大熊猫是否可以转换如下所示的列....
for i, a in enumerate(l):
if ...:
...
...
:(我认为没有Date
的ISO格式)
T
仅使用小时和分钟的格式。四舍五入到最接近的分钟就足够了。我可以从2020-06-29 14:04:21.000000
2020-06-29 14:19:26.000000
2020-06-29 14:34:30.000000
2020-06-29 14:49:35.000000
2020-06-29 15:04:39.000000
2020-06-29 15:19:44.000000
2020-06-29 15:34:48.000000
2020-06-29 15:49:53.000000
2020-06-29 16:20:02.000000
2020-06-29 16:35:07.000000
2020-06-29 16:50:11.000000
2020-06-29 17:05:16.000000
2020-06-29 17:20:20.000000
2020-06-29 17:35:25.000000
2020-06-29 17:50:30.000000
2020-06-29 18:05:34.000000
2020-06-29 18:20:39.000000
2020-06-29 18:35:43.000000
2020-06-29 18:50:48.000000
2020-06-29 19:05:52.000000
转到下面的格式吗?感谢您的提示...
time.time.now()
答案 0 :(得分:2)
import pandas as pd
data = {'Date': ['2019-05-30 00:00:44.000000', '2019-05-30 00:00:50.000000','2019-05-30 00:23:26.000000']}
df = pd.DataFrame(data=data)
df['New_Date'] = pd.to_datetime(df['Date']).dt.round('min').dt.strftime('%m/%d/%Y %H:%M')
输出:
Date New_Date
0 2019-05-30 00:00:44 05/30/2019 00:01
1 2019-05-30 00:00:50 05/30/2019 00:01
2 2019-05-30 00:23:26 05/30/2019 00:23
答案 1 :(得分:1)
如果您希望从time.time()
方法转到所需的格式,则可以使用以下方法:
time.strftime('%m/%d/%Y %H:%M')
此外,如果您的pandas变量是有效的Datetime对象,则可以对其应用strftime
。
答案 2 :(得分:0)
def formatTime(origTD):
date = origTD.split(' ')[0]
time = origTD.split(' ')[1]
day = int(date.split('-')[2])
month = int(date.split('-')[1])
year = date.split('-')[0][-2:]
hour = time.split(':')[0]
minute = time.split(':')[1]
newTD = str(month)+'/'+str(day)+'/'+year+' '+hour+':'+minute
return newTD