Python: Sorting List of Lists by a String Element

Question

Lets say I have the following Python list:

mylist = [
    ["name_a", "sex_male", "age_15", "grade_11"],
    ["name_b", "sex_male", "age_18", "grade_9"],
    ["name_c", "sex_male", "age_11", "grade_8"],
    ["name_d", "sex_male", "age_16", "grade_12"],
    ["name_e", "sex_male", "age_19", "grade_13"],
]

I want to call Python's sort() function to sort mylist by age

For instance, I want to do something like this:

mylist.sort(key=...)

to get the output

mylist = [
    ["name_c", "sex_male", "age_11", "grade_8"],
    ["name_a", "sex_male", "age_15", "grade_11"],
    ["name_d", "sex_male", "age_16", "grade_12"],
    ["name_b", "sex_male", "age_18", "grade_9"],
    ["name_e", "sex_male", "age_19", "grade_13"],
]

What is the right way to do this?

P.S

One more question: why sorting algorithm is used in Python's sort() function? (ex. QuickSort?)

possible duplicate of [In Python how do I sort a list of dictionaries by values of the dictionary?](http://stackoverflow.com/questions/72899/in-python-how-do-i-sort-a-list-of-dictionaries-by-values-of-the-dictionary) — shx2, Mar 29 '14 at 22:12
write a function to extract the age from one item. Use that as the key function. This is preferable to using a lambda for anything more than simple cases, because that way you can write unittests for that function. — John La Rooy, Mar 29 '14 at 22:16

score 2 · Answer 1 · answered Mar 29 '14 at 22:18

2

def extract_age(item):
    """age is stored in the 3rd element of item. eg.
      ['name_c', 'sex_male', 'age_11', 'grade_8']
      and should be returned as an int
    """
    return int(item[2].split('_')[1])

mylist.sort(key=extract_age)

answered Mar 29 '14 at 22:18

John La Rooy

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+1 Although probably the lambda is more pythonic, [_A Foolish Consistency is the Hobgoblin of Little Minds_](http://legacy.python.org/dev/peps/pep-0008/#a-foolish-consistency-is-the-hobgoblin-of-little-minds) and the age extracting function is very worthwhile testing it. Why don't you include your comment in the answer? Is a good motivation. – Paulo Bu Mar 29 '14 at 22:27

Paulo Bu · Answer 2 · 2014-03-29T22:18:30.610

1

Assuming all ages look the same you can do:

>>>mylist.sort(key=lambda e: int(e[2].split('_')[1]))
[['name_c', 'sex_male', 'age_11', 'grade_8'],
 ['name_a', 'sex_male', 'age_15', 'grade_11'],
 ['name_d', 'sex_male', 'age_16', 'grade_12'],
 ['name_b', 'sex_male', 'age_18', 'grade_9'],
 ['name_e', 'sex_male', 'age_19', 'grade_13']]

The lambda expressions just parses the age field by splitting the string on the _ character and converting to int the result of the split's second element.

Python uses TimSort algorithm, a tailored search algorithm for Python created by Tim Peters.

edited Mar 29 '14 at 22:18

answered Mar 29 '14 at 22:12

Paulo Bu

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Note that OP wants to use `mylist.sort(key=...)` – Christian Tapia Mar 29 '14 at 22:15
1

@Christian right, I'm just used to `sorted`, I'll edit, thanks :) – Paulo Bu Mar 29 '14 at 22:16

Python: Sorting List of Lists by a String Element

2 Answers2