In Python, what is the difference between ".append()" and "+= []"?

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What is the difference between:

some_list1 = []


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()
>>> timeit.Timer('s += ["something"]', 's = []').timeit()

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()
>>> timeit.Timer('s += ["something"]', 's = []').timeit()

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()

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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)
>>> a.append("hello")
>>> id(a)

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

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.

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>>> 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.

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+= 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

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

However in reality:

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

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.

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 some_list2 += ["something"]

is actually


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])

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  • append if for a single item. maybe you mean extend.
  • For the more interesting case of += vs extend:…
  • += vs append
  • +1: This is very interesting. I use append anyway, because it results in clearer code. But I didn't realize there was a performance difference. If anything, I would have expected append to be slower, since it's a guaranteed function call, while I presumed += would be optimized further.
  • Isn't there also a functional difference? For instance let a = [], b = [4,5,6], here if you do c = a.append(b) then c would be a list of list [[4,5,6]] while c += b; would lead to a simple list c = [4,5,6].
  • just to set things straight: += gives a better performance than extend or append as long as your input is in the right format. What takes time in the current example is the creation of the ['something'] list. += is about 15% faster
  • @Joe If you're comparing append vs +=, then you must include creation of the list as part of the measurement. Otherwise it'd be a different question (extend vs +=).
  • @jamesdlin yup! But it's easy to be mistaken unless you already know this. A little additional precision has never hurt anyone, right?
  • There is difference between append and +=.
  • There's the fact that append adds one entry to the list, while += adds as many as there are in the other list (i.e. aliases to extend). But he/she knows that already, judging by the way the question was written. Is there some other difference I'm missing?
  • There's a difference because an augmented assignment introduces rebinding (explanation in my answer).
  • Voting this up because this is an important distinction between the two. Good work.
  • s.append takes an arbitrary type and adds it to the list; It's a true append. s.extend takes an iterable (usually a list), and merges the iterable into s, modfiying the memory addresses of s. These are not the same.
  • still, the += tests in that page uses += [one_var]. If we omit creating lists, += becomes the fastest option.