Python中基于列值减去行的最佳方法

我需要根据不同列的值减去行的值。我的数据框如下所示：

Id |    col1    |  col2    | col3 |
1  | 2016-01-02 |  7:00:00 |  Yes |
1  | 2016-01-02 |  7:05:00 |  No  |
1  | 2016-01-02 |  7:10:00 |  Yes |
1  | 2016-01-02 |  8:00:00 |  No  |
2  | 2016-01-02 |  7:10:00 |  Yes | 
2  | 2016-01-02 |  7:50:00 |  No  | 
2  | 2016-01-02 |  9:00:00 |  No  | 
2  | 2016-01-02 |  9:10:00 |  No  | 
2  | 2016-01-02 |  9:15:00 |  No  | 
3  | 2016-01-02 |  6:05:00 |  Yes | 
3  | 2016-01-02 |  6:10:00 |  Yes | 
3  | 2016-01-02 |  6:20:00 |  Yes | 
3  | 2016-01-02 |  6:45:00 |  No  | 

我需要根据

col3

的值计算

col1

和

col2

组合中的平均时间差。规则如下：

每当

col3

do

下一行中有
Yes
时

到目前为止，我所做的一个简化版本是循环遍历数据帧中的所有值，并执行以下操作：

for i in range(len(df)):
    if df['col3'][i] == 'Yes':
        date1 = datetime.combine(df['col1'][i], df['col2'][i])
        date2 = datetime.combine(df['col1'][i+1], df['col2'][i+1])
        dict[df['Id'][i]] = date1-date2

变量
dict
只是一个字典，它保存每个不同
Id
的结果

因为我有超过6毫米的行，循环需要很多时间才能完成，所以我想知道是否有人能想出一个更高效、更优雅的解决方案

谢谢
 我认为您可以使用：

#datetime column - add time to dates
df['datetime'] = pd.to_datetime(df['col1']) + pd.to_timedelta(df['col2'])
#get difference of all values, filter by multiple mask only if `Yes`
#convert to ns for numeric for aggregate
df['dif']=df['datetime'].diff(-1).mul(df['col3'] == 'Yes').fillna(0).values.astype(np.int64)
print (df)
    Id        col1     col2 col3            datetime            dif
0    1  2016-01-02  7:00:00  Yes 2016-01-02 07:00:00  -300000000000
1    1  2016-01-02  7:05:00   No 2016-01-02 07:05:00              0
2    1  2016-01-02  7:10:00  Yes 2016-01-02 07:10:00 -3000000000000
3    1  2016-01-02  8:00:00   No 2016-01-02 08:00:00              0
4    2  2016-01-02  7:10:00  Yes 2016-01-02 07:10:00 -2400000000000
5    2  2016-01-02  7:50:00   No 2016-01-02 07:50:00              0
6    2  2016-01-02  9:00:00   No 2016-01-02 09:00:00              0
7    2  2016-01-02  9:10:00   No 2016-01-02 09:10:00              0
8    2  2016-01-02  9:15:00   No 2016-01-02 09:15:00              0
9    3  2016-01-02  6:05:00  Yes 2016-01-02 06:05:00  -300000000000
10   3  2016-01-02  6:10:00  Yes 2016-01-02 06:10:00  -600000000000
11   3  2016-01-02  6:20:00  Yes 2016-01-02 06:20:00 -1500000000000
12   3  2016-01-02  6:45:00   No 2016-01-02 06:45:00              0

d = pd.to_timedelta(df.groupby('Id')['dif'].mean()).to_dict()
print (d)
{1: Timedelta('-1 days +23:46:15'), 
 2: Timedelta('-1 days +23:52:00'), 
 3: Timedelta('-1 days +23:50:00')}

什么是相同的：

datetime = pd.to_datetime(df['col1']) + pd.to_timedelta(df['col2'])
diff = datetime.diff(-1).mul(df['col3'] == 'Yes').fillna(0).values.astype(np.int64)
d = pd.to_timedelta(pd.Series(diff, index=df.index).groupby(df['Id']).mean()).to_dict()
print (d)
{1: Timedelta('-1 days +23:46:15'), 
 2: Timedelta('-1 days +23:52:00'), 
 3: Timedelta('-1 days +23:50:00')}

但是如果需要删除负时间增量的绝对值，则添加
numpy.abs
：

df['datetime'] = pd.to_datetime(df['col1']) + pd.to_timedelta(df['col2'])
df['dif'] = np.abs(df['datetime'].diff(-1)
                                 .mul(df['col3'] == 'Yes')
                                 .fillna(0)
                                 .values
                                 .astype(np.int64))
print (df)
    Id        col1     col2 col3            datetime            dif
0    1  2016-01-02  7:00:00  Yes 2016-01-02 07:00:00   300000000000
1    1  2016-01-02  7:05:00   No 2016-01-02 07:05:00              0
2    1  2016-01-02  7:10:00  Yes 2016-01-02 07:10:00  3000000000000
3    1  2016-01-02  8:00:00   No 2016-01-02 08:00:00              0
4    2  2016-01-02  7:10:00  Yes 2016-01-02 07:10:00  2400000000000
5    2  2016-01-02  7:50:00   No 2016-01-02 07:50:00              0
6    2  2016-01-02  9:00:00   No 2016-01-02 09:00:00              0
7    2  2016-01-02  9:10:00   No 2016-01-02 09:10:00              0
8    2  2016-01-02  9:15:00   No 2016-01-02 09:15:00              0
9    3  2016-01-02  6:05:00  Yes 2016-01-02 06:05:00   300000000000
10   3  2016-01-02  6:10:00  Yes 2016-01-02 06:10:00   600000000000
11   3  2016-01-02  6:20:00  Yes 2016-01-02 06:20:00  1500000000000
12   3  2016-01-02  6:45:00   No 2016-01-02 06:45:00              0

d = pd.to_timedelta(df.groupby('Id')['dif'].mean()).to_dict()
print (d)
{1: Timedelta('0 days 00:13:45'), 
 2: Timedelta('0 days 00:08:00'), 
 3: Timedelta('0 days 00:10:00')}

是的，更干净、更快。你能解释一下mul的功能吗？我正在看文档，但似乎不明白它在做什么。您可以检查，如果
Yes
它只需乘以1，如果不是
Yes
，它只需乘以0即可。