为什么.loc具有片的包容性行为?
问题说明
由于某些原因，以下两个对iloc/loc的调用会产生不同的行为:
For some reason, the following 2 calls to iloc / loc produce different behavior:
>>> import pandas as pd
>>> df = pd.DataFrame(dict(A=range(3), B=range(3)))
>>> df.iloc[:1]
A B
0 0 0
>>> df.loc[:1]
A B
0 0 0
1 1 1
我知道loc考虑行标签，而iloc考虑行的基于整数的索引.但是为什么loc调用的上限被认为是包含的，而iloc调用的上限却被认为是排他的呢?
I understand that loc considers the row labels, while iloc considers the integer-based indices of the rows. But why is the upper bound for the loc call considered inclusive, while the iloc bound is considered exclusive?
正确答案
快速解答:
使用标签时，进行端到端切片通常更有意义，因为它需要有关DataFrame中其他行的知识较少.
It often makes more sense to do end-inclusive slicing when using labels, because it requires less knowledge about other rows in the DataFrame.
每当您关心标签而不是位置时，末端排他的标签切片都会以一种不方便的方式引入位置依赖性.
Whenever you care about labels instead of positions, end-exclusive label slicing introduces position-dependence in a way that can be inconvenient.
更长的答案:
任何函数的行为都是一个权衡:您偏爱某些用例而不是其他用例.最终，.iloc的操作是熊猫开发人员的主观设计决定(正如@ALlollz的评论所指出的，此行为 @Willz的答案:
That said, maybe there are use cases where you really do want end-exclusive label-based slicing. If so, you can use @Willz's answer in this question:
df.loc[start:end].iloc[:-1]
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