a =
[
[['test', 21, 28], ['test', 61, 68], ['test', 111, 118], ['test', 118, 125]],
[['Columns', 21, 28], ['Columns', 61, 68], ['Columns', 111, 118], ['Columns', 118, 125], ['Columns', 128, 135]],
[['Delete', 14, 20], ['Delete', 47, 53], ['Delete', 54, 60], ['Delete', 78, 84], ['Delete', 95, 101]],
[['rw', 21, 28], ['rw', 61, 68], ['rw', 111, 118]]
]
我想要做的是按照子列表的长度反转来对嵌套列表进行排序
但如果子列表的长度相同,则保持原始顺序.
预期结果：
[
[['Columns', 21, 28], ['Columns', 61, 68], ['Columns', 111, 118], ['Columns', 118, 125], ['Columns', 128, 135]],
[['Delete', 14, 20], ['Delete', 47, 53], ['Delete', 54, 60], ['Delete', 78, 84], ['Delete', 95, 101]],
[['test', 21, 28], ['test', 61, 68], ['test', 111, 118], ['test', 118, 125]],
[['rw', 21, 28], ['rw', 61, 68], ['rw', 111, 118]]
]
我知道我可以使用它按反向长度对列表进行排序,但它不会保持原始顺序：
a = sorted(a, key=len, reverse=True)
(它会颠倒子列表“删除”和“列”)
如何按子列表的反向长度对列表进行排序并保持原始顺序？
或者,摆脱reverse = True并使用返回子列表长度为负数的键函数.
a = [
[['test', 21, 28], ['test', 61, 68], ['test', 111, 118], ['test', 118, 125]],
[['Columns', 21, 28], ['Columns', 61, 68], ['Columns', 111, 118], ['Columns', 118, 125], ['Columns', 128, 135]],
[['Delete', 14, 20], ['Delete', 47, 53], ['Delete', 54, 60], ['Delete', 78, 84], ['Delete', 95, 101]],
[['rw', 21, 28], ['rw', 61, 68], ['rw', 111, 118]]
]
a.sort(key=lambda x: -len(x))
for row in a:
print(row)
输出
[['Columns', 21, 28], ['Columns', 61, 68], ['Columns', 111, 118], ['Columns', 118, 125], ['Columns', 128, 135]]
[['Delete', 14, 20], ['Delete', 47, 53], ['Delete', 54, 60], ['Delete', 78, 84], ['Delete', 95, 101]]
[['test', 21, 28], ['test', 61, 68], ['test', 111, 118], ['test', 118, 125]]
[['rw', 21, 28], ['rw', 61, 68], ['rw', 111, 118]]
如果我用
a.sort(key=len, reverse=True)
我得到与上述相同的输出.如果您也没有得到该输出,则说明很奇怪.请注意,此版本比上面的版本更有效,因为它直接调用len(在C中实现,并且执行快速属性查找以确定对象的长度),并且Python函数调用相对较慢,无论它们是否为def函数.或lambdas.
顺便说一句,当您想就地对列表进行排序时,应始终调用其.sort方法.内置的sorted函数实际上会创建一个新列表,将原始数据复制到该列表,然后对其调用.sort,然后将新列表绑定到目标.因此,避免创建该副本并在原始列表上自行调用.sort会更有效率.