我试图将计算矩阵的结果附加到我的df中。我对如何为计算设计迭代有一个问题。我有下面的代码,应该举例说明我正在尝试做什么。你知道吗
import pandas as pd
from pandas import DataFrame
import numpy as np
np_all = np.array([[1, 'vws.co', 1],
[1, 'nflx', 3],
[1, 'aapl', 2],
[2, 'vws.co', 1],
[2, 'nflx', 2],
[2, 'aapl', 1],
[3, 'vws.co', 1],
[3, 'nflx', 3],
[3, 'aapl', 1]])
df_all = pd.DataFrame(data=np_all, columns=['Date', 'Ticker', 'Close'])
df_all = df_all.sort(['Ticker','Date'], ascending=[1,1])
df_kpi_list = []
stocklist = ['vws.co','nflx','aapl']
print (df_all)
def screener(df_all,ticker):
# Copy df_all to df for single ticker operations
df = df_all
# filter to only relevant ticker
df = df[df['Ticker'] == ticker]
df = df[df.Ticker == ticker.lower()]
def kpi1_calc(df,ticker):
# do some KPI calculation that are appended to new columns of df
pass
def kpi2_calc(df,ticker):
# do more KPI calculation that are appended to new columns of df
pass
def kpi3_calc(df,ticker):
# example of more KPI calculation that are appended to new columns of df
# Add content to df - RSI
rsi = 3 # stupid example of a constant that is stored in df column
r = rsi
# add a RSI column
r['RSI'] = rsi
df_kpi_list.append(r)
return df
return df
return df
# concatenate all the ticker-iteration dfs from df_kpi_list into one df_all
df_all = pd.concat(df_kpi_list)
return df_all
if __name__ == '__main__':
for ticker in stocklist:
df_data = screener(df_all, ticker)
print (df_data)
我有几层更复杂的数据:
什么是最聪明的方式来处理这些信息被计算,并最终总结成一个包罗万象的dfèu all?你知道吗
确保
df = df_all
复制内容,而不是仅仅做一个参考。它会把你的电脑搞砸的。你知道吗
一般来说有两种方法:
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