使用new_list=my_list时,对new_list的任何修改都会每次更改my_list。为什么会出现这种情况,以及如何克隆或复制列表以防止出现这种情况?
当前回答
deepcopy选项是唯一适用于我的方法:
from copy import deepcopy
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = deepcopy(a)
b[0][1]=[3]
print('Deep:')
print(a)
print(b)
print('-----------------------------')
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = a*1
b[0][1]=[3]
print('*1:')
print(a)
print(b)
print('-----------------------------')
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = a[:]
b[0][1]=[3]
print('Vector copy:')
print(a)
print(b)
print('-----------------------------')
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = list(a)
b[0][1]=[3]
print('List copy:')
print(a)
print(b)
print('-----------------------------')
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = a.copy()
b[0][1]=[3]
print('.copy():')
print(a)
print(b)
print('-----------------------------')
a = [ [ list(range(1, 3)) for i in range(3) ] ]
b = a
b[0][1]=[3]
print('Shallow:')
print(a)
print(b)
print('-----------------------------')
导致输出:
Deep:
[[[1, 2], [1, 2], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
*1:
[[[1, 2], [3], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
Vector copy:
[[[1, 2], [3], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
List copy:
[[[1, 2], [3], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
.copy():
[[[1, 2], [3], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
Shallow:
[[[1, 2], [3], [1, 2]]]
[[[1, 2], [3], [1, 2]]]
-----------------------------
其他回答
Python的习惯用法是newList=oldList[:]
菲利克斯已经给出了一个很好的答案,但我想我应该对各种方法进行速度比较:
10.59秒(105.9µs/itn)-copy.depcopy(旧列表)10.16秒(101.6µs/itn)-纯Python Copy()方法使用deepcopy复制类1.488秒(14.88µs/itn)-纯Python Copy()方法不复制类(仅dicts/lists/tuples)0.325秒(3.25µs/itn)-对于old_list:new_list.append(项目)中的项目0.217秒(2.17µs/itn)-[i代表old_list](列表理解)0.186秒(1.86µs/itn)-复制副本(old_list)0.075秒(0.75µs/itn)-列表(旧列表)0.053秒(0.53µs/itn)-新列表=[];新列表扩展(旧列表)0.039秒(0.39µs/itn)-old_list[:](列表切片)
所以最快的是列表切片。但请注意,与copy.deepcopy()和python版本不同,copy.copy()、list[:]和list(list)不会复制列表中的任何列表、字典和类实例,因此如果原始列表发生变化,它们也会在复制的列表中发生变化,反之亦然。
(如果有人感兴趣或想提出任何问题,以下是脚本:)
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
对每种复制模式的简短解释:
浅层副本构造一个新的复合对象,然后(在可能的范围内)向其中插入对原始对象的引用-创建浅层副本:
new_list = my_list
深度副本构造一个新的复合对象,然后递归地将原始对象的副本插入其中,从而创建一个深度副本:
new_list = list(my_list)
list()适用于简单列表的深度复制,例如:
my_list = ["A","B","C"]
但是,对于复杂的列表,如。。。
my_complex_list = [{'A' : 500, 'B' : 501},{'C' : 502}]
…使用deepcopy():
import copy
new_complex_list = copy.deepcopy(my_complex_list)
在Python中,请记住:
list1 = ['apples','bananas','pineapples']
list2 = list1
List2没有存储实际的列表,而是对list1的引用。因此,当您对list1执行任何操作时,list2也会发生变化。使用copy模块(非默认,在pip上下载)制作列表的原始副本(对于简单列表,copy.copy();对于嵌套列表,copy。deepcopy())。这将生成一个不会随第一个列表而更改的副本。
在已经给出的答案中,缺少了一个独立于python版本的非常简单的方法,您可以在大多数时间使用(至少我这样做):
new_list = my_list * 1 # Solution 1 when you are not using nested lists
但是,如果my_list包含其他容器(例如,嵌套列表),则必须按照复制库中上述答案中的其他建议使用deepcopy。例如:
import copy
new_list = copy.deepcopy(my_list) # Solution 2 when you are using nested lists
。奖励:如果您不想复制元素,请使用(AKA浅层复制):
new_list = my_list[:]
让我们了解解决方案#1和解决方案#2之间的区别
>>> a = range(5)
>>> b = a*1
>>> a,b
([0, 1, 2, 3, 4], [0, 1, 2, 3, 4])
>>> a[2] = 55
>>> a,b
([0, 1, 55, 3, 4], [0, 1, 2, 3, 4])
正如您所看到的,当我们不使用嵌套列表时,解决方案#1工作得很好。让我们检查一下当我们将解决方案#1应用于嵌套列表时会发生什么。
>>> from copy import deepcopy
>>> a = [range(i,i+4) for i in range(3)]
>>> a
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]]
>>> b = a*1
>>> c = deepcopy(a)
>>> for i in (a, b, c): print i
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]]
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]]
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]]
>>> a[2].append('99')
>>> for i in (a, b, c): print i
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5, 99]]
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5, 99]] # Solution #1 didn't work in nested list
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]] # Solution #2 - DeepCopy worked in nested list
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