如何在pandas loc API中使用*和*

Question

我尝试在.loc API ::

中使用和

df = pd.DataFrame(dict(age=[99, 33, 33, 22, 33, 44],
                       aa2=[199, 3, 43, 22, 23, 54],
                       nom=['a', 'z', 'f', 'b', 'p', 'a'],))
df.loc[df.age>30]
#    aa2  age nom
# 0  199   99   a
# 1    3   33   z
# 2   43   33   f
# 4   23   33   p
# 5   54   44   a

但是我收到了这个错误::

df.loc[df.age>30 and (df.age > df.aa2)]
# ---------------------------------------------------------------------------
# ValueError                                Traceback (most recent call last)
# <ipython-input-13-930dff789922> in <module>()
# ----> 1 df.loc[df.age>30 and (df.age > df.aa2)]
#
# /site-packages/pandas/core/generic.pyc in __nonzero__(self)
#     729         raise ValueError("The truth value of a {0} is ambiguous. "
#     730                          "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
# --> 731                          .format(self.__class__.__name__))
#     732
#     733     __bool__ = __nonzero__
#
# ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

现在我这样做;（::

df.loc[df.age>30].loc[(df.age > df.aa2)]

#    aa2  age nom
# 1    3   33   z
# 4   23   33   p

Answer 1

>>> df.loc[(df.age>30) & (df.age > df.aa2)]
   aa2  age nom
1    3   33   z
4   23   33   p

Answer 2

我更喜欢更好，更易读的query()方法：

In [3]: df.query('age > 30 and age > aa2')
Out[3]:
   aa2  age nom
1    3   33   z
4   23   33   p

PS好吧，它没有直接回答你的问题（@ M.Klugerford has already shown you how to do this using .loc[]），但它给你一个更好的（在我个人看来）替代