python sort dictionary by key=lambda

Sorting dictionary keys in python [duplicate]

stackoverflow.com › questions › 575819 › sorting-dictionary-keys-in-python

I like this one:

sorted(d, key=d.get)

Answer from dF. on Stack Overflow

GeeksforGeeks

geeksforgeeks.org › python › ways-sort-list-dictionaries-values-python-using-lambda-function

Ways to sort list of dictionaries by values in Python - Using lambda function - GeeksforGeeks

November 14, 2025 - ... Explanation: (sorted(dic, key=lambda x: (x['age'], x['name'])): Sorts the list of dictionaries first by "age" in ascending order and if ages are equal then by "name" in alphabetic order.

reddit.com › r/learnpython › sorting with key=lambda

r/learnpython on Reddit: sorting with key=lambda

March 14, 2023 -

pairs= [(1,'one'),(2,'two'),(3,'three'),(4,"four"),(5,"five")]
pairs.sort(key=lambda pair: pair[1])

>>> pairs
[(5, 'five'), (4, 'four'), (1, 'one'), (3, 'three'), (2, 'two')]

why is it sorted like that I didn't understand? Why can't I write a number greater than 1 in pair[1]?

Top answer

1 of 4

6

pair is a tuple consisting of two items. Index 0 is the first item, index 1 is the second item. There is no other item in the tuple for any other index value to be valid.

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Hi, Firstly, its important to know that Python uses a 0-based indexing system. Basically we count 0 as the first number. So to index a list of 5 numbers looks like this: # So for this list lst = [1, 2, 3, 4, 5] # There are 5 elements total, but we start counting from 0. # So the item at index 0 is the first item in the list. print(lst[0]) # First Item print(lst[1]) # Second Item print(lst[2]) # Third Item print(lst[3]) # Fourth Item print(lst[4]) # Fifth Item Secondly, .sort() uses the key function to identify how to sort the list. You can think of it like the sort function looks at the list, and runs the function against it before sorting. So the first input to the lambda is (1, 'one') and the output is the string one This is because of the 0-indexing I mentioned before and the reason a number higher than 1 is not allowed. There are only 2 indexes available, 0 and 1. Due to the fact index 1 is used, sort() is using the strings to sort the items in the list, not the numbers. Which means you get a weird ordering as its not ordering numerically, but alphabetically. To get a numerical ordering, use the pair[0] in the lambda instead of pair[1]

Discussions

How to sort a dictionary where keys have multiple values?

Use a lambda as your sort function.

# lambda r: r[1][2]
# r == ('John Adams', ('111223333', 'A', 91.0))
# r[1] == ('111223333', 'A', 91.0)
# r[1][2] == 91.0
#
d = {
    'John Adams': ('111223333', 'A', 91.0),
    'Willy Smith Jr.': ('222114444', 'C', 77.55),
    'Phil Jordan': ('777886666', 'F', 59.5)
}

for key, value in sorted(d.items(), key=lambda r: r[1][2]):
    print(key, value)

Produces:

('Phil Jordan', ('777886666', 'F', 59.5))
('Willy Smith Jr.', ('222114444', 'C', 77.55))
('John Adams', ('111223333', 'A', 91.0))

By the way, if you end up with multiple entries sharing the same sort value, you can build a compound sort (secondary, tertiary, etc). Your lambda should just return a tuple: lambda r: (r[1][2], r[0]), which would sort by that last value, and then by name if multiple values equal each other.

Here's without lambda:

def key_func(key_value_tuple):
    name = key_value_tuple[0]
    long_str_num, a_to_f, number = key_value_tuple[1]
    return (number, name)

d = {
    'John Adams': ('111223333', 'A', 91.0),
    'Willy Smith Jr.': ('222114444', 'C', 77.55),
    'Phil Jordan': ('777886666', 'F', 59.5)
}

for key, value in sorted(d.items(), key=key_func):
    print(key, value)

Edit: Added non-lambda option.

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