家庭作业问题需要帮助解决: 策略:每当价格高于50天移动平均线时买入,然后在3个交易日后卖出。我们平均能赚多少利润(以百分比为单位)?在交易日 x 上,如果 (1) 交易日 x-1 的价格低于移动平均线,并且 (2) 交易日 x 的价格高于移动平均线,则我们说价格"高于"50 天移动平均线。
如何将条件 [1] 和 [2] 与 50 天移动平均线合并到当前代码中:
rol=stock.rolling(50).mean()
profitMade=((stock.shift(-3)-stock)/stock)
stock>rol
profitMade[stock>rol].mean()
样本数据集:股票和相应日期:
Date
2002-05-23 1.196429
2002-05-24 1.210000
2002-05-28 1.157143
2002-05-29 1.103571
2002-05-30 1.071429
2002-05-31 1.076429
2002-06-03 1.128571
2002-06-04 1.117857
2002-06-05 1.147143
2002-06-06 1.182143
2002-06-07 1.118571
2002-06-10 1.156429
2002-06-11 1.153571
2002-06-12 1.092857
2002-06-13 1.082857
2002-06-14 0.986429
2002-06-17 0.922143
2002-06-18 0.910714
2002-06-19 0.951429
好的,这就是我所拥有的。不优雅,但我认为它可以完成工作。
import datetime
import pandas as pd
import random
# Sample dataset
df = pd.DataFrame({'Date':sorted([datetime.date.today() - datetime.timedelta(days=i) for i in range(80)]), 'Price':[i for i in range(80)]})
df['Price'] = df['Price'].apply(lambda x: random.uniform(80,100))
>>>df
Date Price
0 2018-11-11 83.894664
1 2018-11-12 86.472656
2 2018-11-13 92.566676
3 2018-11-14 94.145586
.. ... ...
77 2019-01-27 86.338076
78 2019-01-28 95.374173
79 2019-01-29 98.829077
[80 rows x 2 columns]
# MA calculation - borrowing from @jezrael's solution at https://stackoverflow.com/questions/54176642/how-do-i-perform-a-moving-average-in-panda-with-a-column-that-needs-to-be-unique/54195910#54195910
seq = df.set_index('Date')['Price'].rolling(50).mean().rename('MA')
df = df.join(seq, on='Date')
# Set buy signals if current price is higher than 50-day MA
df['Buy'] = df['Price'] - df['MA'] > 0
# Lookbehind check to see that it was below MA at T-1
df['BelowMA'] = df['Price'] - df['MA'] <= 0
# Profit calculations
df['Profit'] = (df['Price'] - df['Price'].shift(-3))/df['Price']
最后,满足给定约束的行
>>> df.loc[(df['BelowMA'].shift(1)==True) & (df['Buy'] == True)]
Date Price MA Buy Profit BelowMA
54 2019-01-04 91.356371 89.634529 True 0.000697 False
56 2019-01-06 94.116616 89.623909 True 0.114375 False
60 2019-01-10 89.383369 89.254519 True -0.082870 False
63 2019-01-13 96.790575 88.977660 True 0.029314 False
69 2019-01-19 92.193939 89.052057 True 0.071259 False
74 2019-01-24 91.392465 89.302293 True 0.020673 False
76 2019-01-26 95.369195 89.292715 True 0.158250 False