pandas 尝试访问数据框列显示“<bound method NDFrame.xxx...”
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Attempt to access dataframe column displays "<bound method NDFrame.xxx..."
提问by Andrew D
I create DataFrame object in Jupyter notebook:
我在 Jupyter notebook 中创建了 DataFrame 对象：
data = {'state':['Ohio','Ohio','Ohio','Nevada','Nevada'],
'year':[2000, 2001, 2002, 2000, 2001],
'pop':[1.5, 2.0, 3.6, 2.4, 2.9]}
frame = DataFrame(data)
When I'm extracting column 'year', it's ok:
当我提取列“年份”时，没关系：
In [30]: frame.year
Out[30]: 0 2000
1 2001
2 2002
3 2000
4 2001
Name: year, dtype: int64
But when I'm extracting column 'pop' (frame.pop), result is:
但是当我提取列“pop”（frame.pop）时，结果是：
Out[31]:
<bound method NDFrame.pop of pop state year
0 1.5 Ohio 2000
1 2.0 Ohio 2001
2 3.6 Ohio 2002
3 2.4 Nevada 2000
4 2.9 Nevada 2001>
Why the result is not the same as for "frame.year"?
为什么结果与“frame.year”不同？
采纳答案by cs95
popis a dataframe level function. The takeaway here is try to avoid using the .accessor to access columns. There are many dataframe attributes and functions that might clash with your column names, in which case the .will invoke those instead of your columns.
pop是数据帧级别的函数。这里的要点是尽量避免使用.访问器来访问列。有许多数据框属性和函数可能与您的列名发生冲突，在这种情况下，.将调用这些而不是您的列。
You want to use the [..]dict accessor instead:
您想改用[..]dict 访问器：
frame['pop']
0 1.5
1 2.0
2 3.6
3 2.4
4 2.9
Name: pop, dtype: float64
If you want to use pop, you can. This is how:
如果你想使用pop，你可以。这是如何：
frame.pop('pop')
0 1.5
1 2.0
2 3.6
3 2.4
4 2.9
Name: pop, dtype: float64
frame
state year
0 Ohio 2000
1 Ohio 2001
2 Ohio 2002
3 Nevada 2000
4 Nevada 2001
Do note that this modifies the original dataframe, so don't do it unless you're trying to remove columns.
请注意，这会修改原始数据框，因此除非您尝试删除列，否则不要这样做。
回答by Daniel Severo
popis a method of pandas.DataFrame. You need to use frame['pop']
pop是一种方法pandas.DataFrame。你需要使用frame['pop']
回答by YOBEN_S
The way I am using ...eval
我使用的方式...eval
frame.eval('pop')
Out[108]:
0 1.5
1 2.0
2 3.6
3 2.4
4 2.9
dtype: float64
