a=[1,2,3]
b=[4,5,6]
c=[]
d=[]
这两种说法有什么区别?
c[:]=a
d=b[:]
但是两者给出相同的结果。
c 为 [1,2,3],d 为 [4,5,6]
功能方面有什么不同吗?
c[:] = a
表示将c中的所有元素替换为a中的元素
>>> l = [1,2,3,4,5]
>>> l[::2] = [0, 0, 0] #you can also replace only particular elements using this
>>> l
[0, 2, 0, 4, 0]
>>> k = [1,2,3,4,5]
>>> g = ['a','b','c','d']
>>> g[:2] = k[:2] # only replace first 2 elements
>>> g
[1, 2, 'c', 'd']
>>> a = [[1,2,3],[4,5,6],[7,8,9]]
>>> c[:] = a #creates a shallow copy
>>> a[0].append('foo') #changing a mutable object inside a changes it in c too
>>> a
[[1, 2, 3, 'foo'], [4, 5, 6], [7, 8, 9]]
>>> c
[[1, 2, 3, 'foo'], [4, 5, 6], [7, 8, 9]]
d = b[:]
表示创建 b 的浅拷贝并将其分配给 d ,类似于 d = list(b)
>>> l = [1,2,3,4,5]
>>> m = [1,2,3]
>>> l = m[::-1]
>>> l
[3,2,1]
>>> l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> m = l[:] #creates a shallow copy
>>> l[0].pop(1) # a mutable object inside l is changed, it affects both l and m
2
>>> l
[[1, 3], [4, 5, 6], [7, 8, 9]]
>>> m
[[1, 3], [4, 5, 6], [7, 8, 9]]
阿什维尼所说的。 :) 我会详细说明一下:
In [1]: a=[1,2,3]
In [2]: b = a
In [3]: c = a[:]
In [4]: b, c
Out[4]: ([1, 2, 3], [1, 2, 3])
In [5]: a is b, a is c
Out[5]: (True, False)
另一种方式:
In [1]: a = [1,2,3]
In [2]: aold = a
In [3]: a[:] = [4,5,6]
In [4]: a, aold
Out[4]: ([4, 5, 6], [4, 5, 6])
In [5]: a = [7,8,9]
In [6]: a, aold
Out[6]: ([7, 8, 9], [4, 5, 6])
看看会发生什么?
Ashwini 的回答准确地描述了正在发生的事情,以下是两种方法之间差异的一些示例:
a=[1,2,3]
b=[4,5,6]
c=[]
c2=c
d=[]
d2=d
c[:]=a # replace all the elements of c by elements of a
assert c2 is c # c and c2 should still be the same list
c2.append(4) # modifying c2 will also modify c
assert c == c2 == [1,2,3,4]
assert c is not a # c and a are not the same list
d=b[:] # create a copy of b and assign it to d
assert d2 is not d # d and d2 are no longer the same list
assert d == [4,5,6] and d2 == [] # d2 is still an empty list
assert d is not b # d and b are not the same list
没有太大区别。
c[:]=a
更新 c
引用的列表。 d=b[:]
创建一个新列表,它是 b 的副本(忘记在第四行创建的旧列表)。在大多数应用程序中,除非您有其他对周围数组的引用,否则您不太可能看到差异。当然,对于 c[:]=...
版本,您必须已经拥有一个列表 c
。
我想如果你改变类型,你就能更清楚地看到差异。
a = [1, 2]
b = (3, 4)
a1 = [5, 6]
b1 = (7, 8)
a[:] = b # now a is [3, 4]
a1 = b1[:] # now a1 is (7, 8)