2017答案- pandas 0.20: .ix已弃用。使用.loc

请参阅文档中的弃用部分

.loc使用基于标签的索引来选择行和列。标签是索引或列的值。使用.loc进行切片包括最后一个元素。

让我们假设我们有一个包含以下列的DataFrame: Foo, bar, quz, ant, cat, sit, dat。

# selects all rows and all columns beginning at 'foo' up to and including 'sat'
df.loc[:, 'foo':'sat']
# foo bar quz ant cat sat

.loc接受与Python列表对行和列所做的相同的切片符号。切片符号为start:stop:step

# slice from 'foo' to 'cat' by every 2nd column
df.loc[:, 'foo':'cat':2]
# foo quz cat

# slice from the beginning to 'bar'
df.loc[:, :'bar']
# foo bar

# slice from 'quz' to the end by 3
df.loc[:, 'quz'::3]
# quz sat

# attempt from 'sat' to 'bar'
df.loc[:, 'sat':'bar']
# no columns returned

# slice from 'sat' to 'bar'
df.loc[:, 'sat':'bar':-1]
sat cat ant quz bar

# slice notation is syntatic sugar for the slice function
# slice from 'quz' to the end by 2 with slice function
df.loc[:, slice('quz',None, 2)]
# quz cat dat

# select specific columns with a list
# select columns foo, bar and dat
df.loc[:, ['foo','bar','dat']]
# foo bar dat

您可以按行和列进行切片。例如，如果你有5行，标签是v w x y z

# slice from 'w' to 'y' and 'foo' to 'ant' by 3
df.loc['w':'y', 'foo':'ant':3]
#    foo ant
# w
# x
# y

你可以通过引用列表中每一列的名称来切片DataFrame的列，如下所示:

data = pandas.DataFrame(np.random.rand(10,5), columns = list('abcde'))
data_ab = data[list('ab')]
data_cde = data[list('cde')]

其等效

 >>> print(df2.loc[140:160,['Relevance','Title']])
 >>> print(df2.ix[140:160,[3,7]])

如果你来这里寻找两个范围的列切片并将它们组合在一起(像我一样)，你可以做这样的事情

op = df[list(df.columns[0:899]) + list(df.columns[3593:])]
print op

这将创建一个具有前900列和(所有)列> 3593的新数据框架(假设您的数据集中有4000列)。

下面介绍如何使用不同的方法进行选择性列切片，包括基于选择标签的、基于索引的和基于选择范围的列切片。

In [37]: import pandas as pd    
In [38]: import numpy as np
In [43]: df = pd.DataFrame(np.random.rand(4,7), columns = list('abcdefg'))

In [44]: df
Out[44]: 
          a         b         c         d         e         f         g
0  0.409038  0.745497  0.890767  0.945890  0.014655  0.458070  0.786633
1  0.570642  0.181552  0.794599  0.036340  0.907011  0.655237  0.735268
2  0.568440  0.501638  0.186635  0.441445  0.703312  0.187447  0.604305
3  0.679125  0.642817  0.697628  0.391686  0.698381  0.936899  0.101806

In [45]: df.loc[:, ["a", "b", "c"]] ## label based selective column slicing 
Out[45]: 
          a         b         c
0  0.409038  0.745497  0.890767
1  0.570642  0.181552  0.794599
2  0.568440  0.501638  0.186635
3  0.679125  0.642817  0.697628

In [46]: df.loc[:, "a":"c"] ## label based column ranges slicing 
Out[46]: 
          a         b         c
0  0.409038  0.745497  0.890767
1  0.570642  0.181552  0.794599
2  0.568440  0.501638  0.186635
3  0.679125  0.642817  0.697628

In [47]: df.iloc[:, 0:3] ## index based column ranges slicing 
Out[47]: 
          a         b         c
0  0.409038  0.745497  0.890767
1  0.570642  0.181552  0.794599
2  0.568440  0.501638  0.186635
3  0.679125  0.642817  0.697628

### with 2 different column ranges, index based slicing: 
In [49]: df[df.columns[0:1].tolist() + df.columns[1:3].tolist()]
Out[49]: 
          a         b         c
0  0.409038  0.745497  0.890767
1  0.570642  0.181552  0.794599
2  0.568440  0.501638  0.186635
3  0.679125  0.642817  0.697628

if数据帧是这样的:

group         name      count
fruit         apple     90
fruit         banana    150
fruit         orange    130
vegetable     broccoli  80
vegetable     kale      70
vegetable     lettuce   125

OUTPUT可以是

   group    name  count
0  fruit   apple     90
1  fruit  banana    150
2  fruit  orange    130

如果使用逻辑运算符np.logical_not

df[np.logical_not(df['group'] == 'vegetable')]

更多关于

https://docs.scipy.org/doc/numpy-1.13.0/reference/routines.logic.html

其他逻辑运算符

logical_and(x1, x2， /[， out, where，…])计算的真值 x1和x2元素。 Logical_or (x1, x2， /[， out, where, casting， 计算x1或x2元素的真值。 logical_not(x， /[， out, where, casting，…])计算真值 NOT x元素的值。 logical_xor(x1, x2， /[， out, where， ..])按元素计算x1 XOR x2的真值。