具有MultiIndex的DataFrame的.loc和.iloc [英] `.loc` and `.iloc` with MultiIndex'd DataFrame

问题描述

在为MultiIndex-ed DataFrame编制索引时，似乎.iloc假设您正在引用索引的内部级别"，而.loc则在外部级别.

When indexing a MultiIndex-ed DataFrame, it seems like .iloc assumes you're referencing the "inner level" of the index while .loc looks at the outer level.

例如:

np.random.seed(123)
iterables = [['bar', 'baz', 'foo', 'qux'], ['one', 'two']]
idx = pd.MultiIndex.from_product(iterables, names=['first', 'second'])
df = pd.DataFrame(np.random.randn(8, 4), index=idx)

# .loc looks at the outer index:

print(df.loc['qux'])
# df.loc['two'] would throw KeyError
              0        1        2        3
second                                    
one    -1.25388 -0.63775  0.90711 -1.42868
two    -0.14007 -0.86175 -0.25562 -2.79859

# while .iloc looks at the inner index:

print(df.iloc[-1])
0   -0.14007
1   -0.86175
2   -0.25562
3   -2.79859
Name: (qux, two), dtype: float64

两个问题:

首先，这是为什么?这是故意设计的决定吗?

Firstly, why is this? Is it a deliberate design decision?

第二，我可以使用.iloc引用索引的外部级别，以产生以下结果吗?我知道我可以先用get_level_values找到索引的最后一个成员，然后再用.loc -index找到索引的最后一个成员，但是徘徊着是否可以直接使用时髦的.iloc语法或某些设计的现有函数来完成它专门针对这种情况.

Secondly, can I use .iloc to reference the outer level of the index, to yield the result below? I'm aware I could first find the last member of the index with get_level_values and then .loc-index with that, but wandering if it can be done more directly, either with funky .iloc syntax or some existing function designed specifically for the case.

# df.iloc[-1]
qux   one     0.89071  1.75489  1.49564  1.06939
      two    -0.77271  0.79486  0.31427 -1.32627

推荐答案

是的，这是精心设计的决定:

.iloc是严格的位置索引器，它不考虑结构 完全只有第一种实际行为. ... .loc 做了 帐户级别的行为. [加重]

.iloc is a strict positional indexer, it does not regard the structure at all, only the first actual behavior. ... .loc does take into account the level behavior. [emphasis added]

因此使用.iloc不能灵活地在问题中给出期望的结果.在几个类似的问题中使用的最接近的解决方法是

So the desired result given in the question is not possible in a flexible manner with .iloc. The closest workaround, used in several similar questions, is

print(df.loc[[df.index.get_level_values(0)[-1]]])
                    0        1        2        3
first second                                    
qux   one    -1.25388 -0.63775  0.90711 -1.42868
      two    -0.14007 -0.86175 -0.25562 -2.79859

使用双括号将保留第一个索引级别.

Using double brackets will retain the first index level.

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