Question

我有一个数据框timings，如下所示：

    start_ms                     end_ms
0   2020-09-01T08:11:19.336Z     2020-09-01T08:11:19.336Z
1   2020-09-01T08:11:20.652Z     2020-09-01T08:11:20.662Z
2   2020-09-01T08:11:20.670Z     2020-09-01T08:11:20.688Z

我正在尝试计算每行start_ms和end_ms之间的时间差（以毫秒为单位），即我希望得到结果

    start_ms                     end_ms                       diff
0   2020-09-01T08:11:19.336Z     2020-09-01T08:11:19.336Z     0
1   2020-09-01T08:11:20.652Z     2020-09-01T08:11:20.662Z     10
2   2020-09-01T08:11:20.670Z     2020-09-01T08:11:20.688Z     18

我可以逐列将时间戳转换为日期时间，但是我不确定是否保留值的顺序。

start_ms_time = pd.to_datetime(timings['start_ms'])
end_ms_time = pd.to_datetime(timings['end_ms'])

是否可以在timings中将时间戳转换为日期时间，并添加时差列？我什至需要转换以获取差异？如何计算以毫秒为单位的时差？

Answer 1

用Series.sub减去列，然后使用Series.dt.components：

start_ms_time = pd.to_datetime(timings['start_ms'])
end_ms_time = pd.to_datetime(timings['end_ms'])

timings['diff'] = end_ms_time.sub(start_ms_time).dt.components.milliseconds
print (timings)
                   start_ms                    end_ms  diff
0  2020-09-01T08:11:19.336Z  2020-09-01T08:11:19.336Z     0
1  2020-09-01T08:11:20.652Z  2020-09-01T08:11:20.662Z    10
2  2020-09-01T08:11:20.670Z  2020-09-01T08:11:20.688Z    18

或将Series.dt.total_seconds乘以1000并乘以整数：

timings['diff'] = end_ms_time.sub(start_ms_time).dt.total_seconds().mul(1000).astype(int)
print (timings)
                   start_ms                    end_ms  diff
0  2020-09-01T08:11:19.336Z  2020-09-01T08:11:19.336Z     0
1  2020-09-01T08:11:20.652Z  2020-09-01T08:11:20.662Z    10
2  2020-09-01T08:11:20.670Z  2020-09-01T08:11:20.688Z    18

使用熊猫计算以毫秒为单位的时差

1 个答案: