将秒设置为0

Question

我有以下日期，我尝试了以下代码，

df['start_date_time'] = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]
df['start_date_time'] = pd.to_datetime([df['start_date_time']).replace(second = 0)

我收到以下错误：

TypeError: replace() got an unexpected keyword argument 'second'

Answer 1

将秒设置为0

pd.to_datetime会返回datetime个对象，其中second为属性：您可以做的事情不多。您可以将second设置为0，但该属性仍然在此处，标准表示仍将包含尾随':00'。

您需要对replace的每个元素应用df：

import pandas as pd

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])
df['start_date_time'] = df['start_date_time'].apply(lambda t: t.replace(second=0))

print(df)
#       start_date_time
# 0 2016-05-19 08:25:00
# 1 2016-05-19 16:00:00
# 2 2016-05-20 07:45:00
# 3 2016-05-24 12:50:00
# 4 2016-05-25 23:00:00
# 5 2016-05-26 19:45:00

第一次

:23和:45已被:00取代，但仍会打印出来。

从字符串

中删除':00'

如果你只想要那些时间的字符串表示并且只将字符串解析为datetime个对象以便删除字符串末尾的':00'，你可以删除最后3个字符：

>>> "2016-05-19 08:25:00"[:-3]
'2016-05-19 08:25'

在初始化df['start_date_time']之前，您可以将其应用于列表中的每个元素：

>>> start_date_time = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]
>>> map(lambda s: s[:-3], start_date_time)
['2016-05-19 08:25', '2016-05-19 16:00', '2016-05-20 07:45', '2016-05-24 12:50', '2016-05-25 23:00', '2016-05-26 19:45']

显示没有秒的日期时间

如果您想使用datetime个对象，但又不想显示秒数：

print(df['start_date_time'].apply(lambda t: t.strftime('%Y-%m-%d %H:%M')))
# 0    2016-05-19 08:25
# 1    2016-05-19 16:00
# 2    2016-05-20 07:45
# 3    2016-05-24 12:50
# 4    2016-05-25 23:00
# 5    2016-05-26 19:45
# Name: start_date_time, dtype: object

Answer 2

您可以使用timedelta减去秒数：

import datetime    
d = datetime.datetime.now() #datetime including seconds
without_seconds = d - datetime.timedelta(seconds=d.second)

Answer 3

您可以先使用转换为numpy values，然后通过强制转换为seconds来截断<M8[m]：

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
print (df)
       start_date_time
0  2016-05-19 08:25:23
1  2016-05-19 16:00:45

df['start_date_time'] = pd.to_datetime(df['start_date_time'])
df['start_date_time'] = df['start_date_time'].values.astype('<M8[m]')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

另一种解决方案是从second创建timedelta系列，并使用

print (pd.to_timedelta(df['start_date_time'].dt.second, unit='s'))
0   00:00:23
1   00:00:45
Name: start_date_time, dtype: timedelta64[ns]

df['start_date_time'] = df['start_date_time'] - 
                        pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

<强>计时：

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])

#20000 rows
df = pd.concat([df]*10000).reset_index(drop=True)


In [181]: %timeit df['start_date_time'] = df['start_date_time'] - pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
100 loops, best of 3: 5.83 ms per loop

In [182]: %timeit df['start_date_time1'] = df['start_date_time'].values.astype('<M8[m]')
The slowest run took 4.21 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 638 µs per loop

In [183]: %timeit df['start_date_time2'] = df['start_date_time'].apply(lambda t: t.replace(second=0))
10 loops, best of 3: 83.4 ms per loop

Answer 4

HTML代码：

<输入type =“ time” class =“ form-control” value =“” name =“ meeting_time”必需/>

Python Django代码：

meeting_time = request.POST ['meeting_time']＃像your_time =“ 12:35:00”

get_time = Meeting_time.strftime（“％H：％M”）

结果是：

get_time =“ 12:35”

