How do operator.itemgetter() and sort() work?

Question

I have the following code:

# initialize
a = []

# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John",  8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])    

# sort the table by age
import operator
a.sort(key=operator.itemgetter(1))    

# print the table
print(a)

It creates a 4x3 table and then it sorts it by age. My question is, what exactly key=operator.itemgetter(1) does? Does the operator.itemgetter function return the item's value? Why can't I just type something like key=a[x][1] there? Or can I? How could with operator print a certain value of the form like 3x2 which is 22?

1. How does exactly Python sort the table? Can I reverse-sort it?

2. How can I sort it based on two columns like first age, and then if age is the same b name?

3. How could I do it without operator?

Solution 1

Looks like you're a little bit confused about all that stuff.

operator is a built-in module providing a set of convenient operators. In two words operator.itemgetter(n) constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it.

So, you can't use key=a[x][1] there, because python has no idea what x is. Instead, you could use a lambda function (elem is just a variable name, no magic there):

a.sort(key=lambda elem: elem[1])

Or just an ordinary function:

def get_second_elem(iterable):
    return iterable[1]

a.sort(key=get_second_elem)

So, here's an important note: in python functions are first-class citizens, so you can pass them to other functions as a parameter.

Other questions:

1. Yes, you can reverse sort, just add reverse=True: a.sort(key=..., reverse=True)
2. To sort by more than one column you can use itemgetter with multiple indices: operator.itemgetter(1,2), or with lambda: lambda elem: (elem[1], elem[2]). This way, iterables are constructed on the fly for each item in list, which are than compared against each other in lexicographic(?) order (first elements compared, if equal - second elements compared, etc)
3. You can fetch value at [3,2] using a[2,1] (indices are zero-based). Using operator... It's possible, but not as clean as just indexing.

Refer to the documentation for details:

1. operator.itemgetter explained
2. Sorting list by custom key in Python

Solution 2

Answer for Python beginners

In simpler words:

1. The key= parameter of sort requires a key function (to be applied to be objects to be sorted) rather than a single key value and
2. that is just what operator.itemgetter(1) will give you: A function that grabs the first item from a list-like object.

(More precisely those are callables, not functions, but that is a difference that can often be ignored.)

Solution 3

You are asking a lot of questions that you could answer yourself by reading the documentation, so I'll give you a general advice: read it and experiment in the python shell. You'll see that itemgetter returns a callable:

>>> func = operator.itemgetter(1)
>>> func(a)
['Paul', 22, 'Car Dealer']
>>> func(a[0])
8

To do it in a different way, you can use lambda:

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

And reverse it:

a.sort(key=operator.itemgetter(1), reverse=True)

Sort by more than one column:

a.sort(key=operator.itemgetter(1,2))

See the sorting How To.

Solution 4

#sorting first by age then profession,you can change it in function "fun".
a = []

def fun(v):
    return (v[1],v[2])

# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John",  8, "Student"])
a.append(["Paul",  8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])

a.sort(key=fun)


print a

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