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python - 在 Pandas 中使用滚动的滑动窗口迭代器

转载 作者:太空狗 更新时间:2023-10-30 02:27:59 28 4
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如果是单行,我可以得到迭代器如下

import pandas as pd
import numpy as np

a = np.zeros((100,40))
X = pd.DataFrame(a)

for index, row in X.iterrows():
print index
print row

现在我希望每个迭代器都返回一个子集 X[0:9, :], X[5:14, :], X[10 :19, :] 等。如何通过滚动 ( pandas.DataFrame.rolling ) 实现此目的?

最佳答案

我将试验以下数据框。

设置

import pandas as pd
import numpy as np
from string import uppercase

def generic_portfolio_df(start, end, freq, num_port, num_sec, seed=314):
np.random.seed(seed)
portfolios = pd.Index(['Portfolio {}'.format(i) for i in uppercase[:num_port]],
name='Portfolio')
securities = ['s{:02d}'.format(i) for i in range(num_sec)]
dates = pd.date_range(start, end, freq=freq)
return pd.DataFrame(np.random.rand(len(dates) * num_sec, num_port),
index=pd.MultiIndex.from_product([dates, securities],
names=['Date', 'Id']),
columns=portfolios
).groupby(level=0).apply(lambda x: x / x.sum())


df = generic_portfolio_df('2014-12-31', '2015-05-30', 'BM', 3, 5)

df.head(10)

enter image description here

我现在将介绍一个函数来滚动多行并将其连接到一个数据框中,我将在列索引中添加一个顶层以指示滚动中的位置。

解决方案第 1 步

def rolled(df, n):
k = range(df.columns.nlevels)
_k = [i - len(k) for i in k]
myroll = pd.concat([df.shift(i).stack(level=k) for i in range(n)],
axis=1, keys=range(n)).unstack(level=_k)
return [(i, row.unstack(0)) for i, row in myroll.iterrows()]

虽然它隐藏在函数中,myroll 看起来像这样

enter image description here

现在我们可以像使用迭代器一样使用它。

解决方案第 2 步

for i, roll in rolled(df.head(5), 3):
print roll
print

0 1 2
Portfolio
Portfolio A 0.326164 NaN NaN
Portfolio B 0.201597 NaN NaN
Portfolio C 0.085340 NaN NaN

0 1 2
Portfolio
Portfolio A 0.278614 0.326164 NaN
Portfolio B 0.314448 0.201597 NaN
Portfolio C 0.266392 0.085340 NaN

0 1 2
Portfolio
Portfolio A 0.258958 0.278614 0.326164
Portfolio B 0.089224 0.314448 0.201597
Portfolio C 0.293570 0.266392 0.085340

0 1 2
Portfolio
Portfolio A 0.092760 0.258958 0.278614
Portfolio B 0.262511 0.089224 0.314448
Portfolio C 0.084208 0.293570 0.266392

0 1 2
Portfolio
Portfolio A 0.043503 0.092760 0.258958
Portfolio B 0.132221 0.262511 0.089224
Portfolio C 0.270490 0.084208 0.293570

关于python - 在 Pandas 中使用滚动的滑动窗口迭代器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38509107/

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