Question

我有一个数据框，其中最后两列指的是时钟时间。两列的数据类型均为String。

样本数据如下：

PROCESS_N   VAL     DATE        TIME_1      TIME_2
35324399    74.95   02/11/18    12:45:26    12:30:36
35324399    74.95   02/11/18    12:45:26    12:35:36
35334154    74.95   02/11/18    12:45:27    12:36:22
35338698    74.95   02/11/18    12:45:28    12:38:30
35338698    74.95   02/11/18    12:45:28    12:38:32
35347905    74.95   02/11/18    12:45:30    12:39:36
35367939    74.95   02/11/18    12:45:30    12:39:39
35371892    74.95   02/11/18    12:45:31    12:39:55

我想在Dataframe的末尾填充另一列，其中TIME_1和TIME_2的差异为Second。

Answer 1

您可以使用pd.to_timedelta，然后访问pd.Series.dt.seconds：

time_cols = ['TIME_1', 'TIME_2']
df[time_cols] = df[time_cols].apply(pd.to_timedelta)

df['Diff'] = (df['TIME_1'] - df['TIME_2']).dt.seconds

print(df)

   PROCESS_N    VAL      DATE   TIME_1   TIME_2  Diff
0   35324399  74.95  02/11/18 12:45:26 12:30:36   890
1   35324399  74.95  02/11/18 12:45:26 12:35:36   590
2   35334154  74.95  02/11/18 12:45:27 12:36:22   545
3   35338698  74.95  02/11/18 12:45:28 12:38:30   418
4   35338698  74.95  02/11/18 12:45:28 12:38:32   416
5   35347905  74.95  02/11/18 12:45:30 12:39:36   354
6   35367939  74.95  02/11/18 12:45:30 12:39:39   351
7   35371892  74.95  02/11/18 12:45:31 12:39:55   336

Answer 2

您可以简单地使用to_datetime()方法进行转换，然后进行区别：

df['diff'] = pd.to_datetime(df['TIME_1']) - pd.to_datetime(df['TIME_2'])

如果您希望将其存储在几秒钟内，则可以这样转换：

df['diff'] = df['diff'].astype('timedelta64[s]')

输出：

  PROCESS_N    VAL      DATE    TIME_1    TIME_2   diff
0   35324399  74.95  02/11/18  12:45:26  12:30:36  890.0
1   35324399  74.95  02/11/18  12:45:26  12:35:36  590.0
2   35334154  74.95  02/11/18  12:45:27  12:36:22  545.0
3   35338698  74.95  02/11/18  12:45:28  12:38:30  418.0
4   35338698  74.95  02/11/18  12:45:28  12:38:32  416.0
5   35347905  74.95  02/11/18  12:45:30  12:39:36  354.0
6   35367939  74.95  02/11/18  12:45:30  12:39:39  351.0
7   35371892  74.95  02/11/18  12:45:31  12:39:55  336.0

Answer 3

使用timedelta-

的秒数差异

df['Diff'] = (df['TIME_1']-df['TIME_2']).astype('timedelta64[s]')

输出

   PROCESS_N    VAL      DATE              TIME_1              TIME_2   Diff
0   35324399  74.95  02/11/18 1900-01-01 12:45:26 1900-01-01 12:30:36  890.0
1   35324399  74.95  02/11/18 1900-01-01 12:45:26 1900-01-01 12:35:36  590.0
2   35334154  74.95  02/11/18 1900-01-01 12:45:27 1900-01-01 12:36:22  545.0
3   35338698  74.95  02/11/18 1900-01-01 12:45:28 1900-01-01 12:38:30  418.0
4   35338698  74.95  02/11/18 1900-01-01 12:45:28 1900-01-01 12:38:32  416.0
5   35347905  74.95  02/11/18 1900-01-01 12:45:30 1900-01-01 12:39:36  354.0
6   35367939  74.95  02/11/18 1900-01-01 12:45:30 1900-01-01 12:39:39  351.0
7   35371892  74.95  02/11/18 1900-01-01 12:45:31 1900-01-01 12:39:55  336.0

如果有str-

，请先计算日期，然后再计算差异

df['TIME_1'] = pd.to_datetime(df['TIME_1'],format='%H:%M:%S')
df['TIME_2'] = pd.to_datetime(df['TIME_2'],format='%H:%M:%S')

Answer 4

如果您知道如何使用数据框，则此代码将为您提供良好的帮助：

import datetime
import time

x1 = time.strptime('12:45:26,000'.split(',')[0], '%H:%M:%S')
x2 = time.strptime('12:30:36,000'.split(',')[0], '%H:%M:%S')
delta_x = datetime.timedelta(hours=x1.tm_hour - x2.tm_hour, minutes=x1.tm_min - 
x2.tm_min, seconds=x1.tm_sec - x2.tm_sec).total_seconds()
print(delta_x)

Answer 5

看看将两个数字都收敛到datetime objects，然后将它们相减，就像这样：

import datetime as dt

time1 = dt.datetime(2018,12,30,23,59,59)
time2 = dt.datetime(2018,12,31,23,59,59)

(time2-time1).total_seconds()

（source）

计算以秒为单位的时差

5 个答案: