Python| a += b 并不总是 a = a + b
在Python中 a += b 的行为方式并不总是与 a = a + b 相同,相同的操作数在不同的条件下可能会给出不同的结果。但要了解它们为何表现出不同的行为,您必须深入研究变量的工作原理。
所以首先,你需要知道幕后发生了什么。
创建新变量:
Python3
a = 10
print(" id of a : ", id(10) ," Value : ", a )
Python3
a = 10 # Assigning value to variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
a = a + 10 # Modifying value of variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
a += 10 # Modifying value of variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
Python3
a = [0, 1] # stores this array in memory and assign its reference to a
print("id of a: ",id(a) , "Value : ", a )
a = a + [2, 3] # this will also behave same store data in memory and assign ref. to variable
print("id of a: ",id(a) , "Value : ", a )
a += [4, 5]
print("id of a: ",id(a) , "Value : ", a )
#But now this will now create new ref. instead this will modify the current object so
# all the other variable pointing to a will also gets changes
Python3
list1 = [5, 4, 3, 2, 1]
list2 = list1
list1 += [1, 2, 3, 4] # modifying value in current reference
print(list1)
print(list2) # as on line 4 it modify the value without creating new object
# variable list2 which is pointing to list1 gets changes
Python3
list1 = [5, 4, 3, 2, 1]
list2 = list1
list1 = list1 + [1, 2, 3, 4]
# Contents of list1 are same as above
# program, but contents of list2 are
# different.
print(list1)
print(list2)
输出 :
id of a : 11094592 Value : 10
在上面的示例中,值 10 被存储在内存中,并且它的引用被分配给 a。
修改变量:
Python3
a = 10 # Assigning value to variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
a = a + 10 # Modifying value of variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
a += 10 # Modifying value of variable creats new object
print(" id of a : ", id(a) ," Value : ", a )
输出 :
id of a : 11094592 Value : 10
id of a : 11094912 Value : 20
id of a : 11095232 Value : 30
每当我们创建或修改 int、float、char、 字符串时,它们都会创建新对象并将新创建的引用分配给它们各自的变量。
但是列表中没有看到相同的行为
Python3
a = [0, 1] # stores this array in memory and assign its reference to a
print("id of a: ",id(a) , "Value : ", a )
a = a + [2, 3] # this will also behave same store data in memory and assign ref. to variable
print("id of a: ",id(a) , "Value : ", a )
a += [4, 5]
print("id of a: ",id(a) , "Value : ", a )
#But now this will now create new ref. instead this will modify the current object so
# all the other variable pointing to a will also gets changes
输出:
id of a: 140266311673864 Value : [0, 1]
id of a: 140266311673608 Value : [0, 1, 2, 3]
id of a: 140266311673608 Value : [0, 1, 2, 3, 4, 5]
此时您可以看到 a = a + b 与 a += b 有时不同的原因。
考虑这些用于列表操作的示例:
示例 1:
Python3
list1 = [5, 4, 3, 2, 1]
list2 = list1
list1 += [1, 2, 3, 4] # modifying value in current reference
print(list1)
print(list2) # as on line 4 it modify the value without creating new object
# variable list2 which is pointing to list1 gets changes
输出:
[5, 4, 3, 2, 1, 1, 2, 3, 4]
[5, 4, 3, 2, 1, 1, 2, 3, 4]
示例 2
Python3
list1 = [5, 4, 3, 2, 1]
list2 = list1
list1 = list1 + [1, 2, 3, 4]
# Contents of list1 are same as above
# program, but contents of list2 are
# different.
print(list1)
print(list2)
输出:
[5, 4, 3, 2, 1, 1, 2, 3, 4]
[5, 4, 3, 2, 1]
- 表达式list1 += [1, 2, 3, 4]就地修改列表,这意味着它扩展了列表,使得“list1”和“list2”仍然具有对同一个列表的引用。
- 表达式list1 = list1 + [1, 2, 3, 4]创建一个新列表并将“list1”引用更改为该新列表,“list2”仍引用旧列表。