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I am working with Python list sort.

I have two lists: one is a list of integers, the other is a list of objects, and the second object list has the attribute id which is also an integer, I want to sort the object list based on the id attribute, in the order of the same id appears in the first list, well, this is an example:

I got a = [1,2,3,4,5]

and b = [o,p,q,r,s], where o.id = 2, p.id = 1, q.id = 3, r.id = 5, s.id = 4

and I want my list b to be sorted in the order of its id appears in list a, which is like this:

sorted_b = [p, o, q, s, r]

Of course, I can achieve this by using nested loops:

sorted_b = []
for i in a:
    for j in b:
        if j.id == i:
            sorted_b.append(j)
            break

but this is a classic ugly and non-Python way to solve a problem, I wonder if there is a way to solve this in a rather neat way, like using the sort method, but I don't know how.

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4 Answers 4

Reset to default
8 
>>> from collections import namedtuple
>>> Foo = namedtuple('Foo', 'name id') # this represents your class with id attribute
>>> a = [1,2,3,4,5]
>>> b = [Foo(name='o', id=2), Foo(name='p', id=1), Foo(name='q', id=3), Foo(name='r', id=5), Foo(name='s', id=4)]
>>> sorted(b, key=lambda x: a.index(x.id))
[Foo(name='p', id=1), Foo(name='o', id=2), Foo(name='q', id=3), Foo(name='s', id=4), Foo(name='r', id=5)]
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Comments

2

This is a simple way to do it:

# Create a dictionary that maps from an ID to the corresponding object
object_by_id = dict((x.id, x) for x in b)

sorted_b = [object_by_id[i] for i in a]

If the list gets big, it's probably the fastest way, too.

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1 Comment

Cool! I like your way to tackle with this problem, many thanks!
1

You can do it with a list comprehension, but in general is it the same.

sorted_b = [ y for x in a for y in b if y.id == x ]
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1 Comment

I think jamylak's answer is what I'm looking for, but thanks anyway:)
0

There is a sorted function in Python. It takes optional keyword argument cmp. You can pass there your customized function for sorting.

cmp definition from the docs:

custom comparison should return a negative, zero or positive number depending on whether the first argument is considered smaller than, equal to, or larger than the second argument

a = [1,2,3,4,5]
def compare(el1, el2):
   if a.index(el1.id) < a.index(el2.id): return -1
   if a.index(el1.id) > a.index(el2.id): return 1
   return 0

sorted(b, cmp=compare)

This is more straightforward however I would encourage you to use the key argument as jamylak described in his answer, because it's more pythonic and in Python 3 the cmp is not longer supported.

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2 Comments

But I got a little bit confused with the two if statements in the compare function, can you explain that to me? I'm not familiar with the cmp argument, thanks.
Then I think the second if statement should be if a.index(el1.id) > a.index(el2.id). Now I got it, thank you fellow.

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