在数据框上滚动函数

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），fix价值c。此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，我想将所有这 6 个计算行放入d并运行函数调用，该函数调用又将返回one值，应该存储在另一列中，e say:

              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

然后这个过程将被迭代到next窗口（同样是 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， 氮

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)

yields

              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

The args and kwargs参数为添加到rolling_apply在 Pandas 版本 0.14.0 中.

因为在我上面的例子中df是一个全局变量，其实没有必要 将其传递给foo作为一个论点。你可以简单地删除df来自def foo行并省略args=(df,)在通话中rolling_apply.

然而，有时df可能未定义在可访问的范围内foo。在这种情况下，有一个简单的解决方法——创建一个闭包：

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))