Answer 5

这是从日期时间中删除秒的一种非常简单的方法：

from datetime import datetime 
print(str(datetime.today())[:16])

输出：

2021-02-14 21:30

它有效地将时间戳转换为文本并且只留下前 16 个符号。只是不要在所有这些括号中迷失自己;)

Answer 6

用以下内容拍摄：

df.index = df.index.map(lambda t: t.strftime('%Y-%m-%d %H:%M'))

如其中一条评论中所述，以上内容适用于日期不是字符串的情况。但是，如果它们是字符串，则只需对列表中每个列表中的最后三个字符进行切片：

import pandas as pd

df = pd.DataFrame({'date': ["2016-05-19 08:25:00"]})

print(df['date'].map(lambda t: t[:-3]))

以上将输出：

0    2016-05-19 08:25
Name: date, dtype: object

Answer 7

正如你提到的那样删除所以我假设你不希望结果中的秒或微秒。如果是这种情况，那么下面的内容可能有所帮助：

datetime_variable.strftime("'%Y-%m-%d %H:%M'")

如果你的日期时间是字符串，那么你可以在datetime obj中转换它：

from dateutil import parser

datetime_variable = parser.parse(str_datetime_var)

datetime_variable.strftime("'%Y-%m-%d %H:%M'")

Answer 8

首先将String转换为datetime对象，然后可以使用replace方法。

from _datetime import *


df = dict()
df['start_date_time'] = ["2016-05-19 08:25:00",
                         "2016-05-19 16:00:00",
                         "2016-05-20 07:45:00",
                         "2016-05-24 12:50:00",
                         "2016-05-25 23:00:00",
                         "2016-05-26 19:45:00"]

for dt in df['start_date_time']:
    cur_dt = datetime.strptime(dt, '%Y-%m-%d %H:%M:%S')
    cur_dt = cur_dt.replace(second=0)
    print(cur_dt)

    cur_dt_without_second = cur_dt.strftime('%Y-%m-%d %H:%M')
    print(cur_dt_without_second)

-------------------
2016-05-19 08:25:00
2016-05-19 08:25
2016-05-19 16:00:00
2016-05-19 16:00
2016-05-20 07:45:00
2016-05-20 07:45
2016-05-24 12:50:00
2016-05-24 12:50
2016-05-25 23:00:00
2016-05-25 23:00
2016-05-26 19:45:00
2016-05-26 19:45

Answer 9

将字符串转换为datetime对象，然后操作

>>> x = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]
>>> for i in x:
...  y = datetime.datetime.strptime(i, '%Y-%m-%d %H:%M:%S')
...  z = datetime.datetime.strftime(y, '%Y-%m-%d %H:%M')
...  print (y, type(y))
...  print (z, type(z))
... 
(datetime.datetime(2016, 5, 19, 8, 25), <type 'datetime.datetime'>)
('2016-05-19 08:25', <type 'str'>)
(datetime.datetime(2016, 5, 19, 16, 0), <type 'datetime.datetime'>)
('2016-05-19 16:00', <type 'str'>)
(datetime.datetime(2016, 5, 20, 7, 45), <type 'datetime.datetime'>)
('2016-05-20 07:45', <type 'str'>)
(datetime.datetime(2016, 5, 24, 12, 50), <type 'datetime.datetime'>)
('2016-05-24 12:50', <type 'str'>)
(datetime.datetime(2016, 5, 25, 23, 0), <type 'datetime.datetime'>)
('2016-05-25 23:00', <type 'str'>)
(datetime.datetime(2016, 5, 26, 19, 45), <type 'datetime.datetime'>)
('2016-05-26 19:45', <type 'str'>)

Answer 10

这是时间模块而不是日期时间模块，但可能是您正在寻找的......

import time
X = time.strftime("%H:%M")
print(X)

或你的秒数

import time
X = time.strftime("%H:%M:%S")
print(X)

如果您要投票，请解释原因

如何从datetime删除秒？

10 个答案:

将秒设置为0

从字符串

显示没有秒的日期时间