如何克隆列表,使其在分配后不会意外更改?
使用 时,每次都会对更改进行任何修改。为什么会这样,我该如何克隆或复制列表以防止它?new_list = my_list
new_list
my_list
使用 时,每次都会对更改进行任何修改。为什么会这样,我该如何克隆或复制列表以防止它?new_list = my_list
new_list
my_list
new_list = my_list
实际上不会创建第二个列表。分配只是将引用复制到列表,而不是实际列表,因此两者都在分配后引用同一列表。new_list
my_list
要实际复制列表,您有几种选择:
您可以使用内置的list.copy()
方法(自Python 3.3起可用):
new_list = old_list.copy()
您可以将其切片:
new_list = old_list[:]
Alex Martelli对此的看法(至少在2007年是这样)是,这是一种奇怪的语法,使用它没有任何意义。;)(在他看来,下一个更具可读性)。
您可以使用内置的 list()
构造函数:
new_list = list(old_list)
您可以使用通用的 copy.copy()
:
import copy
new_list = copy.copy(old_list)
这比因为它必须首先找出数据类型要慢一些。list()
old_list
如果还需要复制列表的元素,请使用 generic 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。这是递归的;它将处理任意数量的嵌套列表(或其他容器)级别。
例:
import copy
class Foo(object):
def __init__(self, val):
self.val = val
def __repr__(self):
return f'Foo({self.val!r})'
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(f'original: {a}\nlist.copy(): {b}\nslice: {c}\nlist(): {d}\ncopy: {e}\ndeepcopy: {f}')
结果:
original: ['foo', Foo(5), 'baz']
list.copy(): ['foo', Foo(5)]
slice: ['foo', Foo(5)]
list(): ['foo', Foo(5)]
copy: ['foo', Foo(5)]
deepcopy: ['foo', Foo(1)]
Felix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:
复印.深度复印(old_list)
Copy()
Copy()
for item in old_list: new_list.append(item)
[i for i in old_list]
)复制(old_list)
list(old_list)
new_list = []; new_list.extend(old_list)
old_list[:]
)所以最快的是列表切片。但请注意,与 python 版本不同,和 和 不会复制列表中的任何列表、字典和类实例,因此,如果原始文件发生更改,它们也会在复制的列表中更改,反之亦然。copy.copy()
list[:]
list(list)
copy.deepcopy()
(如果有人感兴趣或想提出任何问题,这里是脚本:)
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