python - 将时间字符串格式化为 100 毫秒 - Python

Question

我有一个函数，可以将时间字符串格式化为最接近的 0.1 秒。但是，在四舍五入到最接近的秒时，它不会格式化时间戳。我将在下面附上一个示例:

d = ({                            
    'Time' : ['2019-08-02 09:50:19.880','2019-08-02 09:50:19.970','2019-08-02 09:50:19.980','2019-08-02 09:50:20.070'],
    })

df = pd.DataFrame(data = d)

def format_time(col):
    col = pd.to_datetime(col)
    t = col    
    s = t.strftime('%Y-%m-%d %H:%M:%S.%f')
    tail = s[-8:]
    f = round(float(tail), 3)
    temp = "%.1f" % f
    return "%s%s" % (s[:-8], temp[0:])

df['Time'] = df['Time'].apply(format_time)

输出:

                     Time
0   2019-08-02 09:50:19.9
1  2019-08-02 09:50:110.0
2  2019-08-02 09:50:110.0
3   2019-08-02 09:50:20.1

预期:

                    Time
0   2019-08-02 09:50:19.9
1   2019-08-02 09:50:20.0
2   2019-08-02 09:50:20.0
3   2019-08-02 09:50:20.1

Answer 1

您可以根据日期舍入 timestamp() 方法返回的值，并从中实例化一个新的 datetime。我添加了一个示例，需要更新从第二到年份的所有字段。

import datetime

d = ({                            
    'Time' : ['2019-08-02 09:50:19.880',
              '2019-08-02 09:50:19.970',
              '2019-08-02 09:50:19.980',
              '2019-08-02 09:50:20.070',
              '2019-08-02 09:50:20.320',
              '2019-12-31 23:59:59.990'
             ],
    })

df = pd.DataFrame(data = d)

def format_time(col):
    t = pd.to_datetime(col)      
    t = datetime.datetime.fromtimestamp(round(t.timestamp(), 1))
    return t.strftime('%Y-%m-%d %H:%M:%S.%f')[:-5]

df['Time'] = df['Time'].apply(format_time)
print(df)

结果是:

                    Time
0  2019-08-02 09:50:19.9
1  2019-08-02 09:50:20.0
2  2019-08-02 09:50:20.0
3  2019-08-02 09:50:20.1
4  2019-08-02 09:50:20.3
5  2020-01-01 00:00:00.0