How to sort one list based on the value of another list in Python?

We can sort a list by the values ​​in another list by setting the second list to the index number of the value in the first list in sorted order .

Sort a list by values ​​in another list

Example

In this example, we will sort the list based on the values ​​in another list, i.e. the indexes of the second list will be in the order they are placed in the sorted order -

# Two Lists
list1 = ['BMW', 'Toyota', 'Audi', 'Tesla', 'Hyundai']
list2 = [2, 5, 1, 4, 3]

print("List1 = \n",list1)
print("List2 (indexes) = \n",list2)

# Sorting the List1 based on List2
res = [val for (_, val) in sorted(zip(list2, list1), key=lambda x: x[0])]
print("\nSorted List = ",res)

Output

List1 = 
 ['BMW', 'Toyota', 'Audi', 'Tesla', 'Hyundai']
List2 (indexes) = 
 [2, 5, 1, 4, 3]

Sorted List =  ['Audi', 'BMW', 'Hyundai', 'Tesla', 'Toyota']

Use a dictionary to sort a list by values ​​in another list

Example

In this example, we will sort one list by the value of another list. We add these two lists to a dictionary and then sort them using the key values ​​and lambdas in the dictionary.

# Two Lists
list1 = ['BMW', 'Toyota', 'Audi', 'Tesla', 'Hyundai']
list2 = [2, 5, 1, 4, 3]

print("List1 = \n",list1)
print("List2 (indexes) = \n",list2)

# Blank dictionary
newDict = {}

# Blank list
outList = []

# Both the lists in our dictionary newDict
newDict = {list1[i]: list2[i] for i in range(len(list2))}

# Sorting using the sorting() based on key and values
# Using lambda as well
s = {k: v for k, v in sorted(newDict.items(), key=lambda item: item[1])}

# Element addition in the list
for i in s.keys():
   outList.append(i)

print("\nSorted List = \n",outList)

Output

List1 = 
 ['BMW', 'Toyota', 'Audi', 'Tesla', 'Hyundai']
List2 (indexes) = 
 [2, 5, 1, 4, 3]

Sorted List = 
['Audi', 'BMW', 'Hyundai', 'Tesla', 'Toyota']

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