python：如何对两个不同键的复杂列表进行排序

Question

我有一个按以下方式构建的奇怪列表：

[[name_d, 5], [name_e, 10], [name_a, 5]] 

我想先按数字（desc）对其进行排序，然后，如果数字相同，则按名称（asc）。所以我想要的结果是：

[[name_e, 10], [name_a, 5], [name_d, 5]]

我试着想出一个可以在 sort 方法中使用的 lambda 函数，但我不确定我能不能做到。

Answer 1

python 中的排序函数允许将函数作为排序键传递：

l = [[name_d, 5], [name_e, 10], [name_a, 5]]
# copy
l_sorted = sorted(l, key=lambda x: (x[1] * -1, x[0]))
# in place
l.sort(key=lambda x: (x[1] * -1, x[0]))

编辑：
1.排序顺序
2. 演示复制和就地排序

Answer 2

您可以对列表进行两次排序以获得结果，只需颠倒顺序即可：

import operator

l = [[name_d, 5], [name_e, 10], [name_a, 5]]

l.sort(operator.itemgetter(1))
l.sort(operator.itemgetter(0), reverse=True)

然后你会得到预期的排序列表。

Answer 3

    这是我精心设计的（用于解决相同类型的问题）。我只用我安装的最新版本的 Python (OS X) 检查了它。下面的导入部分是（命名笨拙的）排序键：sortKeyWithTwoListOrders 和 sortKeyWith2ndThen1stListValue

---

#Tested under Python 2.7.1 & Python 3.2.3:

import random # Just to shuffle for demo purposes

# Our two lists to sort
firstCol=['abc','ghi','jkl','mno','bcd','hjk']
secondCol=[5,4,2,1]

# Build 2 dimensional list [[firstCol,secondCol]...]
myList = []
for firstInd in range(0, len(firstCol)):
  for secondInd in range(0, len(secondCol)):
    myList = myList + [[firstCol[firstInd],secondCol[secondInd]]]

random.shuffle(myList)

print ("myList (shuffled):")
for i in range(0,len(myList)):
  print (myList[i])

def sortKeyWithTwoListOrders(item):
  return secondCol.index(item[1]), firstCol.index(item[0])

myList.sort(key=sortKeyWithTwoListOrders)
print ("myList (sorted according to strict list order, second column then first column):")
for i in range(0,len(myList)):
  print (myList[i])

random.shuffle(myList)

print ("myList (shuffled again):")
for i in range(0,len(myList)):
  print (myList[i])

def sortKeyWith2ndThen1stListValue(item):
  return item[1], item[0]

myList.sort(key=sortKeyWith2ndThen1stListValue)
print ("myList (sorted according to *values*, second column then first column):")
for i in range(0,len(myList)):
  print (myList[i])

---

myList (shuffled):
['ghi', 5]
['abc', 2]
['abc', 1]
['abc', 4]
['hjk', 5]
['bcd', 4]
['jkl', 5]
['jkl', 2]
['bcd', 1]
['ghi', 1]
['mno', 5]
['ghi', 2]
['hjk', 2]
['jkl', 4]
['mno', 4]
['bcd', 2]
['bcd', 5]
['ghi', 4]
['hjk', 4]
['mno', 2]
['abc', 5]
['mno', 1]
['hjk', 1]
['jkl', 1]
myList (sorted according to strict list order, second column then first column):
['abc', 5]
['ghi', 5]
['jkl', 5]
['mno', 5]
['bcd', 5]
['hjk', 5]
['abc', 4]
['ghi', 4]
['jkl', 4]
['mno', 4]
['bcd', 4]
['hjk', 4]
['abc', 2]
['ghi', 2]
['jkl', 2]
['mno', 2]
['bcd', 2]
['hjk', 2]
['abc', 1]
['ghi', 1]
['jkl', 1]
['mno', 1]
['bcd', 1]
['hjk', 1]
myList (shuffled again):
['hjk', 4]
['ghi', 1]
['abc', 5]
['bcd', 5]
['ghi', 4]
['mno', 1]
['jkl', 1]
['abc', 1]
['hjk', 1]
['jkl', 2]
['hjk', 5]
['mno', 2]
['jkl', 4]
['ghi', 5]
['bcd', 1]
['bcd', 2]
['jkl', 5]
['abc', 2]
['hjk', 2]
['abc', 4]
['mno', 4]
['mno', 5]
['bcd', 4]
['ghi', 2]
myList (sorted according to *values*, second column then first column):
['abc', 1]
['bcd', 1]
['ghi', 1]
['hjk', 1]
['jkl', 1]
['mno', 1]
['abc', 2]
['bcd', 2]
['ghi', 2]
['hjk', 2]
['jkl', 2]
['mno', 2]
['abc', 4]
['bcd', 4]
['ghi', 4]
['hjk', 4]
['jkl', 4]
['mno', 4]
['abc', 5]
['bcd', 5]
['ghi', 5]
['hjk', 5]
['jkl', 5]
['mno', 5]

Answer 4

让我们给你的列表起个名字 - “arr”

arr=sort(key=lambda x:x[1],reverse=True)

输出 - [[name_e,10],[name_d,5],[name_a,5]]

Answer 5

它不需要是你传递给sort方法的lambda函数，你实际上可以提供一个真正的函数，因为它们是python中的一等对象。

L.sort(my_comparison_function)

应该没问题