使用new_list = my_list
,您实际上没有两个列表。赋值仅复制对列表的引用,而不是实际列表,因此new_list
和my_list
在赋值后引用相同的列表。
要实际复制列表,您有各种可能性:
你可以使用内置的list.copy()
方法(从 python 3.3 开始提供):
new_list = old_list.copy()
你可以切片:
new_list = old_list[:]
Alex Martelli对此的看法(至少在 2007 年 )是这样的, 它是一种奇怪的语法,并且永远使用它是没有意义的 。 ;)(在他看来,下一个更具可读性)。
您可以使用内置的list()
函数:
new_list = list(old_list)
您可以使用通用copy.copy()
:
import copy
new_list = copy.copy(old_list)
这比list()
慢一点,因为它必须首先找出old_list
的数据类型。
如果列表包含对象并且您也想要复制它们,请使用通用copy.deepcopy()
:
import copy
new_list = copy.deepcopy(old_list)
显然是最慢和最需要内存的方法,但有时是不可避免的。
例:
import copy
class Foo(object):
def __init__(self, val):
self.val = val
def __repr__(self):
return str(self.val)
foo = Foo(1)
a = ['foo', foo]
b = a.copy()
c = a[:]
d = list(a)
e = copy.copy(a)
f = copy.deepcopy(a)
# edit orignal list and instance
a.append('baz')
foo.val = 5
print('original: %r\n list.copy(): %r\n slice: %r\n list(): %r\n copy: %r\n deepcopy: %r'
% (a, b, c, d, e, f))
结果:
original: ['foo', 5, 'baz']
list.copy(): ['foo', 5]
slice: ['foo', 5]
list(): ['foo', 5]
copy: ['foo', 5]
deepcopy: ['foo', 1]
菲利克斯已经提供了一个很好的答案,但我想我会对各种方法进行速度比较:
copy.deepcopy(old_list)
Copy()
方法Copy()
方法不复制类(只有 dicts / lists / tuples) for item in old_list: new_list.append(item)
[i for i in old_list]
( 列表理解 ) copy.copy(old_list)
list(old_list)
new_list = []; new_list.extend(old_list)
old_list[:]
( 列表切片 ) 所以最快的是列表切片。但请注意, copy.copy()
, list[:]
和list(list)
与copy.deepcopy()
和 python 版本不同,不会复制列表中的任何列表,字典和类实例,因此如果原件发生更改,他们也会在复制的列表中进行更改,反之亦然。
(这是脚本,如果有人有兴趣或想提出任何问题:)
from copy import deepcopy
class old_class:
def __init__(self):
self.blah = 'blah'
class new_class(object):
def __init__(self):
self.blah = 'blah'
dignore = {str: None, unicode: None, int: None, type(None): None}
def Copy(obj, use_deepcopy=True):
t = type(obj)
if t in (list, tuple):
if t == tuple:
# Convert to a list if a tuple to
# allow assigning to when copying
is_tuple = True
obj = list(obj)
else:
# Otherwise just do a quick slice copy
obj = obj[:]
is_tuple = False
# Copy each item recursively
for x in xrange(len(obj)):
if type(obj[x]) in dignore:
continue
obj[x] = Copy(obj[x], use_deepcopy)
if is_tuple:
# Convert back into a tuple again
obj = tuple(obj)
elif t == dict:
# Use the fast shallow dict copy() method and copy any
# values which aren't immutable (like lists, dicts etc)
obj = obj.copy()
for k in obj:
if type(obj[k]) in dignore:
continue
obj[k] = Copy(obj[k], use_deepcopy)
elif t in dignore:
# Numeric or string/unicode?
# It's immutable, so ignore it!
pass
elif use_deepcopy:
obj = deepcopy(obj)
return obj
if __name__ == '__main__':
import copy
from time import time
num_times = 100000
L = [None, 'blah', 1, 543.4532,
['foo'], ('bar',), {'blah': 'blah'},
old_class(), new_class()]
t = time()
for i in xrange(num_times):
Copy(L)
print 'Custom Copy:', time()-t
t = time()
for i in xrange(num_times):
Copy(L, use_deepcopy=False)
print 'Custom Copy Only Copying Lists/Tuples/Dicts (no classes):', time()-t
t = time()
for i in xrange(num_times):
copy.copy(L)
print 'copy.copy:', time()-t
t = time()
for i in xrange(num_times):
copy.deepcopy(L)
print 'copy.deepcopy:', time()-t
t = time()
for i in xrange(num_times):
L[:]
print 'list slicing [:]:', time()-t
t = time()
for i in xrange(num_times):
list(L)
print 'list(L):', time()-t
t = time()
for i in xrange(num_times):
[i for i in L]
print 'list expression(L):', time()-t
t = time()
for i in xrange(num_times):
a = []
a.extend(L)
print 'list extend:', time()-t
t = time()
for i in xrange(num_times):
a = []
for y in L:
a.append(y)
print 'list append:', time()-t
t = time()
for i in xrange(num_times):
a = []
a.extend(i for i in L)
print 'generator expression extend:', time()-t
编辑 :在基准测试中添加了新式,旧式类和 dicts,并使 python 版本更快,并添加了更多方法,包括列表表达式和extend()
。
我被告知 Python 3.3+ 添加了list.copy()
方法,它应该像切片一样快:
newlist = old_list.copy()