基於Python中的列值減去行的最佳方式

Question

我需要根據不同列的值減去行的值。我的數據框看起來是這樣的：

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 | 

我需要計算的col1和col2基礎上col3值組合的平均時間差。規則如下：

每當有在col3一個Yes做row-next row

的是我迄今所做的簡化版本是遍歷所有的數據幀和值做到這一點：

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 

可變dict僅僅是保存每個不同Id結果的字典。

因爲我有超過6MM的行，循環需要花費很多時間才能完成，所以我想知道是否有人能夠提出更高效和優雅的解決方案。

謝謝！

Answer 1

我認爲你可以使用：

#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')} 

什麼是一樣的：

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')} 

但如果需要的絕對值爲刪除負面timedelta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')}