python:如何在两个不同的键上对复杂列表进行排序

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:
一。排序顺序

这是我(为了解决同一类问题)想出的办法。我只用我最新安装的Python(OS X)版本检查过它。下面的导入部分是(笨拙地命名)排序键: sortKeyWithTwoListOrders排序键 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]

您可以对列表进行两次排序以获得结果,只需颠倒顺序:

import operator

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

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

它不需要是lambda函数 sort 方法,实际上可以提供一个真正的函数,因为它们是python中的一级对象。

L.sort(my_comparison_function)