Why does the following code give 'None'? How can I resolve this?
为什么以下代码给出“无”？我该如何解决这个问题？
def f1(list1):
f2(list1.append(2))
def f2(list1):
print(list1)
f1([1])
What also doesn't work:
什么也行不通：
def f1(list1):
arg1 = list1.append(2)
f2(arg1)
2 个解决方案
#1
6
It depends on what you want to do. If you want list1 to have changed after a call to f1, use
这取决于你想做什么。如果您希望在调用f1后更改list1，请使用
def f1(list1):
list1.append(2)
f2(list1)
See what happens:
走着瞧吧：
>>> l = [1]
>>> f1(l) # Modifies l in-place!
[1, 2]
>>> l
[1, 2]
If you don't want list1 to be changed:
如果您不想更改list1：
def f1(list1):
f2(list1 + [2])
Now see this:
现在看到这个：
>>> l = [1]
>>> f1(l) # Leaves l alone!
[1, 2]
>>> l
[1]
#2
7
In general, Python methods that mutate an object (such as list.append, list.extend, or list.sort) return None.
通常，改变对象的Python方法（例如list.append，list.extend或list.sort）将返回None。
If you wish to print out the new list:
如果您想打印新列表：
def f1(list1):
list1.append(2)
f2(list1)
#1
6
It depends on what you want to do. If you want list1 to have changed after a call to f1, use
这取决于你想做什么。如果您希望在调用f1后更改list1，请使用
def f1(list1):
list1.append(2)
f2(list1)
See what happens:
走着瞧吧：
>>> l = [1]
>>> f1(l) # Modifies l in-place!
[1, 2]
>>> l
[1, 2]
If you don't want list1 to be changed:
如果您不想更改list1：
def f1(list1):
f2(list1 + [2])
Now see this:
现在看到这个：
>>> l = [1]
>>> f1(l) # Leaves l alone!
[1, 2]
>>> l
[1]
#2
7
In general, Python methods that mutate an object (such as list.append, list.extend, or list.sort) return None.
通常，改变对象的Python方法（例如list.append，list.extend或list.sort）将返回None。
If you wish to print out the new list:
如果您想打印新列表：
def f1(list1):
list1.append(2)
f2(list1)