在数据框上滚动函数
Rolling a function on a data frame
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
,我在固定的 window(此处为 6)上应用滚动函数,我以某种方式为每一行(或日期) ), 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
在此之后 "loop" 我想在 d
和 运行 中获取所有这 6 个计算行的函数调用,这反过来将 return 一个值,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
然后这个过程将被迭代到 next window(再次 6 长),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
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
参数为 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,所以没有必要
将其作为参数传递给 foo
。您可以简单地从 def
foo
行中删除 df
并在对 rolling_apply
.
的调用中省略 args=(df,)
但是,有时 df
可能未在 foo
可访问的范围内定义。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))
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
,我在固定的 window(此处为 6)上应用滚动函数,我以某种方式为每一行(或日期) ), 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
在此之后 "loop" 我想在 d
和 运行 中获取所有这 6 个计算行的函数调用,这反过来将 return 一个值,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
然后这个过程将被迭代到 next window(再次 6 长),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
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
参数为 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,所以没有必要
将其作为参数传递给 foo
。您可以简单地从 def
foo
行中删除 df
并在对 rolling_apply
.
args=(df,)
但是,有时 df
可能未在 foo
可访问的范围内定义。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))