我需要根据不同列的值减去行的值。我的数据框看起来像这样：
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col1的值计算col2和col3组合的平均时差。规则如下：
只要
Yescol3中有row-next rowfor 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
到目前为止，我所做的简化版本是循环遍历数据框中的所有值并执行此操作：
dict变量Id只是一个字典，用于保存每个不同{{1}}的结果。
由于我的行数超过6MM，因此循环需要花费大量时间才能完成，因此我想知道是否有人能够提出更高效，更优雅的解决方案。
谢谢！
答案 0 :(得分：1)
我认为你可以使用：
#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')}
与...相同：
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')}
但是如果需要绝对值来消除负时间，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')}