python，按每个元素的子字符串的键对列表进行排序
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python, sorting a list by a key that's a substring of each element
提问by user33061
Part of a programme builds this list,
程序的一部分建立了这个列表，
[u'1 x Affinity for war', u'1 x Intellect', u'2 x Charisma', u'2 x Perception', u'3 x Population growth', u'4 x Affinity for the land', u'5 x Morale']
I'm currently trying to sort it alphabetically by the name of the evolution rather than by the number. Is there any way I can do this without just changing the order the two things appear in the list (as in 'intellect x 1)?
我目前正在尝试按进化的名称而不是数字的字母顺序对其进行排序。有什么方法可以做到这一点，而无需更改两件事出现在列表中的顺序（如在“智力 x 1”中）？
回答by S.Lott
You have to get the "key" from the string.
您必须从字符串中获取“密钥”。
def myKeyFunc( aString ):
stuff, x, label = aString.partition(' x ')
return label
aList.sort( key= myKeyFunc )
回答by dF.
How about:
怎么样：
lst.sort(key=lamdba s: s.split(' x ')[1])
回答by pboucher
Not knowing if your items are standardized at 1 digit, 1 space, 1 'x', 1 space, multiple words I wrote this up:
不知道您的项目是否标准化为 1 位数字、1 个空格、1 个“x”、1 个空格、多个单词，我写了这个：
mylist = [u'1 x Affinity for war', u'1 x Intellect', u'2 x Charisma', u'2 x Perception', u'3 x Population growth', u'4 x Affinity for the land', u'5 x Morale']
def sort(a, b):
return cmp(" ".join(a.split()[2:]), " ".join(b.split()[2:]))
mylist.sort(sort)
You can edit the parsing inside the sortmethod but you probably get the idea.
您可以编辑sort方法内部的解析，但您可能已经明白了。
Cheers, Patrick
干杯，帕特里克
回答by Justin Standard
To do so, you need to implement a custom compare:
为此，您需要实现自定义比较：
def myCompare(x, y):
x_name = " ".join(x.split()[2:])
y_name = " ".join(y.split()[2:])
return cmp(x_name, y_name)
Then you use that compare definition as the input to your sort function:
然后您使用该比较定义作为排序函数的输入：
myList.sort(myCompare)
回答by Teifion
As you are trying to sort what is essentially custom data, I'd go with a custom sort.
当您尝试对本质上是自定义数据的内容进行排序时，我会使用自定义排序。
