How to sort with lambda in Python

Use

a = sorted(a, key=lambda x: x.modified, reverse=True)#             ^^^^

On Python 2.x, the sorted function takes its arguments in this order:

sorted(iterable, cmp=None, key=None, reverse=False)

so without the key=, the function you pass in will be considered a cmp function which takes 2 arguments.

lst = [('candy','30','100'), ('apple','10','200'), ('baby','20','300')]lst.sort(key=lambda x:x[1])print(lst)

It will print as following:

[('apple', '10', '200'), ('baby', '20', '300'), ('candy', '30', '100')]

You're trying to use key functions with lambda functions.

Python and other languages like C# or F# use lambda functions.

Also, when it comes to key functions and according to the documentation

Both list.sort() and sorted() have a key parameter tospecify a function to be called on each list element prior to makingcomparisons.

...

The value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes. This technique is fast because the key function is called exactly once for each input record.

So, key functions have a parameter key and it can indeed receive a lambda function.

In Real Python there's a nice example of its usage. Let's say you have the following list

ids = ['id1', 'id100', 'id2', 'id22', 'id3', 'id30']

and want to sort through its "integers". Then, you'd do something like

sorted_ids = sorted(ids, key=lambda x: int(x[2:])) # Integer sort

and printing it would give

['id1', 'id2', 'id3', 'id22', 'id30', 'id100']

In your particular case, you're only missing to write key= before lambda. So, you'd want to use the following

a = sorted(a, key=lambda x: x.modified, reverse=True)