Question

我有一个数据框，其中有一个usage_duration列（这是datetime格式中另外两列的差异）。它看起来如下：

processid, userid, usage_duration 
17613,root,0 days 23:41:03.000000000
17641,root,2 days 04:05:26.000000000
13848,acs,0 days 00:00:50.000000000
3912,acs,0 days 06:07:38.000000000
6156,acs,0 days 17:22:43.000000000

现在我想把它转换成几分钟。它应如下所示：

processid, userid, usage_duration_min
17613,root,1421
17641,root,3125
13848,acs,0
3912,acs,367
6156,acs,1042

有人能告诉我怎么可能吗？

非常感谢您的支持

Answer 1

使用total_seconds或seconds除以60，最后转换为integer s：

#if necessary converting to timedelta
#df['usage_duration'] = pd.to_timedelta(df['usage_duration'])

df['new'] = df['usage_duration'].dt.total_seconds().div(60).astype(int)

或者：

 df['new'] = (df['usage_duration'].dt.seconds.div(60).astype(int) 
             + df['usage_duration'].dt.days.multiply(1440).astype(int) )

print (df)
   processid userid  usage_duration   new
0      17613   root 0 days 23:41:03  1421
1      17641   root 2 days 04:05:26  3125
2      13848    acs 0 days 00:00:50     0
3       3912    acs 0 days 06:07:38   367
4       6156    acs 0 days 17:22:43  1042

Answer 2

这是一种方式：

s = pd.Series(['0 days 23:41:03.000000000', '2 days 04:05:26.000000000',
               '0 days 00:00:50.000000000', '0 days 06:07:38.000000000',
               '0 days 17:22:43.000000000'])

s = pd.to_timedelta(s).astype('timedelta64[m]').astype(int)

print(s)

0    1421
1    3125
2       0
3     367
4    1042
dtype: int32

如何将日期时间格式转换为分钟 - 熊猫

2 个答案: