Python 按位置选择熊猫列
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Selecting pandas column by location
提问by Jason Strimpel
I'm simply trying to access named pandas columns by an integer.
我只是想通过一个整数访问命名的熊猫列。
You can select a row by location using df.ix[3].
您可以使用 来按位置选择一行df.ix[3]。
But how to select a column by integer?
但是如何按整数选择列？
My dataframe:
我的数据框：
df=pandas.DataFrame({'a':np.random.rand(5), 'b':np.random.rand(5)})
采纳答案by DSM
Two approaches that come to mind:
想到的两种方法：
>>> df
A B C D
0 0.424634 1.716633 0.282734 2.086944
1 -1.325816 2.056277 2.583704 -0.776403
2 1.457809 -0.407279 -1.560583 -1.316246
3 -0.757134 -1.321025 1.325853 -2.513373
4 1.366180 -1.265185 -2.184617 0.881514
>>> df.iloc[:, 2]
0 0.282734
1 2.583704
2 -1.560583
3 1.325853
4 -2.184617
Name: C
>>> df[df.columns[2]]
0 0.282734
1 2.583704
2 -1.560583
3 1.325853
4 -2.184617
Name: C
Edit: The original answer suggested the use of df.ix[:,2]but this function is now deprecated. Users should switch to df.iloc[:,2].
编辑：原始答案建议使用，df.ix[:,2]但现在不推荐使用此功能。用户应切换到df.iloc[:,2].
回答by Adrian
You can also use df.icol(n)to access a column by integer.
您还可以使用df.icol(n)整数访问列。
Update: icolis deprecated and the same functionality can be achieved by:
更新：icol已弃用，可以通过以下方式实现相同的功能：
df.iloc[:, n] # to access the column at the nth position
回答by Stefano Fedele
The method .transpose() converts columns to rows and rows to column, hence you could even write
方法 .transpose() 将列转换为行，将行转换为列，因此您甚至可以编写
df.transpose().ix[3]
回答by Surya
You could use label based using .loc or index based using .iloc method to do column-slicing including column ranges:
您可以使用基于标签的 .loc 或基于索引的使用 .iloc 方法进行列切片，包括列范围：
In [50]: import pandas as pd
In [51]: import numpy as np
In [52]: df = pd.DataFrame(np.random.rand(4,4), columns = list('abcd'))
In [53]: df
Out[53]:
a b c d
0 0.806811 0.187630 0.978159 0.317261
1 0.738792 0.862661 0.580592 0.010177
2 0.224633 0.342579 0.214512 0.375147
3 0.875262 0.151867 0.071244 0.893735
In [54]: df.loc[:, ["a", "b", "d"]] ### Selective columns based slicing
Out[54]:
a b d
0 0.806811 0.187630 0.317261
1 0.738792 0.862661 0.010177
2 0.224633 0.342579 0.375147
3 0.875262 0.151867 0.893735
In [55]: df.loc[:, "a":"c"] ### Selective label based column ranges slicing
Out[55]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244
In [56]: df.iloc[:, 0:3] ### Selective index based column ranges slicing
Out[56]:
a b c
0 0.806811 0.187630 0.978159
1 0.738792 0.862661 0.580592
2 0.224633 0.342579 0.214512
3 0.875262 0.151867 0.071244
回答by Safwan
You can access multiple columns by passing a list of column indices to dataFrame.ix.
您可以通过将列索引列表传递给 dataFrame.ix 来访问多个列。
For example:
例如：
>>> df = pandas.DataFrame({
'a': np.random.rand(5),
'b': np.random.rand(5),
'c': np.random.rand(5),
'd': np.random.rand(5)
})
>>> df
a b c d
0 0.705718 0.414073 0.007040 0.889579
1 0.198005 0.520747 0.827818 0.366271
2 0.974552 0.667484 0.056246 0.524306
3 0.512126 0.775926 0.837896 0.955200
4 0.793203 0.686405 0.401596 0.544421
>>> df.ix[:,[1,3]]
b d
0 0.414073 0.889579
1 0.520747 0.366271
2 0.667484 0.524306
3 0.775926 0.955200
4 0.686405 0.544421
