What's "better" the reverse method or the reversed built-in function?

Question

What is typically regarded as more Pythonic/better/faster to use, the reverse method or the reversed built-in function?

Both in action:

_list = list(xrange(4))

print _list

rlist = list(reversed(_list))

print rlist

_list.reverse()

print _list

Solution 1

Depends on whether you want to reverse the list in-place (i.e. change the list) or not. No other real difference.

_{Often using reversed leads to nicer code.}

Solution 2

foo.reverse() actually reverses the elements in the container. reversed() doesn't actually reverse anything, it merely returns an object that can be used to iterate over the container's elements in reverse order. If that's what you need, it's often faster than actually reversing the elements.

Solution 3

There seems to be a great difference. I really thought it's the other way round. Why is rearranging the values in a list faster than creating a new one from an iterator ?

from decorators import bench

_list = range(10 ** 6)

@ bench
def foo():
  list(reversed(_list))

@ bench
def bar():
  _list.reverse()

foo()
bar()

print foo.time
print bar.time

0.167278051376
0.0122621059418

Solution 4

Without knowing real stats about performance, _list.reverse() modifies the list itself, whereas reversed(_list) returns an iterator ready to traverse the list in reversed order. That's a big difference itself.

If that's not a problem, object.reverse() seems more readable to me, but maybe you have specific speed requirements. And if reverse() does not belong to 80% of software that's consuming resources, I wouldn't bother (as a general rule of thumb).

Solution 5

• _list.reverse() does an in-place reversal and does not return a value
• reversed(_list) does not change _list, but returns a reverse iterable object
• _list[::-1] does not change _list, but returns reversed slice

example:

_list = [1,2,3]
ret1 = list(reversed(_list))
ret2 = _list[::-1] #reverse order slice
ret3 = _list.reverse() #no value set in ret3
print('ret1,ret2,ret3,_list:',ret1,ret2,ret3,_list)

_list = [1,2,3]
for x in reversed(_list):
    print(x)

output:

ret1,ret2,ret3,_list: [3, 2, 1] [3, 2, 1] None [3, 2, 1]
3
2
1

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Comments

[Admin]

reversed doesn't copy, it returns an iterator.

[Karl Knechtel]

The memory allocation itself shouldn't take too long; IIRC the list is a single contiguous allocation (PyObject**). More likely the difference is because list(reversed(_list)), even though the list constructor is C code, must go through the reverseiterator's API instead of just swapping pointers in a tight C loop.

[Karl Knechtel]

If they have to check for listreverseiterator then they have to check for everything... eventually it gets kind of unwieldy, and slows down short copies. Besides, the intended-to-be-common case for reversed is when you're going to iterate anyway and don't need to create a temporary.

[Weidenrinde]

If you do something with the lists, the picture changes. Comparing for i in reversed(_list):... and _list.reverse();for i in _list:..., the former is a little faster.