如何对字符串和 int 值数组、按字母顺序和 "reversed"-字母顺序的多个属性进行排序
How to sort array of string and int values, multiple attributes with alphabetical and "reversed"-alphabetical order
我有一个包含 [name, surname, int1, int2]
个元素的数组,我需要按以下顺序对它进行排序:
按int1
(递减)。
如果int1
相同,按name
按"reversed"字母顺序排序。
如果name
相同,按surname
字母顺序排列。
所以我有这个:
print(sorted(a, key = lambda x: [-int(x[2]), x[0], x[1]]))
而且我不知道如何按逆字母顺序对 x[0]
进行排序
-x[0], x[0][::-1]
对我不起作用。
示例:
[('Petia', 'Anja', 3, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Katia', 3, 0),
('Kolia', 'Alexey', 10, 0),
('Yana', 'Anja', 10, 0)]
至
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
您可以创建一个 class 实现 <
(<
是 CPythons sorted
所需要的 - 如果您正在使用另一个 Python 实现,您可能需要额外的比较运算符)。这允许完全控制 "ordering"。例如:
class Sorter(object):
def __init__(self, tup):
self.name, self.surname, self.int1, self.int2 = tup
def __lt__(self, other):
# Just to make the logic clearer, in practise you could do nest the ifs
# to avoid computing self.int1 == other.int1 twice
if self.int1 == other.int1 and self.name == other.name:
return self.surname < other.surname
elif self.int1 == other.int1:
return self.name > other.name
else:
return self.int1 > other.int1
然后将其用作 key
for sorted
:
>>> sorted(a, key=Sorter)
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
>>> intab='abcdefghijklmnopqrstuvwxyz'
>>> tab = string.maketrans(intab+intab.upper(), intab[::-1]+intab.upper()[::-1])
>>>
>>> slst = sorted(lst, key = lambda x: [-int(x[2]), x[0].translate(tab), x[1]])
>>> pprint(slst)
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
>>>
我有一个包含 [name, surname, int1, int2]
个元素的数组,我需要按以下顺序对它进行排序:
按
int1
(递减)。如果
int1
相同,按name
按"reversed"字母顺序排序。如果
name
相同,按surname
字母顺序排列。
所以我有这个:
print(sorted(a, key = lambda x: [-int(x[2]), x[0], x[1]]))
而且我不知道如何按逆字母顺序对 x[0]
进行排序
-x[0], x[0][::-1]
对我不起作用。
示例:
[('Petia', 'Anja', 3, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Katia', 3, 0),
('Kolia', 'Alexey', 10, 0),
('Yana', 'Anja', 10, 0)]
至
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
您可以创建一个 class 实现 <
(<
是 CPythons sorted
所需要的 - 如果您正在使用另一个 Python 实现,您可能需要额外的比较运算符)。这允许完全控制 "ordering"。例如:
class Sorter(object):
def __init__(self, tup):
self.name, self.surname, self.int1, self.int2 = tup
def __lt__(self, other):
# Just to make the logic clearer, in practise you could do nest the ifs
# to avoid computing self.int1 == other.int1 twice
if self.int1 == other.int1 and self.name == other.name:
return self.surname < other.surname
elif self.int1 == other.int1:
return self.name > other.name
else:
return self.int1 > other.int1
然后将其用作 key
for sorted
:
>>> sorted(a, key=Sorter)
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
>>> intab='abcdefghijklmnopqrstuvwxyz'
>>> tab = string.maketrans(intab+intab.upper(), intab[::-1]+intab.upper()[::-1])
>>>
>>> slst = sorted(lst, key = lambda x: [-int(x[2]), x[0].translate(tab), x[1]])
>>> pprint(slst)
[('Yana', 'Anja', 10, 0),
('Kolia', 'Alexey', 10, 0),
('Vasia', 'Katia', 3, 0),
('Petia', 'Anja', 3, 0),
('Petia', 'Katia', 3, 0)]
>>>