我试图将计算矩阵的结果附加到我的df。我有一个问题是如何设计我的迭代计算的大局。我有以下代码,应该举例说明我想要做的事情。
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)
我有几层增加的数据复杂性:
处理这些计算信息的最聪明方法是什么,最后总结为全包的df_all?
答案 0 :(得分:0)
确保
df = df_all
复制内容而不仅仅是参考。它可能会在以后搞砸你的计算机。
一般有两种方法: