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Signals.py
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import pandas as pd
pd.set_option('expand_frame_repr', False) # 当列太多时不换行
pd.set_option('display.max_rows', 1000)
import numpy as np
# ===移动平均线策略
# 简单移动均线策略,仅作为案例不具交易价值
def signal_moving_average(df, para=[50, 500, 30, 300]):
"""
简单的移动平均线策略
当短期均线由下向上穿过长期均线的时候,买入;然后由上向下穿过的时候,卖出。
:param df: 原始数据
:param para: 参数,[ma_short, ma_long]
:return:
"""
#signal:1 买入
#signal: -1 卖出
#signal: 平仓 0
# ===计算指标
ma_long_fast = para[0]
ma_long_slow = para[1]
ma_short_fast = para[1]
ma_short_slow = para[2]
# 计算均线
df['ma_long_fast'] = df['close'].rolling(ma_long_fast, min_periods=1).mean()
df['ma_long_slow'] = df['close'].rolling(ma_long_slow, min_periods=1).mean()
df['ma_short_fast'] = df['close'].rolling(ma_short_fast, min_periods=1).mean()
df['ma_short_slow'] = df['close'].rolling(ma_short_slow, min_periods=1).mean()
# ===找出买入信号
condition1 = df['ma_long_fast'] > df['ma_long_slow'] # 短期均线 > 长期均线
condition2 = df['ma_long_fast'].shift(1) <= df['ma_long_slow'].shift(1) # 之前的短期均线 <= 长期均线
df.loc[condition1 & condition2, 'signal'] = 1 # 将产生做多信号的那根K线的signal设置为1,1代表做多
# ===找出买入平仓信号
condition1 = df['ma_long_fast'] < df['ma_long_slow'] # 短期均线 < 长期均线
condition2 = df['ma_long_fast'].shift(1) >= df['ma_long_slow'].shift(1) # 之前的短期均线 >= 长期均线
df.loc[condition1 & condition2, 'signal'] = 0 # 将产生平仓信号当天的signal设置为0,0代表平仓
# ===找出卖出信号
condition1 = df['ma_short_slow'] > df['ma_short_fast'] # 短期均线 < 长期均线
condition2 = df['ma_short_slow'].shift(1) <= df['ma_short_fast'].shift(1) # 之前的短期均线 >= 长期均线
df.loc[condition1 & condition2, 'signal'] = -1 # 将产生做空信号的那根K线的signal设置为-1,1代表做多
# ===找出卖出平仓信号
condition1 = df['ma_short_slow'] < df['ma_short_fast'] # 短期均线 < 长期均线
condition2 = df['ma_short_slow'].shift(1) >= df['ma_short_fast'].shift(1) # 之前的短期均线 >= 长期均线
df.loc[condition1 & condition2, 'signal'] = 0 # 将产生平仓信号当天的signal设置为0,0代表平仓
df.drop(['ma_long_fast', 'ma_long_fast', 'ma_short_slow', 'ma_short_fast'], axis=1, inplace=True)
# ===由signal计算出实际的每天持有仓位
# signal的计算运用了收盘价,是每根K线收盘之后产生的信号,到第二根开盘的时候才买入,仓位才会改变。
df['pos'] = df['signal'].shift()
df['pos'].fillna(method='ffill', inplace=True)
df['pos'].fillna(value=0, inplace=True) # 将初始行数的position补全为0
return df