apply在后台使用循环，因此，如果需要更好的性能，最好的和最快的方法是最好的选择。
没有循环，只有链2条件向量化解决方案：
m1 = all_actions['Lower'] <= all_actions['Mid']
m2 = all_actions['Mid'] <= all_actions['Upper']
qualified_actions = m1 & m2
感谢on Clements提供的另一种解决方案：
all_actions.Mid.between(all_actions.Lower, all_actions.Upper)
：
np.random.seed(2017)
N = 45000
all_actions=pd.DataFrame(np.random.randint(50, size=(N,3)),columns=['Lower','Mid','Upper'])
#print (all_actions)
In [85]: %%timeit
...: qualified_actions = []
...: for row in all_actions.index:
...: if all_actions.ix[row,'Lower'] <= all_actions.ix[row, 'Mid'] <= all_actions.ix[row,'Upper']:
...: qualified_actions.append(True)
...: else:
...: qualified_actions.append(False)
...:
...:
__main__:259: DeprecationWarning:
.ix is deprecated. Please use
.loc for label based indexing or
.iloc for positional indexing
See the documentation here:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#ix-indexer-is-deprecated
1 loop, best of 3: 579 ms per loop
In [86]: %%timeit
...: (all_actions.apply(lambda row: row['Lower'] <= row['Mid'] <= row['Upper'], axis=1))
...:
1 loop, best of 3: 1.17 s per loop
In [87]: %%timeit
...: ((all_actions['Lower'] <= all_actions['Mid']) & (all_actions['Mid'] <= all_actions['Upper']))
...:
1000 loops, best of 3: 509 µs per loop
In [90]: %%timeit
...: (all_actions.Mid.between(all_actions.Lower, all_actions.Upper))
...:
1000 loops, best of 3: 520 µs per loop
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