简单的条件:

if any(str(elem) in ['a','b'] for elem in df['column'].tolist()):

你也可以使用pandas.Series.isin，尽管它比s.values中的'a'长一点:

In [2]: s = pd.Series(list('abc'))

In [3]: s
Out[3]: 
0    a
1    b
2    c
dtype: object

In [3]: s.isin(['a'])
Out[3]: 
0    True
1    False
2    False
dtype: bool

In [4]: s[s.isin(['a'])].empty
Out[4]: False

In [5]: s[s.isin(['z'])].empty
Out[5]: True

但是，如果您需要为一个DataFrame同时匹配多个值，这种方法可以更加灵活。

>>> df = DataFrame({'A': [1, 2, 3], 'B': [1, 4, 7]})
>>> df.isin({'A': [1, 3], 'B': [4, 7, 12]})
       A      B
0   True  False  # Note that B didn't match 1 here.
1  False   True
2   True   True

Series的in检查该值是否在索引中:

In [11]: s = pd.Series(list('abc'))

In [12]: s
Out[12]: 
0    a
1    b
2    c
dtype: object

In [13]: 1 in s
Out[13]: True

In [14]: 'a' in s
Out[14]: False

一种选择是查看它是否为唯一值:

In [21]: s.unique()
Out[21]: array(['a', 'b', 'c'], dtype=object)

In [22]: 'a' in s.unique()
Out[22]: True

或者一个python集合:

In [23]: set(s)
Out[23]: {'a', 'b', 'c'}

In [24]: 'a' in set(s)
Out[24]: True

正如@DSM所指出的，直接在值上使用in可能更有效(特别是如果你只对一个值执行此操作):

In [31]: s.values
Out[31]: array(['a', 'b', 'c'], dtype=object)

In [32]: 'a' in s.values
Out[32]: True

使用query()查找符合条件的行，并获得形状为[0]的行数。如果存在至少一个条目，则此语句为True:

df.query('id == 123').shape[0] > 0

简单的条件:

if any(str(elem) in ['a','b'] for elem in df['column'].tolist()):

我做了一些简单的测试:

In [10]: x = pd.Series(range(1000000))

In [13]: timeit 999999 in x.values
567 µs ± 25.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [24]: timeit 9 in x.values
666 µs ± 15.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [16]: timeit (x == 999999).any()
6.86 ms ± 107 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [21]: timeit x.eq(999999).any()
7.03 ms ± 33.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [22]: timeit x.eq(9).any()
7.04 ms ± 60 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [15]: timeit x.isin([999999]).any()
9.54 ms ± 291 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [17]: timeit 999999 in set(x)
79.8 ms ± 1.98 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

有趣的是，不管你是查找9还是999999，似乎使用in语法所花费的时间是相同的(一定是使用了某种向量化计算)

In [24]: timeit 9 in x.values
666 µs ± 15.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [25]: timeit 9999 in x.values
647 µs ± 5.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [26]: timeit 999999 in x.values
642 µs ± 2.11 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [27]: timeit 99199 in x.values
644 µs ± 5.31 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [28]: timeit 1 in x.values
667 µs ± 20.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

似乎使用x.values是最快的，但也许在pandas中有更优雅的方法?