英文:
Remove None from generator output
问题
我如何从这个生成器中删除所有的None值：
(i.get("x") for i in l)
我认为可以使用赋值表达式来完成。我尝试过：
(a:=i.get("x") for i in l if a)
但它什么都没有返回。
英文:
How do I remove all the None values from this generator:
(i.get("x") for i in l)
I believe to can be done using assignment expressions. I tried:
(a:=i.get("x") for i in l if a)
But it returned nothing.
答案1
得分: 4
如果您使用支持海象运算符（Python 3.8+）的Python版本，您可以在您的推导式中使用它：
l = [
{"x": 1},
{"x": 0},
{"y": 2},
]
print([
clean
for i in l
if (clean := i.get("x")) is not None
])
应该会得到：
[1, 0]
英文:
If you have a version of python that supports the walrus operator (3.8+) then you can use it in your comprehension:
l = [
{"x": 1},
{"x": 0},
{"y": 2},
]
print([
clean
for i in l
if (clean := i.get("x")) is not None
])
Should give you:
[1, 0]
答案2
得分: 1
没有河马操作符，我们可以只是迭代一个从 i.get('x') 创建的单例元组。这会创建一个名为 x 的变量，其绑定到 i.get('x') 返回的值。
>>> lst = [{'d': 42}, {'x': 27}]
>>> [i.get('x') for i in lst]
[None, 27]
>>> [x
... for i in lst
... for x in (i.get('x'),)
... if x is not None]
[27]
这不考虑 x 是其中一个字典中的键且其关联值为 None 的特殊情况。
英文:
Without the walrus operator, we can just iterate over a singleton tuple created from i.get('x'). This has the effect of creating a variable x bound to the value returned by i.get('x').
>>> lst = [{'d': 42}, {'x': 27}]
>>> [i.get('x') for i in lst]
[None, 27]
>>> [x
... for i in lst
... for x in (i.get('x'),)
... if x is not None]
[27]
This does not account for the corner case of 'x' being a key in one of the dictionaries with the associated value None.
答案3
得分: 0
你可以使用海象操作符，但对于列表推导或生成器，赋值必须在if部分完成：
(a for i in l if (a:=i.get('x')) is not None)
英文:
Yes, you can use the walrus operator, but for a comprehension or generator the assignment has to be done in the if part:
(a for i in l if (a:=i.get('x')) is not None)
答案4
得分: -1
可以过滤生成器结果，仅返回不为None的结果。
filter(lambda x: x is not None, (i.get("x") for i in l))
或者
(x.get("x") for x in l if x.get("x") is not None)
英文:
You can filter the generator result to return only the results that are not None.
filter(lambda x: x is not None, (i.get("x") for i in l))
or
(x.get("x") for x in l if x.get("x") is not None)
答案5
得分: -1
你可以首先检查字典中是否包含 "x"：
l = [{'x': 1}, {'x': 0}, {'y': 2}]
result = [d["x"] for d in l if "x" in d]
结果：
[1, 0]
如果你想要过滤掉值为 None 的情况，你可以使用沃尔鲁斯方法：
result = [x for d in l if (x := d.get("x")) is not None]
英文:
You could just check if "x" is in the dict first:
l = [{'x': 1}, {'x': 0}, {'y': 2}]
result = [d["x"] for d in l if "x" in d]
Result:
[1, 0]
In the event you want to filter out values that are None as well then you could use the walrus approch:
result = [x for d in l if (x := d.get("x")) is not None]
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