可以使用pd.to_datetime的格式arg：

In [11]: s = pd.Series(["29062017", "01AUG2017"])

In [12]: pd.to_datetime(s, format="%d%m%Y", errors="coerce")
Out[12]:
0   2017-06-29
1          NaT
dtype: datetime64[ns]

In [13]: pd.to_datetime(s, format="%d%b%Y", errors="coerce")
Out[13]:
0          NaT
1   2017-08-01
dtype: datetime64[ns]

• 注意：coerce参数意味着失败将是NaT。*

并将NaN s从一个填入另一个，例如使用fillna：

In [14]: pd.to_datetime(s, format="%d%m%Y", errors="coerce").fillna(
    ...:     pd.to_datetime(s, format="%d%b%Y", errors="coerce"))
Out[14]:
0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

任何不匹配这两种格式的字符串都将保持为NaT。

另一种方法是使用Map器和replace将月份代码替换为相应的数字：

s = pd.Series(["29062017", "01AUG2017"]); s

0     29062017
1    01AUG2017
dtype: object

m = {'JAN' : '01', ..., 'AUG' : '08', ...}  # you fill in the rest

s = s.replace(m, regex=True); s

0    29062017
1    01082017
dtype: object

现在你只需要一个pd.to_datetime调用：

pd.to_datetime(s, format="%d%m%Y", errors="coerce")

0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

既然你有两种约会时间...

s.apply(lambda x : pd.to_datetime(x, format="%d%m%Y") if x.isdigit() else pd.to_datetime(x, format="%d%b%Y"))

Out[360]: 
0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

我想提出一些建议

设置

m = dict(
    JAN='01', FEB='02', MAR='03', APR='04',
    MAY='05', JUN='06', JUL='07', AUG='08',
    SEP='09', OCT='10', NOV='11', DEC='12'
)

m2 = m.copy()
m2.update({v: v for v in m.values()})

f = lambda x: m.get(x, x)

备选案文1

列表理解

pd.Series(
    pd.to_datetime(
        [x[:2] + f(x[2:5]) + x[5:] for x in s.values.tolist()],
        format='%d%m%Y'),
    s.index)

0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

备选案文2

创建 Dataframe

pd.to_datetime(
    pd.DataFrame(dict(
        day=s.str[:2],
        year=s.str[-4:],
        month=s.str[2:-4].map(m2)
    )))

0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

方案2B

创建 Dataframe

pd.to_datetime(
    pd.DataFrame(dict(
        day=s.str[:2],
        year=s.str[-4:],
        month=s.str[2:-4].map(f)
    )))

0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

选项2C

创建 Dataframe
我估计这是最快的

pd.to_datetime(
    pd.DataFrame(dict(
        day=s.str[:2].astype(int),
        year=s.str[-4:].astype(int),
        month=s.str[2:-4].map(m2).astype(int)
    )))

0   2017-06-29
1   2017-08-01
dtype: datetime64[ns]

测试

s = pd.Series(["29062017", "01AUG2017"] * 100000)

%timeit pd.to_datetime(s.replace(m, regex=True), format='%d%m%Y')
%timeit pd.to_datetime(s.str[:2] + s.str[2:5].replace(m) + s.str[5:], format='%d%m%Y')
%timeit pd.to_datetime(s.str[:2] + s.str[2:5].map(f) + s.str[5:], format='%d%m%Y')
%timeit pd.to_datetime(s, format='%d%m%Y', errors='coerce').fillna(pd.to_datetime(s, format='%d%b%Y', errors='coerce'))
%timeit pd.Series(pd.to_datetime([x[:2] + f(x[2:5]) + x[5:] for x in s.values.tolist()], format='%d%m%Y'), s.index)
%timeit pd.to_datetime(pd.DataFrame(dict(day=s.str[:2], year=s.str[-4:], month=s.str[2:-4].map(m2))))
%timeit pd.to_datetime(pd.DataFrame(dict(day=s.str[:2], year=s.str[-4:], month=s.str[2:-4].map(f))))
%timeit pd.to_datetime(pd.DataFrame(dict(day=s.str[:2].astype(int), year=s.str[-4:].astype(int), month=s.str[2:-4].map(m2).astype(int))))

1.39 s ± 24 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
690 ms ± 17.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
613 ms ± 13.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
533 ms ± 14.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
529 ms ± 8.04 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
557 ms ± 13 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
607 ms ± 26.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
328 ms ± 31.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

下面是我对这个问题的解决方案：

def set_date(col):
    # date_formates = ["21 June, 2018", "12/11/2018 09:15:32", "April-21" ]
    date_formats = ["%d %B, %Y", "%d/%m/%Y %H:%M:%S", "%B-%y", "%d %B, %Y", "%m/%d/Y"] # Can add different date formats to this list to test
    for x in date_formats:
        col = pd.to_datetime(col, errors="ignore", format= f"{x}")

    col = pd.to_datetime(col, errors="coerce") # To remove errors in the columns like strings or numbers
    return col