python deepcopy和浅拷贝和传递引用

use*_*288 2 python linux reference deep-copy shallow-copy

关于python深度复制和浅拷贝的问题.

帖子在 深拷贝和浅拷贝之间有什么区别?

无法帮助我.

为什么1的总和是6而不是10?

eg1:

kvps = { '1' : 1, '2' : 2 }
theCopy = kvps.copy()  # both point to the same mem location ? 
kvps['1'] = 5
sum = kvps['1'] + theCopy['1']
print sum
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输出和是6

eg2:

aList = [1,2]
bList = [3,4]
kvps = { '1' : aList, '2' : bList }

theCopy = kvps.copy()  # both point to the same mem location ? 
kvps['1'][0] = 5
sum = kvps['1'][0] + theCopy['1'][0]

print sum
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输出总和是10

eg3:

import copy

aList = [1,2]
bList = [3,4]
kvps = { '1' : aList, '2' : bList }

theCopy = copy.deepcopy(kvps)
kvps['1'][0] = 5
sum = kvps['1'][0] + theCopy['1'][0]

print sum
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输出和是6.

另外,例如4

kvps = { '1' : 1, '2' : 2 }    
theCopy = dict(kvps)  #  theCopy hold a reference to kvps ?     
kvps['1'] = 5  # should also change theCopy , right ?    
sum = kvps['1'] + theCopy['1']    
print kvps    
print theCopy    
print sum
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它的总和是6,如果复制是对kvps的引用,它应该是10.

lun*_*chs 6

浅拷贝在顶级容器中创建可变对象的副本.深层复制在数据结构中创建所有可变容器的新实例.

"eg 2"导致10,因为你在外面复制了dict,但是里面的两个列表仍然是旧列表,并且列表可以就地更改(它们是可变的).

深层复制使运行aList.copy(),bList.copy()并用它们的副本替换dict中的值.


例如1解释:

kvps = {'1': 1, '2': 2}
theCopy = kvps.copy()

# the above is equivalent to:
kvps = {'1': 1, '2': 2}
theCopy = {'1': 1, '2': 2}
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当你将它应用于例如2时:

kvps = {'1': aList, '2': bList}
theCopy = {'1': aList, '2': bList}
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两个dicts中的列表对象是相同的对象,因此修改其中一个列表将反映在两个dicts中.


执行深层复制(例如3)会导致:

kvps = {'1': aList, '2': bList}
theCopy = {'1': [1, 2], '2': [3, 4]}
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这意味着两个dicts具有完全不同的内容,并且修改一个不会修改另一个.


例如4 via dict()等同于浅拷贝.