sort list based on dictionary values even with missing keys

March 2, 2023

I am trying to sort a list based on dictionary values. The only problem is that if a list contains something that doesn’t exist in the dictionary it wont sort it.

here is what I have

d = {
    "hello0" : 0,
    "hello1" : 1,
    "hello2" : 2,
    "hello3" : 3
          
}


l1 = ["hello1","hello3","hello0","hello2"] #list 1
l2 = ["hello4","hello1","hello3","hello0","hello2"] #list 2



try:
    sorted_l1 = sorted(l1,key=lambda x : d[x])
    sorted_l2 = sorted(l2,key=lambda x : d[x])
    #print(sorted_l1)
    print(sorted_l2)
except Exception as e:      
    print(f"Keyerror {e}")

List 1 will get sorted just fine because everything in that list exists in the dictionary but list 2 will not get sorted and will get a keyerror.

How do I sort list 2 so that the missing key gets added to the end of the sorted list 2 ?
Do I use {}.get ? or is there some another way ?

>Solution :

max_value_dict = max(d.values())
sorted_l2 = sorted(l2,key=lambda x : d.get(x, max_value_dict + 1))

Whenever x is not a key in d, the value associated to x by d.get in the sort is strictly greater than every value in d. Thus x is placed at the end.