人生苦短
早用Python
这是工作中遇到的一个坑,首先看一下问题
raw_list = [["百度", "CPY"], ["京东", "CPY"], ["黄轩", "PN"], ["百度", "CPY"]]
列表嵌套了列表,并且有一个重复列表["百度", "CPY"]
,现在要求将这个重复元素进行去重(重复是指嵌套的列表内两个元素都相同),并且保证元素顺序不变,输出还是嵌套列表,即最后结果应该长这样:[["百度", "CPY"], ["京东", "CPY"], ["黄轩", "PN"]]
正常Python去重都是使用set
,所以我这边也是用这种思想处理一下
In [8]: new_list = [list(t) for t in set(tuple(_) for _ in raw_list)]
In [9]: new_list
Out[9]: [['京东', 'CPY'], ['百度', 'CPY'], ['黄轩', 'PN']]
=。=以为大功告成,结果发现嵌套列表顺序变了
好吧一步步找一下是从哪边顺序变了的
In [10]: s = set(tuple(_) for _ in raw_list)
In [11]: s
Out[11]: {('京东', 'CPY'), ('百度', 'CPY'), ('黄轩', 'PN')}
恍然大悟关于set
的两个关键词:无序 和 不重复 =。=
所以从set
解决排序问题基本无望了,然而我还没有放弃,现在问题就变成了对于new_list怎么按照raw_list元素顺序排序,当然肯定要通过sort
实现
翻一下Python文档找到以下一段话
文档地址
sort(*, key=None, reverse=False)
This method sorts the list in place, using only < comparisons between
items. Exceptions are not suppressed - if any comparison operations
fail, the entire sort operation will fail (and the list will likely be left in a
partially modified state).
[`sort()`](https://docs.python.org/3/library/stdtypes.html?highlight=sort#list.sort "list.sort")
accepts two arguments that can only be passed by keyword ( [keyword-only arguments](https://docs.python.org/3/glossary.html#keyword-only-parameter) ):
key specifies a function of one argument that is used to extract a
comparison key from each list element (for example, key=str.lower).
The key corresponding to each item in the list is calculated once and then used for the entire sorting process. The default value of None
means that list items are sorted directly without calculating a separate
key value.
开始划重点:
sort方法通过参数key
指定一个方法,换句话说,key参数的值是函数。
这个函数和new_list上的每个元素会产生一个结果,sort
通过这个结果进行排序。
于是这里就想到求出new_list里的每一个元素在raw_list里的索引,根据这个索引进行排序。
代码实现如下:
In [13]: new_list.sort(key=raw_list.index)
In [14]: new_list
Out[14]: [['百度', 'CPY'], ['京东', 'CPY'], ['黄轩', 'PN']]
结果和期望一样 =。=