CODEWITHSUNDEEP
pythonlist
HighAllegiant
HighAllegiant

Reputation: 71

Python Sorting Lists in Lists

Given: lst = [['John',3],['Blake',4],['Ted',3]]

Result: lst = [['John',3],['Ted',3],['Blake',4]]

I'm looking for a way to sort lists in lists first numerically then alphabetically without the use of the "itemgetter" syntax.

Upvotes: 0

Views: 1342

Answers (5)

John La Rooy
John La Rooy

Reputation: 304175

In Python2, you can use this

>>> lst = [['John',3],['Blake',4],['Ted',3]]
>>> lst.sort(key=sorted)
>>> lst
[['John', 3], ['Ted', 3], ['Blake', 4]]

This works because ints are always "less than" strings in Python2

You can no longer sort str and int objects in Python3 though

Upvotes: 0

ozdrgnaDiies
ozdrgnaDiies

Reputation: 1929

I think this may have been asked before in the following question:

Sorting a list of lists in Python

Here is the explanation given by Dave Webb:

The key argument to sort specifies a function of one argument that is used to extract a comparison key from each list element. So we can create a simple lambda that returns the last element from each row to be used in the sort:

c2.sort(key = lambda row: row[2])

A lambda is a simple anonymous function. It's handy when you want to create a simple single use function like this. The equivalent code not using a lambda would be:

def sort_key(row):
    return row[2]

c2.sort(key = sort_key)

If you want to sort on more entries, just make the key function return a tuple containing the values you wish to sort on in order of importance. For example:

c2.sort(key = lambda row: (row[2],row[1]))

or:

c2.sort(key = lambda row: (row[2],row[1],row[0]))

Upvotes: 2

juliomalegria
juliomalegria

Reputation: 24921

You could use the argument key from the built-in sorted function.

In this key argument, you pass a function with one parameter, that returns something that will be sorted instead of sorting the list by its elements.

def my_func(elem):
    # return a tuple (second element, first element)
    return (elem[1], elem[0])

>>> lst = [['John',3],['Blake',4],['Ted',3]]
>>> sorted(lst, key=my_func)
[['John', 3], ['Ted', 3], ['Blake', 4]]

Or even shorter:

>>> sorted(lst, key=lambda x: (x[1],x[0]))
[['John', 3], ['Ted', 3], ['Blake', 4]]

In both ways, you sort first numerically, then alphabetically.

Upvotes: 2

Miki Tebeka
Miki Tebeka

Reputation: 13850

What's wrong with itemgetter?

lst.sort(key=lambda l: list(reversed(l)) should do the trick

Upvotes: 1

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798676

Since you insist:

lst.sort(key=lambda x: x[::-1])

Upvotes: 5

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