关于 .apply() 和通过函数传递整个列的问题

Question

我有一个数据框，代码为 headers，行中有过去一年的每日调整收盘价，我想计算年化波动率，但我不确定如何通过这些列.当我运行此代码时出现异常：TypeError: 'Series' object is not callable

import pandas as pd
import datetime as datetime
from datetime import timedelta
import yfinance as yf

df = pd.read_excel('C:/Users/Jacob/Downloads/Benchmark Tickers.xlsx', sheet_name='Sheet1')

tickers_list = df['Ticker'].tolist()

data = pd.DataFrame(columns=tickers_list)
    
for ticker in tickers_list:
    data[ticker] = yf.download(ticker, start=datetime.datetime.now()-datetime.timedelta(days=365), end=datetime.date.today()) ["Adj Close"]

def volatility(ticker):
    return data[ticker].pct_change().rolling(252).std()*(252**0.5)

data[ticker].apply(volatility(ticker))
    
export_excel = data.to_excel(r'C:/Users/User/Downloads/testvol.xlsx', sheet_name='Sheet1', index= True)

我如何将此波动率函数应用于每一列？

这里是一个 link 的数据，当您执行 yfinance 拉取时被拉取： https://docs.google.com/spreadsheets/d/11-kS1ah1lP8v6xv2JQZt_0i7YmynQxIw6q0stEvS_nM/edit?usp=sharing

Answer 1

当您在系列上调用 apply 时，您不会像通常调用函数时那样指定参数，而是将其作为 key-word 参数传递：

data[ticker] = data[ticker].apply(volatility, ticker=ticker)

那应该能解决这个问题。

pandas 文档也有一些很好的例子。参见 here。

Answer 2

在数据框上使用 .apply 时，默认情况下 axis=0 会为每一列执行计算 column-wise，因此无需指定每个代码名称。

import pandas as pd
import yfinance as yf
from datetime import datetime, timedelta, date

# given a list of tickers
tickers = ['EFT', 'PPR', 'SRLN']

# create empty dataframe with columns
data = pd.DataFrame(columns=tickers)

# get data
for ticker in tickers:
    data[ticker] = yf.download(ticker, start=datetime.now()-timedelta(days=365), end=date.today()) ["Adj Close"]

# perform calculation for each ticker and add it to the calcs dataframe
# 252 is to many for rolling; that means it needs 252 rows to perform the calculation. 
# to will be used to show that it's working
calcs = data.apply(lambda x: x.pct_change().rolling(10).std()*(10**0.5))

# display(calcs.head(20))
                 EFT       PPR      SRLN
Date                                    
2019-08-20       NaN       NaN       NaN
2019-08-21       NaN       NaN       NaN
2019-08-22       NaN       NaN       NaN
2019-08-23       NaN       NaN       NaN
2019-08-26       NaN       NaN       NaN
2019-08-27       NaN       NaN       NaN
2019-08-28       NaN       NaN       NaN
2019-08-29       NaN       NaN       NaN
2019-08-30       NaN       NaN       NaN
2019-09-03       NaN       NaN       NaN
2019-09-04  0.009594  0.012125  0.004690
2019-09-05  0.009483  0.012122  0.004691
2019-09-06  0.009870  0.009697  0.004736
2019-09-09  0.009037  0.010020  0.004191
2019-09-10  0.009205  0.009544  0.003981
2019-09-11  0.006672  0.009543  0.004084
2019-09-12  0.006492  0.010054  0.003925
2019-09-13  0.005592  0.010049  0.003992
2019-09-16  0.005428  0.012274  0.003367
2019-09-17  0.004926  0.010776  0.002505

关于 .apply() 和通过函数传递整个列的问题

Question about .apply() and passing entire columns through functions

python

apply

pandas

yfinance