Pandas 索引和密钥错误
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Pandas indexing and Key error
提问by Yash
Consider the following:
考虑以下：
d = {'a': 0.0, 'b': 1.0, 'c': 2.0}
e = pd.Series(d, index = ['a', 'b', 'c'])
df = pd.DataFrame({ 'A' : 1.,'B' : e,'C' :pd.Timestamp('20130102')}).
When i try to access the first row of column B in the following way:
当我尝试通过以下方式访问 B 列的第一行时：
>>> df.B[0]
0.0
I get the correct result.
我得到正确的结果。
However, after reading KeyError: 0 when accessing value in pandas series, I was under the assumption that, since I have specified the index as 'a', 'b' and 'c', the correct way to access the first row of column B (using positional arguments) is:
df.B.iloc[0], and df.B[0]should raise a Key Error. I dont know what am I missing. Can someone clarify in which case do I get a Key Error ?
但是，在阅读KeyError: 0 when accessing value in pandas series 之后，我假设，因为我已将索引指定为 'a'、'b' 和 'c'，这是访问列第一行的正确方法B（使用位置参数）是:
df.B.iloc[0]，并且df.B[0]应该引发一个关键错误。我不知道我错过了什么。有人可以澄清在哪种情况下我会收到 Key Error 吗？
回答by Justinas Marozas
Problem in your referenced Question is that index of given dataframe is integer, but does not start from 0.
您引用的问题中的问题是给定数据帧的索引是整数，但不是从 0 开始。
Pandas behaviour when asking for df.B[0]is ambiguous and depends on data type of index and data type of value passed to python slice syntax. It can behave like df.B.loc[0](index label based) or df.B.iloc[0](position based) or probably something else I'm not aware of. For predictable behaviour I recommend using locand iloc.
请求时的 Pandas 行为df.B[0]不明确，取决于传递给 python 切片语法的索引的数据类型和值的数据类型。它可以表现得像df.B.loc[0]（基于索引标签）或df.B.iloc[0]（基于位置）或者可能是我不知道的其他东西。对于可预测的行为，我建议使用loc和iloc。
To illustrate this with your example:
用你的例子来说明这一点：
d = [0.0, 1.0, 2.0]
e = pd.Series(d, index = ['a', 'b', 'c'])
df = pd.DataFrame({'A': 1., 'B': e, 'C': pd.Timestamp('20130102')})
df.B[0] # 0.0 - fall back to position based
df.B['0'] # KeyError - no label '0' in index
df.B['a'] # 0.0 - found label 'a' in index
df.B.loc[0] # TypeError - string index queried by integer value
df.B.loc['0'] # KeyError - no label '0' in index
df.B.loc['a'] # 0.0 - found label 'a' in index
df.B.iloc[0] # 0.0 - position based query for row 0
df.B.iloc['0'] # TypeError - string can't be used for position
df.B.iloc['a'] # TypeError - string can't be used for position
With example from referenced article:
以参考文章中的示例为例：
d = [0.0, 1.0, 2.0]
e = pd.Series(d, index = [4, 5, 6])
df = pd.DataFrame({'A': 1., 'B': e, 'C': pd.Timestamp('20130102')})
df.B[0] # KeyError - label 0 not in index
df.B['0'] # KeyError - label '0' not in index
df.B.loc[0] # KeyError - label 0 not in index
df.B.loc['0'] # KeyError - label '0' not in index
df.B.iloc[0] # 0.0 - position based query for row 0
df.B.iloc['0'] # TypeError - string can't be used for position
回答by xyzjayne
df.Breturns a pandas series which is why you can do positional indexing. If you select column B as a dataframe this will throw an error:
df.B返回一个Pandas系列，这就是您可以进行位置索引的原因。如果您选择 B 列作为数据框，这将引发错误：
df[['B']][0]
回答by NiGiord
df.Bis actually a pandas.Seriesobject (a shortcut for df['B']), which can be iterated. df.B[0]is no longer a "row" but just the first element of df.B, since by writing df.Byou basically create a 1-D object.
df.B实际上是一个pandas.Series对象（ 的快捷方式df['B']），可以迭代。df.B[0]不再是“行”而只是 的第一个元素df.B，因为通过编写df.B您基本上创建了一个一维对象。
More information in the data structure documentation
数据结构文档中的更多信息
You can treat a DataFrame semantically like a dict of like-indexed Series objects.
您可以在语义上将 DataFrame 视为类似索引的 Series 对象的字典。
