Details of shallow copy and deep copy examples of Python basic tutorial

Details of shallow copy and deep copy examples of Python foundation tutorials

There are many articles on the web about deep copy and shallow copy of Python. Here is a comparison of the three copies and attached examples for your reference

One-like replication

#encoding:utf-8
# define 1 A nested set 
lista=[1,2,3,[4,5,6,[7,8,9]]]

listb=lista
# Print out  lista and listb The address of the value 
print id(lista) #4511103096
print id(listb) #4511103096

# Modify the lista The contents of, listb The contents of the 

lista[0]=0
print lista #[0, 2, 3, [4, 5, 6, [7, 8, 9]]]
print listb #[0, 2, 3, [4, 5, 6, [7, 8, 9]]]

As you can see from the above practice, the replication operation is to point lista to the address in memory, and to point listb to it, without creating the address in memory itself.

Shallow copy

#encoding:utf-8
# define 1 A nested set 
import copy
lista=[1,2,3,[4,5,6,[7,8,9]]]
# use copy In the module copy methods 
listb=copy.copy(lista)

print id(lista) #4396231640

print id(listb) #4396231712

# found lista and listb The address in memory is different, 
# Prints the first in the element 1 The address value of the element 
print id(lista[0])#140666751466536

print id(listb[0])#140666751466536

# The address value of the element is 1 Kind of, does that mean, change lista The first of 1 An element, 
# listb Or shall I change it 

lista[0]=0

print lista #[0, 2, 3, [4, 5, 6, [7, 8, 9]]]

print listb #[1, 2, 3, [4, 5, 6, [7, 8, 9]]]

# Turns out, he didn't lista1 Why? Because I used a shallow copy 
# Let's print it out lista[0] and listb[0] The address of the 
print id(lista[0]) #140666751466560

print id(listb[0]) #140666751466536

# Will find lista[0] Redirect to another address in memory 


# This can be modified by the above methods lista[3] . lista[3] Is also 1 a list

print id(lista[3])#4499779240

print id(listb[3])#4499779240

# Modify the lista[3] In the first 1 An element? 

lista[3][0]=0

print lista #[0, 2, 3, [0, 5, 6, [7, 8, 9]]]

print listb #[0, 2, 3, [0, 5, 6, [7, 8, 9]]]

# Found to modify lista[3][0] The value of the element, listb[3][0] It's going to change 

According to the above practice, a shallow copy does not copy objects in an object.

Deep copy

  #encoding:utf-8
  # define 1 A nested set 
  import copy
  lista=[1,2,3,[4,5,6,[7,8,9]]]

  # The deep copy function is an enhancement on the shallow copy, so it has the shallow copy function 
  listb=copy.deepcopy(lista)

  lista[3][0]=0

  print lista #[1, 2, 3, [0, 5, 6, [7, 8, 9]]]

  print listb #[1, 2, 3, [4, 5, 6, [7, 8, 9]]]

  # Deep copy will resolve the changes lista[3][0] Student: The value doesn't affect listb The values in the 

  # Let's modify the first one 3 Is the element in the layer list also copied 

  lista[3][3][0]=0

  print lista #[1, 2, 3, [0, 5, 6, [0, 8, 9]]]

  print listb #[1, 2, 3, [4, 5, 6, [7, 8, 9]]]

  # It's not affected at all 

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