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]]

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.

Does this answer your question? [sort Python list with two keys but only one in reverse order](https://stackoverflow.com/questions/37693373/sort-python-list-with-two-keys-but-only-one-in-reverse-order) — user202729, Aug 14 '21 at 13:24
I do not understand why mine has been marked as duplicate when the other question is 5 years younger... — Giovanni Di Milia, Jul 14 '23 at 16:33
The age has nothing to do with it. They dont want duplicates — Rohit Gupta, Jul 18 '23 at 05:03

score 40 · Accepted Answer · edited Mar 02 '19 at 09:27

40

Sort functions in python allow to pass a function as sort key:

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]))

Edits:
1. Sort order
2. Demonstrate copy and in place sorting

edited Mar 02 '19 at 09:27

dzimiks

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answered Oct 20 '10 at 16:04

tback

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he needs descending sorting for the number – SilentGhost Oct 20 '10 at 16:07
this doesn't work in the way I want, because it sorts also the name in desc (or asc) mode – Giovanni Di Milia Oct 20 '10 at 16:09
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@Giovanni: it's not a black box. Do you see how it is solved? and you cannot modify it to suit your problem? – SilentGhost Oct 20 '10 at 16:10
Corrected the sort order. Like SilentGhost said already - it should have been obvious how to change the order. – tback Oct 20 '10 at 16:15
1

btw: if you don't want to copy the listL `l.sort(key=lambda...)` – tback Oct 20 '10 at 16:15
@SilentGhost, Sorry, I didn't know that I could invert the sorting with the "-" – Giovanni Di Milia Oct 20 '10 at 16:18
1

It solves my problem. But I don't understand what x[1] * -1 means. An explanation would be better. – arshovon Jul 10 '16 at 07:25

score 2 · Answer 2 · answered Nov 29 '16 at 07:06

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.

Mayank Maheshwari · Answer 3 · 2022-05-02T18:19:18.567

0

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]]

edited May 02 '22 at 18:19

answered Sep 06 '21 at 08:40

Mayank Maheshwari

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score -1 · Answer 4 · answered May 21 '12 at 16:21

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]

score -2 · Answer 5 · answered Oct 20 '10 at 16:04

-2

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

answered Oct 20 '10 at 16:04

Daniel DiPaolo

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-1 Comparison function goes away in Python 3, whereas the other answer is future proof. – Steven Rumbalski Oct 20 '10 at 16:12
I thought about this, but how can I write a comparison function that works on two different keys? basically I need a f((x[0],x[1]), (y[0],y[1])) – Giovanni Di Milia Oct 20 '10 at 16:14
@Steven: what are you talking about? this answer might be useless, but not for the reasons you state. [Read the docs](http://docs.python.org/py3k/library/stdtypes.html#mutable-sequence-types) – SilentGhost Oct 20 '10 at 16:14
1

@SilentGhost Maybe I'm misreading, but when I look at the linked docs i see that sort accepts a /key/ function in python 3, but not a /comparison/ function. – Steven Rumbalski Oct 20 '10 at 17:24
@Steven: and what is key function if not a comparison function? – SilentGhost Oct 20 '10 at 17:26
1

@SilentGhost A comparison function is called on pairs of items and "should return a negative integer if self < other, zero if self == other, a positive integer if self > other." Whereas a key function is called once for each item and returns the value(s) you want an item sorted on. See http://wiki.python.org/moin/HowTo/Sorting/ paying attention specifically to sections titled _Key_Functions_ and _The_Old_Way_Using_the_cmp_Parameter_. The key parameter was added in Python 2.4. The cmp parameter was removed in Python 3. – Steven Rumbalski Oct 20 '10 at 18:05
@StevenRumbalski You can pass a function as a key. Doesn't need to be a cmp function. – Apr 17 '19 at 20:02
@Steve: Read the exchange. I'm talking about key functions the whole time. – Steven Rumbalski Apr 17 '19 at 22:41
@StevenRumbalski You are most definitely not talking about key functions the entire time. – Apr 18 '19 at 13:56

How to sort a complex list on two different keys

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