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How to sort a list/tuple of lists/tuples by the element at a given index (11 answers)

Closed 6 years ago.

我有一张图例列表看起来像这样的东西:

[( abc , 121),( abc , 231),( abc , 148), ( abc ,221)]

我想用图普尔斯内部的整数值来按升序排序这个列表。有可能吗?

Answer 1

尝试使用 key key 关键词参数 sorded () , 以默认的递增顺序排序 :

sorted(
    [( abc , 121), ( abc , 231), ( abc , 148), ( abc , 221)], 
    key=lambda x: x[1]
)

key 应该是一个函数,用以识别如何从数据结构中获取可比元素。在您的情况下,它是图普的第二个元素,因此我们访问 [1] 。

关于优化,请参见 jamilak S 使用 < code> operator. ititgetter(1) 的响应,该响应基本上是 lambda x: x[1] 的更快版本。

Answer 2

>>> from operator import itemgetter
>>> data = [( abc , 121),( abc , 231),( abc , 148), ( abc ,221)]
>>> sorted(data,key=itemgetter(1))
[( abc , 121), ( abc , 148), ( abc , 221), ( abc , 231)]

IMO using itemgetter is more readable in this case than the solution by @cheeken. It is also faster since almost all of the computation will be done on the c side (no pun intended) rather than through the use of lambda.

>python -m timeit -s "from operator import itemgetter; data = [( abc , 121),( abc , 231),( abc , 148), ( abc ,221)]" "sorted(data,key=itemgetter(1))"
1000000 loops, best of 3: 1.22 usec per loop

>python -m timeit -s "data = [( abc , 121),( abc , 231),( abc , 148), ( abc ,221)]" "sorted(data,key=lambda x: x[1])"
1000000 loops, best of 3: 1.4 usec per loop

Answer 3

Adding to Cheeken s answer, This is how you sort a list of tuples by the 2nd item in descending order.

sorted([( abc , 121),( abc , 231),( abc , 148), ( abc ,221)],key=lambda x: x[1], reverse=True)

Answer 4

作为匹顿新植物,我只想提一下如果数据真的看起来是这样的:

data = [( abc , 121),( abc , 231),( abc , 148), ( abc ,221)]

然后 sorded () 将自动按图普的第二个元素排序, 因为第一个元素完全相同。

Answer 5

用于某种就地类型,使用

foo = [(list of tuples)]
foo.sort(key=lambda x:x[0]) #To sort by first element of the tuple

Answer 6

Python 维基语:

>>> from operator import itemgetter, attrgetter    
>>> sorted(student_tuples, key=itemgetter(2))
[( dave ,  B , 10), ( jane ,  B , 12), ( john ,  A , 15)]    
>>> sorted(student_objects, key=attrgetter( age ))
[( dave ,  B , 10), ( jane ,  B , 12), ( john ,  A , 15)]

Answer 7

对于羊羔回避方法,首先定义您自己的函数 :

def MyFn(a):
    return a[1]

然后:

sorted([( abc , 121),( abc , 231),( abc , 148), ( abc ,221)], key=MyFn)

Answer 8

将 Python 2.7/code> 改为 Python 2.7/code>,这样可以使被接受的回答略为易读:



sorted([( abc , 121),( abc , 231),( abc , 148), ( abc ,221)], key=lambda (k, val): val)

Answer 9

OP 中的排序值是整数这一事实与问题本身无关。换句话说, 如果排序值是文字, 被接受的答案会有效。我提出这一点还想指出, 在排序中可以修改该分类( 例如, 以计算大小例 ) 。

>>> sorted([(121,  abc ), (231,  def ), (148,  ABC ), (221,  DEF )], key=lambda x: x[1])
[(148,  ABC ), (221,  DEF ), (121,  abc ), (231,  def )]
>>> sorted([(121,  abc ), (231,  def ), (148,  ABC ), (221,  DEF )], key=lambda x: str.lower(x[1]))
[(121,  abc ), (148,  ABC ), (231,  def ), (221,  DEF )]

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