[英]Filter 'None' output Python using filter()
I am getting an error when trying to filter the 'None' output, when using the filter function in Python, here is my code: 尝试过滤“无”输出时出现错误,在Python中使用过滤器功能时,这是我的代码:
def my_function(x):
if (x % 2 == 0):
x=filter(None, x)
return(x)
for x in range(1, 10):
sum=sum+(my_function(x))
print(sum)
and Python cannot make the sum as the None output cannot be removed, because it is "adding" this 并且Python无法计算总和,因为无法删除None输出,因为它正在“添加”此
None
2
None
4
None
6
None
8
None
and not this 不是这个
2
4
6
8
Several errors here. 这里有几个错误。 First, don't use
sum
because is a builtin python function name. 首先,不要使用
sum
因为它是内置的python函数名称。 The, filter
expect an iterable to work with, and a function. filter
期望可迭代的函数和一个函数。 None
is not a function it will use the identity
function (which returns the same value it takes) (tip by @bro-grammer). None
不是一个函数,它将使用identity
函数(返回相同的值)(@ bro-grammer提示)。 Since x is not an iterable you can't use filter on it. 由于x不可迭代,因此无法对其使用过滤器。
You want to check if x
is even: 您要检查
x
是否为偶数:
def my_function(x):
if (x % 2 == 0):
return True
return False
sumation = 0
for x in range(1, 10):
if my_function(x):
sumation += x
print(sumation)
The other option using actual filter
and sum
is : 使用实际
filter
和sum
的另一个选项是:
>>> def my_function(x):
... if (x % 2 == 0):
... return True
... return False
...
>>> sumation = sum(filter(my_function, range(1, 10)))
>>> sumation
20
And the pythonic way of doing this is with a generator and sum
: pythonic的方法是使用generator和
sum
:
>>> sum(x for x in range(1, 10) if x % 2 == 0 )
20
I guess you're trying to use the filter function in a wrong way. 我猜您正在尝试以错误的方式使用过滤器功能。 You could use this as an example with this code:
您可以将以下代码用作示例:
def my_function(x):
if (x % 2 == 0):
return True
else:
return False
alist = filter(my_function,list(range(1,10)))
print(sum(alist))
Hope it helps 希望能帮助到你
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