Python ValueError:时间数据与格式不匹配'；%Y/%m/%d%H:%m:%S'；

我需要将数千条Datetime记录转换为时间戳。当我运行以下代码时：

from time import strptime
import pandas as pd
import numpy as np
import datetime

cepo = str(pd.read_csv('test.csv',sep = ',')) 
print(cepo)
element = datetime.datetime.strptime(cepo,'%Y/%m/%d %H:%M:%S')
tuple = element.timetuple()
timestamp = time.mktime(tuple)
print(timestamp)

我得到以下错误：

     Date      Time
0  2018/02/11  02:36:59
1  2018/02/12  00:47:11
2  2018/02/12  01:36:36
3  2018/02/12  03:27:51
4  2018/02/12  03:48:29
5  2018/02/12  03:50:49
Traceback (most recent call last):
  File "epoch.py", line 9, in <module>
    element = datetime.datetime.strptime(cepo,'%Y/%m/%d %H:%M:%S')
  File "/Users/joaofontiela/opt/anaconda3/envs/pywork/lib/python3.8/_strptime.py", line 568, in _strptime_datetime
    tt, fraction, gmtoff_fraction = _strptime(data_string, format)
  File "/Users/joaofontiela/opt/anaconda3/envs/pywork/lib/python3.8/_strptime.py", line 349, in _strptime
    raise ValueError("time data %r does not match format %r" %
ValueError: time data '         Date      Time\n0  2018/02/11  02:36:59\n1  2018/02/12  00:47:11\n2  2018/02/12  01:36:36\n3  2018/02/12  03:27:51\n4  2018/02/12  03:48:29\n5  2018/02/12  03:50:49' does not match format '%Y/%m/%d %H:%M:%S'

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如何修复“时间数据与格式“%Y/%m/%d%H:%m:%S”不匹配”的错误？

您可以轻松地将csv中的“日期”和“时间”列组合在一起

# load the csv file content into a dataframe
df = pd.read_csv(filename)

# combine date and time columns and parse to datetime
df['datetime'] = pd.to_datetime(df['Date'] + ' ' + df['Time'])

print(df['datetime'])
0   2018-02-11 02:36:59
1   2018-02-12 00:47:11
2   2018-02-12 01:36:36
3   2018-02-12 03:27:51
4   2018-02-12 03:48:29
5   2018-02-12 03:50:49
Name: datetime, dtype: datetime64[ns]

您可以计算Unix时间（自1970-01-01 UTC以来的Posix/秒），如下所示

---

为什么要将数据帧转换为字符串？如果要使用pandas将字符串解析为datetime，请使用。csv转换尝试处理的文件末尾是否有意外字符？假定text.csv是完整显示的，并且在下一行（未显示）中没有一系列格式错误的字符引起问题。感谢您的帮助。但现在又出现了另一个问题。如何将df.datetime转换为posix数据格式？我研究了几个示例，其中大多数都涉及将posix数据格式转换为datetime。任何帮助都是好的，因为我不是一个熟练的pragrammer@Meocomandante当然，我已经编辑过了

.astype（int）

从纪元开始转换为纳秒，因此您可以除以1e9得到秒。非常感谢您的帮助

# load the csv file content into a dataframe
df = pd.read_csv(filename)

# combine date and time columns and parse to datetime
df['datetime'] = pd.to_datetime(df['Date'] + ' ' + df['Time'])

print(df['datetime'])
0   2018-02-11 02:36:59
1   2018-02-12 00:47:11
2   2018-02-12 01:36:36
3   2018-02-12 03:27:51
4   2018-02-12 03:48:29
5   2018-02-12 03:50:49
Name: datetime, dtype: datetime64[ns]

df['posix[s]'] = df['datetime'].astype(int) / 1e9

print(df['posix[s]'])
0    1.518317e+09
1    1.518396e+09
2    1.518399e+09
3    1.518406e+09
4    1.518407e+09
5    1.518407e+09
Name: posix[s], dtype: float64