要从字符串中删除所有的破折号/斜杠/连字符，可以使用replace方法：

df.apply(lambda x: x.str.replace('[/-]',' ',regex=True))

>>>
'''
  INITIATION DATE      DATE CUT    DATE GIVEN
0     1 July 2022  21 July 2022  11 July 2022
1    17 July 2022  16 July 2022  21 July 2022
2    16 July 2022  01 July 2022  09 July 2022
3    19 July 2022  31 July 2022  27 July 2022

如果还需要将字符串转换为日期时间，请尝试以下操作：

df.apply(lambda x: pd.to_datetime(x.str.replace('[/-]',' ',regex=True)))

>>>
'''
  INITIATION DATE   DATE CUT DATE GIVEN
0      2022-07-01 2022-07-21 2022-07-11
1      2022-07-17 2022-07-16 2022-07-21
2      2022-07-16 2022-07-01 2022-07-09
3      2022-07-19 2022-07-31 2022-07-27

您可以使用pd.to_datetime将字符串转换为日期时间对象。该函数采用format参数，该参数使用常用的格式代码指定日期时间字符串的格式

df['INITIATION DATE'] = pd.to_datetime(df['INITIATION DATE'], format='%d-%B-%Y').dt.strftime('%d %B %Y')
df['DATE CUT'] = pd.to_datetime(df['DATE CUT'], format='%d %B %Y').dt.strftime('%d %B %Y')
df['DATE GIVEN'] = pd.to_datetime(df['DATE GIVEN'], format='%d/%B/%Y').dt.strftime('%d %B %Y')

输出

INITIATION DATE      DATE CUT    DATE GIVEN
0    01 July 2022  21 July 2022  11 July 2022
1    17 July 2022  16 July 2022  21 July 2022
2    16 July 2022  01 July 2022  09 July 2022
3    19 July 2022  31 July 2022  27 July 2022

出现该错误是因为日期时间字符串（例如'18 July 2022'）与格式说明符（'%d-%b-%Y'）不匹配，原因是格式说明符中存在额外的连字符。