1

我正在尝试根据给定的购买日期测试不同的卖出简单移动平均线标准。

我有一个如下的购买数据数据库（df_buy）。我想填写NaN值。（FCU = 首次收盘价）

  Symbol                Time  buy_price     LOD  date_FCU_10dMA  price_FCU_10dMA
0    AMD 2019-12-12 09:36:00      39.52   27.43             NaN              NaN
1    AMD 2020-01-16 09:33:00      49.21   27.43             NaN              NaN
2   BITF 2021-08-03 09:47:00       4.26    2.81             NaN              NaN
3   DOCN 2021-06-14 09:32:00      41.76   35.35             NaN              NaN
4   NVDA 2020-07-29 09:38:00     416.81  169.32             NaN              NaN
5   NVDA 2020-09-25 10:34:00     499.72  169.32             NaN              NaN
6   UPST 2021-02-09 09:32:00      76.83   22.61             NaN              NaN
7   UPST 2021-03-18 09:32:00      88.56   22.61             NaN              NaN

我有另一个包含每日库存数据的数据库，如下所示（df_day）

        Symbol                Time  Close   LOD   10MA    20MA
2722244   AEHR 2019-11-25 16:00:00   1.90  1.29  2.005  1.8870
2722289   AEHR 2019-11-26 16:00:00   1.92  1.29  2.032  1.8925
2722383   AEHR 2019-12-02 16:00:00   1.88  1.29  2.056  1.8985
2722435   AEHR 2019-12-03 16:00:00   1.88  1.29  2.046  1.8995
2722471   AEHR 2019-12-04 16:00:00   1.89  1.29  2.020  1.9055
2722569   AEHR 2019-12-06 16:00:00   1.93  1.29  1.993  1.9140

根据该策略，首次收盘必须在买入日期后至少 2 天。

df_filt2['price_FCU_10dMA'] = (df_buy['buy_price'] > df_day['20MA'])
Error: ValueError: Can only compare identically-labeled Series objects


for i, row in df_buy.iterrows():
    # drop unlisted symbols
    filt1 = (df_day['Symbol'] != df_buy['Symbol'].loc[i])
    df_filt1 = df_day.drop(index=df_day[filt1].index)

    # drop trades from before buy date + 2
    filt2 = (pd.to_datetime(df_day['Time']) < (pd.to_datetime(df_buy['Time'].loc[i] + pd.to_timedelta(2, unit='d'))))
    df_filt2 = df_filt1.drop(index=df_filt1[filt2].index)
    # sort values
    df_filt2.sort_values(by=['Symbol', 'Time'], inplace=True)

    # drop rows where Close is above the 10MA
    df_filt2['price_FCU_10dMA'] = (df_buy['buy_price'] > df_day['20MA'])
    filt3 = (df_filt2['price_FCU_10dMA'] == False)

    # trail_sell = df_filt2[filt3].loc[10]
    df_buy['price_FCU_10dMA'].loc[i] = df_filt2.loc[filt3, 'Close'].iloc[0]  # returns the single value of first close under 10ma
    df_buy['date_FCU_10dMA'].loc[i] = df_filt2.loc[filt3, 'Time'].iloc[0]  # returns the single value of first close under 10ma
4

1 回答 1

0

您可以在一个逻辑语句中组合所有 3 个条件：

two_days = pd.to_timedelta('2d')
no_match = pd.Series({'Time': np.nan, 'Close': np.nan})

for i, row in df_buy.iterrows():
    cond = (
        (df_day['Symbol'] == row['Symbol']) &
        (df_day['Time'] > row['Time'] + two_days) &
        (df_day['Close'] < df_day['10MA'])
    )
    match = no_match if not cond.any() else df_day[cond].head(1)
    df_buy.loc[i, ['price_FCU_10dMA', 'date_FCU_10dMA']] = match[['Close', 'Time']]
于 2021-09-11T22:30:53.610 回答