파이썬에서“.append ()”와“+ = []”의 차이점은 무엇입니까?

파이썬에서“.append ()”와“+ = []”의 차이점은 무엇입니까?

---

차이점은 무엇입니까?

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

과

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

---

귀하의 경우 유일한 차이점은 성능입니다. 추가는 두 배 빠릅니다.

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

일반적으로 append목록에 한 항목을 추가 하고 오른쪽 목록의 모든 요소를 왼쪽 목록에 +=복사 합니다.

업데이트 : 성능 분석

바이트 코드를 비교하면 append버전이주기를 LOAD_ATTR+ CALL_FUNCTION, 및 + = version-in으로 낭비 한다고 가정 할 수 있습니다 BUILD_LIST. 분명히 + BUILD_LIST보다 큽니다 .LOAD_ATTRCALL_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

LOAD_ATTR오버 헤드 를 제거하여 성능을 더욱 향상시킬 수 있습니다 .

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

---

당신이 준 예에서, 출력면 사이에는 차이가 없다 append하고 +=. 그러나 append와 +(질문이 원래 묻는) 차이점이 있습니다.

>>> 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

로서 당신은 참조 할 수 append와 +=같은 결과를 가지고, 새 목록을 생성하지 않고 목록에 항목을 추가합니다. 를 사용 +하면 두 목록이 추가되고 새 목록이 생성됩니다.

---

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

append는 목록에 단일 요소를 추가합니다.이 요소는 무엇이든 가능합니다. +=[]목록에 가입합니다.

---

+ =는 과제입니다. 사용하면 실제로 'some_list2 = some_list2 + ['something ']'입니다. 할당에는 리 바인딩이 포함되므로 다음과 같습니다.

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

+ = 연산자는 일반적으로 list + list와 같은 새 목록 객체를 만들어야합니다.

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

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

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

그러나 실제로 :

>>> 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. It doesn't return a new list, rather it modifies the original list.

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

So here the some_list1 will get modified.

Whereas using + to combine the elements of lists returns a new list.

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

So here the some_list2 and ["something"] are the two lists which is combined and a new list is returned which is assigned to some_list2

참고URL : https://stackoverflow.com/questions/725782/in-python-what-is-the-difference-between-append-and