Python 熊猫重新索引数据框问题
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Pandas reindexing data frame issue
提问by Edgar Aroutiounian
Say I have the following data frame,
假设我有以下数据框，
A B
0 1986-87 232131
1 1987-88 564564
2 1988-89 123125
...
And so on.
等等。
I'm trying to reindex, with <myFrame>.set_index('A'), so that I get
我正在尝试使用 重新索引<myFrame>.set_index('A')，以便我得到
B
1986-87 232131
1987-88 564564
1988-89 123125
but I keep getting this instead:
但我一直得到这个：
B
A
1986-87 232131
1987-88 564564
1988-89 123125
and its annoying as heck cause I tried the other reindexing methods. I'm not sure what the Ais actually representing because it doesn't appear in <myFrame>.columnsor <myFrame>.indexand doing <myFrame>['B'][0]gives me 232131, so what is Ain this reindexed data frame and how can I index correctly from the beginning or get rid of this strange Ain the incorrectly reindex data frame.
这很烦人，因为我尝试了其他重新索引方法。我不确定A实际代表什么，因为它没有出现在<myFrame>.columns或<myFrame>.index正在做<myFrame>['B'][0]给我232131，所以A这个重新索引的数据框中有什么以及我如何从头开始A正确索引或在不正确的重新索引中摆脱这个奇怪的数据框。
采纳答案by Andy Hayden
You need to reset the name/names attribute of the index:
您需要重置索引的 name/names 属性：
df.index.names = [None]
Example:
例子：
In [11]: df = pd.DataFrame([[1, 2], [3, 4]], columns=['A', 'B']).set_index('A')
In [12]: df
Out[12]:
B
A
1 2
3 4
In [13]: df.index.names = [None]
In [14]: df
Out[14]:
B
1 2
3 4
The names describe the index, and give some meaning to the index, and also distinguishes between different levels in the index (in a MultiIndex).
名称描述索引，并赋予索引一些含义，还区分索引中的不同级别（在 MultiIndex 中）。
As @DSM points out, do so at your own peril, this loses info if you want to reset_index back:
正如@DSM 指出的那样，这样做后果自负，如果您想重新设置 reset_index，这会丢失信息：
In [15]: df.reset_index() # col_fill=['A', 'B'])
Out[15]:
index B
0 1 2
1 3 4
However, you can col_fill in the names manually:
但是，您可以手动 col_fill 名称：
In [16]: df.reset_index(col_fill=['A'])
Out[16]:
A B
0 1 2
1 3 4
回答by Marius
I think your main problem is that you need to actually save the result of set_index, or use inplace=True, for the index to be set:
我认为您的主要问题是您需要实际保存set_index或使用的结果inplace=True以设置索引：
# Either
df.set_index('A', inplace=True)
# Or:
# df = df.set_index('A')
The output you were seeing was correct, it was a dataframe indexed by A, but you just hadn't stored it in a variable. Once you have stored it, things should work like you expect:
您看到的输出是正确的，它是一个由 A 索引的数据帧，但您只是没有将其存储在变量中。一旦你存储了它，事情应该像你期望的那样工作：
df.index
Out[6]: Index([u'1986-87', u'1987-88', u'1988-89'], dtype=object)
df.loc[u'1987-88']
Out[8]:
B 564564
Name: 1987-88, dtype: int64
回答by J_yang
I have a dataframe that is generated from appending multiple dataframe together into a long list. As shown in figure, the default index is a loop between 0 ~ 7 because each original df has this index. The total row number is 240. So how can reindex the new df into 0~239 instead of 30 x 0~7.
我有一个数据帧，它是通过将多个数据帧附加到一个长列表中而生成的。如图，默认索引是0~7之间的循环，因为每个原始df都有这个索引。总行数为 240。那么如何将新的 df 重新索引为 0~239 而不是 30 x 0~7。
I tried df.reset_index(drop=True), but it doesn't seem to work. I also tried:df.reindex(np.arange(240))but it returned error
我试过了df.reset_index(drop=True)，但似乎不起作用。我也试过：df.reindex(np.arange(240))但它返回错误
ValueError: cannot reindex from a duplicate axis
