What is the difference between:

some_list1 = []
some_list1.append("something")

and

some_list2 = []
some_list2 += ["something"]

For your case the only difference is performance: append is twice as fast.

Python 3.0 (r30:67507, Dec  3 2008, 20:14:27) [MSC v.1500 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import timeit
>>> timeit.Timer('s.append("something")', 's = []').timeit()
0.20177424499999999
>>> timeit.Timer('s += ["something"]', 's = []').timeit()
0.41192320500000079

Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import timeit
>>> timeit.Timer('s.append("something")', 's = []').timeit()
0.23079359499999999
>>> timeit.Timer('s += ["something"]', 's = []').timeit()
0.44208112500000141

In general case append will add one item to the list, while += will copy all elements of right-hand-side list into the left-hand-side list.

Update: perf analysis

Comparing bytecodes we can assume that append version wastes cycles in LOAD_ATTR + CALL_FUNCTION, and += version — in BUILD_LIST. Apparently BUILD_LIST outweighs LOAD_ATTR + CALL_FUNCTION.

>>> import dis
>>> dis.dis(compile("s = []; s.append('spam')", '', 'exec'))
  1           0 BUILD_LIST               0
              3 STORE_NAME               0 (s)
              6 LOAD_NAME                0 (s)
              9 LOAD_ATTR                1 (append)
             12 LOAD_CONST               0 ('spam')
             15 CALL_FUNCTION            1
             18 POP_TOP
             19 LOAD_CONST               1 (None)
             22 RETURN_VALUE
>>> dis.dis(compile("s = []; s += ['spam']", '', 'exec'))
  1           0 BUILD_LIST               0
              3 STORE_NAME               0 (s)
              6 LOAD_NAME                0 (s)
              9 LOAD_CONST               0 ('spam')
             12 BUILD_LIST               1
             15 INPLACE_ADD
             16 STORE_NAME               0 (s)
             19 LOAD_CONST               1 (None)
             22 RETURN_VALUE

We can improve performance even more by removing LOAD_ATTR overhead:

>>> timeit.Timer('a("something")', 's = []; a = s.append').timeit()
0.15924410999923566

In the example you gave, there is no difference, in terms of output, between append and +=. But there is a difference between append and + (which the question originally asked about).

>>> a = []
>>> id(a)
11814312
>>> a.append("hello")
>>> id(a)
11814312

>>> b = []
>>> id(b)
11828720
>>> c = b + ["hello"]
>>> id(c)
11833752
>>> b += ["hello"]
>>> id(b)
11828720

As you can see, append and += have the same result; they add the item to the list, without producing a new list. Using + adds the two lists and produces a new list.

>>> a=[]
>>> a.append([1,2])
>>> a
[[1, 2]]
>>> a=[]
>>> a+=[1,2]
>>> a
[1, 2]

See that append adds a single element to the list, which may be anything. +=[] joins the lists.

+= is an assignment. When you use it you’re really saying ‘some_list2= some_list2+[‘something’]’. Assignments involve rebinding, so:

l= []

def a1(x):
    l.append(x) # works

def a2(x):
    l= l+[x] # assign to l, makes l local
             # so attempt to read l for addition gives UnboundLocalError

def a3(x):
    l+= [x]  # fails for the same reason

The += operator should also normally create a new list object like list+list normally does:

>>> l1= []
>>> l2= l1

>>> l1.append('x')
>>> l1 is l2
True

>>> l1= l1+['x']
>>> l1 is l2
False

However in reality:

>>> l2= l1
>>> l1+= ['x']
>>> l1 is l2
True

This is because Python lists implement __iadd__() to make a += augmented assignment short-circuit and call list.extend() instead. (It’s a bit of a strange wart this: it usually does what you meant, but for confusing reasons.)

In general, if you’re appending/extended an existing list, and you want to keep the reference to the same list (instead of making a new one), it’s best to be explicit and stick with the append()/extend() methods.

 some_list2 += ["something"]

is actually

 some_list2.extend(["something"])

for one value, there is no difference. Documentation states, that:

s.append(x) same as s[len(s):len(s)] = [x]
s.extend(x) same as s[len(s):len(s)] = x

Thus obviously s.append(x) is same as s.extend([x])

The difference is that concatenate will flatten the resulting list, whereas append will keep the levels intact:

So for example with:

myList = [ ]
listA = [1,2,3]
listB = ["a","b","c"]

Using append, you end up with a list of lists:

>> myList.append(listA)
>> myList.append(listB)
>> myList
[[1,2,3],['a',b','c']]

Using concatenate instead, you end up with a flat list:

>> myList += listA + listB
>> myList
[1,2,3,"a","b","c"]

The performance tests here are not correct:

1. You shouldn’t run the profile only once.
2. If comparing append vs. += [] number of times you should declare append as a local function.
3. time results are different on different python versions: 64 and 32 bit

e.g.

timeit.Timer(‘for i in xrange(100): app(i)’, ‘s = [] ; app = s.append’).timeit()

good tests can be found here: http://markandclick.com/1/post/2012/01/python-list-append-vs.html

In addition to the aspects described in the other answers, append and +[] have very different behaviors when you’re trying to build a list of lists.

>>> list1=[[1,2],[3,4]]
>>> list2=[5,6]
>>> list3=list1+list2
>>> list3
[[1, 2], [3, 4], 5, 6]
>>> list1.append(list2)
>>> list1
[[1, 2], [3, 4], [5, 6]]

list1+[‘5′,’6’] adds ‘5’ and ‘6’ to the list1 as individual elements. list1.append([‘5′,’6’]) adds the list [‘5′,’6’] to the list1 as a single element.

The rebinding behaviour mentioned in other answers does matter in certain circumstances:

>>> a = ([],[])
>>> a[0].append(1)
>>> a
([1], [])
>>> a[1] += [1]
Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment

That’s because augmented assignment always rebinds, even if the object was mutated in-place. The rebinding here happens to be a[1] = *mutated list*, which doesn’t work for tuples.

let’s take an example first

list1=[1,2,3,4]
list2=list1     (that means they points to same object)

if we do 
list1=list1+[5]    it will create a new object of list
print(list1)       output [1,2,3,4,5] 
print(list2)       output [1,2,3,4]

but if we append  then 
list1.append(5)     no new object of list created
print(list1)       output [1,2,3,4,5] 
print(list2)       output [1,2,3,4,5]

extend(list) also do the same work as append it just append a list instead of a 
single variable 

The append() method adds a single item to the existing list

some_list1 = []
some_list1.append("something")

So here the some_list1 will get modified.

Updated:

Whereas using + to combine the elements of lists (more than one element) in the existing list similar to the extend (as corrected by Flux).

some_list2 = []
some_list2 += ["something"]

So here the some_list2 and [“something”] are the two lists that are combined.

“+” does not mutate the list

.append() mutates the old list