Question

我有一个示例数据框（df），如下所示：

              Date_Time      Open      High       Low     Close   UOD  VWB
20  2020-07-01 10:30:00  10298.85  10299.90  10287.85  10299.90    UP    3
21  2020-07-01 10:35:00  10301.40  10310.00  10299.15  10305.75    UP    3
22  2020-07-01 10:40:00  10305.75  10305.75  10285.50  10290.00  DOWN    3
24  2020-07-01 10:45:00  10290.00  10291.20  10277.65  10282.65  DOWN    0
25  2020-07-01 10:50:00  10282.30  10289.80  10278.00  10282.00  DOWN    3
26  2020-07-01 10:55:00  10280.10  10295.00  10279.80  10291.50    UP    3
27  2020-07-01 11:00:00  10290.00  10299.95  10287.30  10297.55    UP    3
28  2020-07-01 11:05:00  10296.70  10306.30  10294.50  10299.40    UP    3
29  2020-07-01 11:10:00  10299.95  10301.10  10291.50  10292.00  DOWN    0
30  2020-07-01 11:15:00  10293.05  10298.70  10286.00  10291.55  DOWN    3
31  2020-07-01 11:20:00  10292.00  10298.70  10286.00  10351.45  DOWN    1

我具有以下条件：

检查df ['VWB'] == 0＆df ['UOD'] ==“ DOWN”并获取相应的 Open 值（在我的示例中为10290.00）

然后在该行之后找到第一个出现的 Close 值大于该 Open 值（10290.00）的情况。

在条件1（df ['VWB'] == 0＆df ['UOD'] ==“ DOWN”）和条件2（首次出现）之间找到两个错误之间的时间差，在另一列（TD）中。

我希望通过有效列按如下所示进行期望的出场

              Date_Time      Open      High       Low     Close   UOD  VWB  Valid    TD
20  2020-07-01 10:30:00  10298.85  10299.90  10287.85  10299.90    UP    3      0
21  2020-07-01 10:35:00  10301.40  10310.00  10299.15  10305.75    UP    3      0
22  2020-07-01 10:40:00  10305.75  10305.75  10285.50  10290.00  DOWN    3      0
23  2020-07-01 10:45:00  10290.00  10291.20  10277.65  10282.65  DOWN    0      0
25  2020-07-01 10:50:00  10282.30  10289.80  10278.00  10282.00  DOWN    3      0
26  2020-07-01 10:55:00  10280.10  10295.00  10279.80  10291.50    UP    3      1    600 <<= first occurrence
27  2020-07-01 11:00:00  10290.00  10299.95  10287.30  10297.55    UP    3      0
28  2020-07-01 11:05:00  10296.70  10306.30  10294.50  10299.40    UP    3      0
29  2020-07-01 11:10:00  10299.95  10301.10  10291.50  10292.00  DOWN    0      0
30  2020-07-01 11:15:00  10293.05  10298.70  10286.00  10291.55  DOWN    3      0
31  2020-07-01 11:20:00  10292.00  10298.70  10286.00  10351.45  DOWN    1      1    600 <<= first occurrence

Answer 1

这是一种方法，不确定这是否是最好的方法，并且可能可以进行优化（内嵌评论）

#gets open value per the condition
open_val = df.loc[(df['VWB'] == 0) & (df['UOD'] == "DOWN"),'Open']

#check where open value > df['Close'] and create groups
c = df['Close'].gt(open_val.reindex(df.index,method='ffill'))
a = np.digitize(df.index,open_val.index)

#get first index in each group and set the Valid column
valid_idx = c.groupby(a).idxmax()
df['Valid'] = c.loc[valid_idx].reindex(df.index,fill_value=False).astype(int)

#calculate time difference and mask where consition matches
TD = (df['Date_Time'] - 
df.loc[open_val.index,'Date_Time'].reindex(df.index,method='ffill')).dt.total_seconds()
df['TD'] = TD.where(df['Valid'].eq(1))

print(df[['Date_Time','Open','Close','UOD','VWB','Valid','TD']])

             Date_Time      Open     Close   UOD  VWB  Valid     TD
20 2020-07-01 10:30:00  10298.85  10299.90    UP    3      0    NaN
21 2020-07-01 10:35:00  10301.40  10305.75    UP    3      0    NaN
22 2020-07-01 10:40:00  10305.75  10290.00  DOWN    3      0    NaN
24 2020-07-01 10:45:00  10290.00  10282.65  DOWN    0      0    NaN
25 2020-07-01 10:50:00  10282.30  10282.00  DOWN    3      0    NaN
26 2020-07-01 10:55:00  10280.10  10291.50    UP    3      1  600.0
27 2020-07-01 11:00:00  10290.00  10297.55    UP    3      0    NaN
28 2020-07-01 11:05:00  10296.70  10299.40    UP    3      0    NaN
29 2020-07-01 11:10:00  10299.95  10292.00  DOWN    0      0    NaN
30 2020-07-01 11:15:00  10293.05  10291.55  DOWN    3      0    NaN
31 2020-07-01 11:20:00  10292.00  10351.45  DOWN    1      1  600.0

Answer 2

样本Df

            Open    Close       UOD   VWB
     0  10298.85    10287.85    UP      3
     1  10301.40    10299.15    UP      3
     2  10305.75    10290.00    DOWN    3
     3  10290.00    10282.65    DOWN    0
     4  10282.30    10282.00    DOWN    3
     5  10280.10    10291.50    UP      3

代码

a = df.index[(df["UOD"] == "DOWN") & (df["VWB"] == 0)].tolist() ##It will return all index which satisfy the above 1st condition

print(a)

输出

[3]

代码

  loc = []
  for i in a:
     for j in range(i+1, len(df)):
        if df.iloc[i]["Open"] < df.iloc[j]["Close"]:
            loc.append(j)    
            break

 valid = [1 if i in loc else 0 for i in range(0, len(df))]
 df["Valid"] = valid

最终输出：

          Open       Close      UOD    VWB   Valid
   0    10298.85    10287.85    UP      3   0
   1    10301.40    10299.15    UP      3   0
   2    10305.75    10290.00    DOWN    3   0
   3    10290.00    10282.65    DOWN    0   0
   4    10282.30    10282.00    DOWN    3   0
   5    10280.10    10291.50    UP      3   1

计算有条件的两行之间的时间差

2 个答案: