python: how to sort a complex list on two different keys

Question

I have a weird list built in the following way:

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

and I want to sort it first by the number (desc) and then, if the number is the same, by the name (asc). So the result I would like to have is:

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

I tried to think to a lambda function that I can use in the sort method, but I'm not sure I can do it.

Answer 1

You may sort the list twice to get the result, just reverse the order:

import operator

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

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

Then you will get the sorted list as expected.

Answer 2

    Here's something I whipped up (to solve the same type of problem). I've only checked it with my latest versions of installed Python (OS X). The import parts below are the (clunkily-named) sort keys: sortKeyWithTwoListOrders and 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 3

Let's give your list a name - "arr"

arr.sort(key=lambda x:(x[0],-x[1]),reverse=True)

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

Answer 4

It doesn't need to be a lambda function you pass into the sort method, you can actually provide a real function since they are first-class objects in python.

L.sort(my_comparison_function)

Should work just fine