如何在Python中将整数（秒）添加到hh:mm:ss格式？

我在python中有以下数据帧，我试图通过在“开始时间”中添加“持续时间”（以秒为单位）来计算列“新时间”

Serial  start_date     start_time     Duration(seconds)  New time
    A   5/22/2017       10:37:24        216 
    A   5/22/2017       10:37:26        213 
    A   5/22/2017       10:37:29         3  
    A   5/22/2017       10:39:55         60 
    A   5/22/2017       10:51:50        380 
    A   5/22/2017       10:51:57        339 

我想在开始时间中添加持续时间。持续时间以秒为单位。 “新时间”应为hh:mm:ss格式

我试图在论坛上寻找类似的问题，但无法回避

以下是信息

data.info()

start_date         13661 non-null object
start_time         13661 non-null object
Duration           13661 non-null int64

我试着从论坛上的一个类似问题中得到提示，使用datetime

data.newtime = data.start_time + datetime.timedelta(data.Duration)

当我执行此命令时，我得到以下错误：TypeError:timedelta-days组件的不支持类型：Series

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-95-fdfac1490ba5> in <module>()
----> 1 data.newtime = data.start_time + datetime.timedelta(data.Duration)

TypeError: unsupported type for timedelta days component: Series

---------------------------------------------------------------------------
TypeError回溯（最近一次调用上次）
在（）
---->1 data.newtime=data.start\u time+datetime.timedelta（data.Duration）
TypeError:timedelta days组件的类型不受支持：Series

我不知道该怎么办。python新手

感谢您的帮助

---

TIA

以下是一个片段，可以帮助您自己解决问题：

from datetime import datetime, timedelta

my_date = datetime.strptime('5/22/2017 10:37:24', '%m/%d/%Y %H:%M:%S')
my_time_diff = timedelta(seconds=216)
my_new_date = my_date + my_time_diff
print(my_new_date.strftime('%m/%d/%Y %H:%M:%S'))

有用资源：

您可以使用，并且输出也是

timedelta

：

df['New time'] = pd.to_timedelta(df['start_time']) + 
                 pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
  Serial start_date start_time  Duration(seconds) New time
0      A  5/22/2017   10:37:24                216 10:41:00
1      A  5/22/2017   10:37:26                213 10:40:59
2      A  5/22/2017   10:37:29                  3 10:37:32
3      A  5/22/2017   10:39:55                 60 10:40:55
4      A  5/22/2017   10:51:50                380 10:58:10
5      A  5/22/2017   10:51:57                339 10:57:36

但如果秒数更多，输出就会改变，因为还有几天：

print (df)
  Serial start_date start_time  Duration(seconds)
0      A  5/22/2017   10:37:24                216
1      A  5/22/2017   10:37:26             213000
2      A  5/22/2017   10:37:29                  3
3      A  5/22/2017   10:39:55                 60
4      A  5/22/2017   10:51:50                380
5      A  5/22/2017   10:51:57                339

df['New time'] = pd.to_timedelta(df['start_time']) + 
                 pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
  Serial start_date start_time  Duration(seconds)        New time
0      A  5/22/2017   10:37:24                216 0 days 10:41:00
1      A  5/22/2017   10:37:26             213000 2 days 21:47:26
2      A  5/22/2017   10:37:29                  3 0 days 10:37:32
3      A  5/22/2017   10:39:55                 60 0 days 10:40:55
4      A  5/22/2017   10:51:50                380 0 days 10:58:10
5      A  5/22/2017   10:51:57                339 0 days 10:57:36

---

也可以添加日期时间：

df['New date'] = pd.to_datetime(df['start_date']) + \
                 pd.to_timedelta(df['start_time']) +  \
                 pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
  Serial start_date start_time  Duration(seconds)            New date
0      A  5/22/2017   10:37:24                216 2017-05-22 10:41:00
1      A  5/22/2017   10:37:26                213 2017-05-22 10:40:59
2      A  5/22/2017   10:37:29                  3 2017-05-22 10:37:32
3      A  5/22/2017   10:39:55                 60 2017-05-22 10:40:55
4      A  5/22/2017   10:51:50                380 2017-05-22 10:58:10
5      A  5/22/2017   10:51:57                339 2017-05-22 10:57:36

---

--- 如果需要将

timedelta

转换为

string

，格式为

HH:MM:SS

和lost

days

（如果存在）：

---

@马克西米利安：我试过引用这个。在这里，我想为数据帧中的每个行时间条目计算这个值。我已经编辑了这个错误，我得到了。谢谢！现在问题很清楚了。请注意，如果您对日志数据执行此操作，则闰秒（Python忽略）的存在可能会影响您的计算。通常只在物理实验和金融交易中重要的时间戳。很高兴能帮上忙！天气真好！我对各种选择投了赞成票。又是一个漂亮的！

df['New date'] = pd.to_datetime(df['start_date']) + \
                 pd.to_timedelta(df['start_time']) +  \
                 pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
  Serial start_date start_time  Duration(seconds)            New date
0      A  5/22/2017   10:37:24                216 2017-05-22 10:41:00
1      A  5/22/2017   10:37:26             213000 2017-05-24 21:47:26
2      A  5/22/2017   10:37:29                  3 2017-05-22 10:37:32
3      A  5/22/2017   10:39:55                 60 2017-05-22 10:40:55
4      A  5/22/2017   10:51:50                380 2017-05-22 10:58:10
5      A  5/22/2017   10:51:57                339 2017-05-22 10:57:36

df['New time'] = pd.to_timedelta(df['start_time']) + 
                 pd.to_timedelta(df['Duration(seconds)'], unit='s')
df['New time'] = pd.to_datetime(df['New time']).dt.strftime('%H:%M:%S')
print (df)
  Serial start_date start_time  Duration(seconds)  New time
0      A  5/22/2017   10:37:24                216  10:41:00
1      A  5/22/2017   10:37:26             213000  21:47:26
2      A  5/22/2017   10:37:29                  3  10:37:32
3      A  5/22/2017   10:39:55                 60  10:40:55
4      A  5/22/2017   10:51:50                380  10:58:10
5      A  5/22/2017   10:51:57                339  10:57:36