011-01-05  164   48 NaN   a + b*4
2011-01-06  205   60   2   a + b*4       X
2011-01-07  246   72   4   a + b*4
2011-01-08  287   84   6   
2011-01-09  328   96   8   
2011-01-10  369  108  10

              a    b   c   d                               e

2011-01-01    0    0 NaN                                  NaN
2011-01-02   41   12 NaN   a + b*4 ---|                   NaN
2011-01-03   82   24 NaN   a + b*4    | These values      NaN
2011-01-04  123   36 NaN   a + b*4    | are input to      NaN
2011-01-05  164   48 NaN   a + b*4    | function          NaN
2011-01-06  205   60   2   a + b*4    | yielding          X
2011-01-07  246   72   4   a + b*4 ---| value Y in        Y
2011-01-08  287   84   6                column 'e'
2011-01-09  328   96   8   
2011-01-10  369  108  10

希望这足够清楚，

谢谢，N

Answer 1

你可以使用pd.rolling_apply:

import numpy as np
import pandas as pd
df = pd.read_table('data', sep='\s+')

def foo(x, df):
    window = df.iloc[x]
    # print(window)
    c = df.ix[int(x[-1]), 'c']
    dvals = window['a'] + window['b']*c
    return bar(dvals)

def bar(dvals):
    # print(dvals)
    return dvals.mean()

df['e'] = pd.rolling_apply(np.arange(len(df)), 6, foo, args=(df,))
print(df)

产量

              a    b   c       e
2011-01-01    0    0 NaN     NaN
2011-01-02   41   12 NaN     NaN
2011-01-03   82   24 NaN     NaN
2011-01-04  123   36 NaN     NaN
2011-01-05  164   48 NaN     NaN
2011-01-06  205   60   2   162.5
2011-01-07  246   72   4   311.5
2011-01-08  287   84   6   508.5
2011-01-09  328   96   8   753.5
2011-01-10  369  108  10  1046.5

---

args 和kwargs 参数是added to rolling_apply in Pandas version 0.14.0 .

因为在我上面的例子中 df 是一个全局变量，所以它并不是真正必要的将其作为参数传递给 foo。您可以简单地从 def 中删除 dffoo 行，并在调用 rolling_apply 时省略 args=(df,)。

但是，有时 df 可能未定义在 foo 可访问的范围内。在这种情况下，有一个简单的解决方法——关闭:

def foo(df):
    def inner_foo(x):
        window = df.iloc[x]
        # print(window)
        c = df.ix[int(x[-1]), 'c']
        dvals = window['a'] + window['b']*c
        return bar(dvals)
    return inner_foo

df['e'] = pd.rolling_apply(np.arange(len(df)), 6, foo(df))