Python 计算特定行的几何平均回报率

我有一个这样的数据帧

       Date       price     mid      std         top             btm
     ..............
    1999-07-21  8.6912  8.504580    0.084923    9.674425    8.334735
    1999-07-22  8.6978  8.508515    0.092034    8.692583    8.324447
    1999-07-23  8.8127  8.524605    0.118186    10.760976   8.288234
    1999-07-24  8.8779  8.688810    0.091124    8.871057    8.506563
     ..............

我想创建一个名为“diff”的新列。 如果在一行中，'price'>'top'，那么我想用这行价格的几何平均回报率和前一行n-5的价格来填充这行的'diff'。5天几何平均数

例如，在第1999-07-22行中，价格大于top，因此我想在这一行中用07-22和07-17的几何平均值填充“diff”，注意日期可能不是连续的，因为不包括节假日。只有一小部分行可以满足需求。因此，“diff”中的大多数值都将丢失

您能告诉我如何在python中执行此操作吗？

与for set NaNs一起使用：

编辑：

我相信你需要：

df['Date'] = pd.to_datetime(df['Date'])
df = df.set_index('Date')

from scipy.stats.mstats import gmean

df['gmean'] = (df['price'].rolling('5d')
                          .apply(gmean, raw=True)
                          .where(df['price'] > df['top']))
print (df)
             price       mid       std        top       btm     gmean
Date                                                                 
1999-07-21  8.6912  8.504580  0.084923   9.674425  8.334735       NaN
1999-07-22  8.6978  8.508515  0.092034   8.692583  8.324447  8.694499
1999-07-23  8.8127  8.524605  0.118186  10.760976  8.288234       NaN
1999-07-24  8.8779  8.688810  0.091124   8.871057  8.506563  8.769546

---

您可以通过取price和top列的差值，然后分配的值来实现这一点。下面是另一个解决方案：

import pandas as pd
from functools import reduce

__name__ = 'RunScript'

ddict = {
    'Date':['1999-07-21','1999-07-22','1999-07-23','1999-07-24',],
    'price':[8.6912,8.6978,8.8127,8.8779],
    'mid':[8.504580,8.508515,8.524605,8.688810],
    'std':[0.084923,0.092034,0.118186,0.091124],
    'top':[9.674425,8.692583,10.760976,8.871057],
    'btm':[8.334735,8.324447,8.288234,8.506563],
    }


data = pd.DataFrame(ddict)


def geo_mean(iter):
    """
        Geometric mean function. Pass iterable
    """
    return reduce(lambda a, b: a * b, iter) ** (1.0 / len(iter))


def set_geo_mean(df):
    # Shift the price row down one period
    data['shifted price'] = data['price'].shift(periods=1)

    # Create a masked expression that evaluates price vs top
    masked_expression = df['price'] > df['top']

    # Return rows from dataframe where masked expression is true
    masked_data = df[masked_expression]

    # Apply our function to the relevant rows
    df.loc[masked_expression, 'geo_mean'] = geo_mean([masked_data['price'], masked_data['shifted price']])

    # Drop the shifted price data column once complete
    df.drop('shifted price', axis=1, inplace=True)


if __name__ == 'RunScript':
    # Call function and pass dataframe argument.
    set_geo_mean(data)

---

对不起，我刚才把问题简单化了，但我觉得我应该更具体一些。请在编辑后查看我的问题。@JAKE-不确定是否理解您的问题，但如果需要每5天的几何平均数，请使用滚动gmean检查解决方案。抱歉，我刚才简化了问题，但我发现我应该更具体一些。请在编辑后查看我的问题。好的，那么是否可以添加10行预期输出的示例数据？我认为应该从示例数据中删除不必要的列。

import pandas as pd
import numpy as np

df = pd.DataFrame(...)

df['diff'] = df['price'] - df['top']

df.loc[df['diff'] <= 0, 'diff'] = np.NaN # or 0

import pandas as pd
from functools import reduce

__name__ = 'RunScript'

ddict = {
    'Date':['1999-07-21','1999-07-22','1999-07-23','1999-07-24',],
    'price':[8.6912,8.6978,8.8127,8.8779],
    'mid':[8.504580,8.508515,8.524605,8.688810],
    'std':[0.084923,0.092034,0.118186,0.091124],
    'top':[9.674425,8.692583,10.760976,8.871057],
    'btm':[8.334735,8.324447,8.288234,8.506563],
    }


data = pd.DataFrame(ddict)


def geo_mean(iter):
    """
        Geometric mean function. Pass iterable
    """
    return reduce(lambda a, b: a * b, iter) ** (1.0 / len(iter))


def set_geo_mean(df):
    # Shift the price row down one period
    data['shifted price'] = data['price'].shift(periods=1)

    # Create a masked expression that evaluates price vs top
    masked_expression = df['price'] > df['top']

    # Return rows from dataframe where masked expression is true
    masked_data = df[masked_expression]

    # Apply our function to the relevant rows
    df.loc[masked_expression, 'geo_mean'] = geo_mean([masked_data['price'], masked_data['shifted price']])

    # Drop the shifted price data column once complete
    df.drop('shifted price', axis=1, inplace=True)


if __name__ == 'RunScript':
    # Call function and pass dataframe argument.
    set_geo_mean(data)