Question

我有一个像这样的列表

[['N_ALA_A0001', [9.25, 24.41, 64.13]]]   
[['O_ALA_A0001', [9.05, 21.51, 64.38]]]   
[['N_ALA_A0001', [9.25, 24.41, 64.13]]]   
[['N_TYR_A0002', [7.84, 21.93, 66.24]]]   
[['N_ALA_A0001', [9.25, 24.41, 64.13]]]   
[['O_TYR_A0002', [5.48, 20.92, 65.6]]]   
[['N_ALA_A0001', [9.25, 24.41, 64.13]]]   
[['N_VAL_A0003', [6.19, 18.75, 65.89]]]   
[['N_ALA_A0001', [9.25, 24.41, 64.13]]]

我想删除冗余列表并打印uniq行

[['N_ALA_A0001', [9.25, 24.41, 64.13]]]   
[['O_ALA_A0001', [9.05, 21.51, 64.38]]]   
[['N_TYR_A0002', [7.84, 21.93, 66.24]]]   
[['O_TYR_A0002', [5.48, 20.92, 65.6]]]   
[['N_VAL_A0003', [6.19, 18.75, 65.89]]]

我尝试过，strip，set，uniq但这不起作用;它给出了错误：

AttributeError：＆＃39; list＆＃39;对象没有属性＆＃39; readline＆＃39;

print set(uniqlist)

TypeError：不可用类型：＆＃39; list＆＃39;

任何帮助都将不胜感激。

Answer 1

试试这个

the_list = [['N_ALA_A0001', [9.25, 24.41, 64.13]],   
        ['O_ALA_A0001', [9.05, 21.51, 64.38]],  
        ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
        ['N_TYR_A0002', [7.84, 21.93, 66.24]],   
        ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
        ['O_TYR_A0002', [5.48, 20.92, 65.6]],   
        ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
        ['N_VAL_A0003', [6.19, 18.75, 65.89]],   
        ['N_ALA_A0001', [9.25, 24.41, 64.13]]]

print map(eval, set(map(str, the_list)))

set(the_list)无效，因为列表不可用。使用str将其转换为可哈希类型的字符串。

Answer 2

您可以将所有内部列表转换为元组，然后进行设置，然后将其转换回来：

a = [['N_ALA_A0001', [9.25, 24.41, 64.13]],['O_ALA_A0001', [9.05, 21.51, 64.38]],['N_ALA_A0001', [9.25, 24.41, 64.13]], ['N_TYR_A0002', [7.84, 21.93, 66.24]], ['N_ALA_A0001', [9.25, 24.41, 64.13]], ['O_TYR_A0002', [5.48, 20.92, 65.6]], ['N_ALA_A0001', [9.25, 24.41, 64.13]], ['N_VAL_A0003', [6.19, 18.75, 65.89]], ['N_ALA_A0001', [9.25, 24.41, 64.13]]]
s = set((k, tuple(v)) for k, v in a)
a = [k, list(v) for k, v in s]

如果您想保留订单，可以使用OrderedDict：

from collections import OrderedDict
s = OrderedDict({(k, tuple(v)):1 for k, v in a})
a = [[k, list(v)] for k, v in s.keys()]

请注意，我的解决方案会检查列表的整个元素是否具有唯一性。如果您只想检查第一个元素（N_ALA_A0001, ...），可以使用OrderedDict，如

s = OrderedDict({k: v for k, v in a})
a = [[k, v] for k, v in s.iteritems()]

Answer 3

首先，删除一行嵌套：

x = [['N_ALA_A0001', [9.25, 24.41, 64.13]],   
     ['O_ALA_A0001', [9.05, 21.51, 64.38]],  
     ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
     ['N_TYR_A0002', [7.84, 21.93, 66.24]],   
     ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
     ['O_TYR_A0002', [5.48, 20.92, 65.6]],   
     ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
     ['N_VAL_A0003', [6.19, 18.75, 65.89]],   
     ['N_ALA_A0001', [9.25, 24.41, 64.13]]]

现在，简单的字典理解将为您提供所需的信息：

d = {z:w for z, w in x}

给出了：

{'O_ALA_A0001': [9.05, 21.51, 64.38], 
 'N_ALA_A0001': [9.25, 24.41, 64.13], 
 'N_VAL_A0003': [6.19, 18.75, 65.89], 
 'N_TYR_A0002': [7.84, 21.93, 66.24], 
 'O_TYR_A0002': [5.48, 20.92, 65.6]}

如果需要，可以很容易地将其转换回列表。

正如@ m01指出的那样，如果你想保持这些顺序，可以使用OrderedDict（以及一个小的语法更改）轻松完成：

from collections import OrderedDict
d = OrderedDict(((z, w) for z, w in x))

Answer 4

如果您希望它基于第一个值是唯一的，只需将其转换为字典并返回。

the_list = [['N_ALA_A0001', [9.25, 24.41, 64.13]],   
            ['O_ALA_A0001', [9.05, 21.51, 64.38]],  
            ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
            ['N_TYR_A0002', [7.84, 21.93, 66.24]],   
            ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
            ['O_TYR_A0002', [5.48, 20.92, 65.6]],   
            ['N_ALA_A0001', [9.25, 24.41, 64.13]],   
            ['N_VAL_A0003', [6.19, 18.75, 65.89]],   
            ['N_ALA_A0001', [9.25, 24.41, 64.13]]]

unique_list = dict(the_list).items()

如何在python中获取唯一列表

4 个答案: