Python 从列表中获取第一个非 None 值
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Getting the first non None value from list
提问by Pran
Given a list, is there a way to get the first non-None value? And, if so, what would be the pythonic way to do so?
给定一个列表，有没有办法获得第一个非 None 值？而且，如果是这样，pythonic 的方法是什么？
For example, I have:
例如，我有：
a = objA.addreses.country.codeb = objB.country.codec = Noned = 'CA'
a = objA.addreses.country.codeb = objB.country.codec = Noned = 'CA'
In this case, if a is None, then I would like to get b. If a and b are both None, the I would like to get d.
在这种情况下，如果 a 是 None，那么我想得到 b。如果 a 和 b 都是 None，我想得到 d。
Currently I am doing something along the lines of (((a or b) or c) or d), is there another way?
目前我正在做一些类似的事情(((a or b) or c) or d)，还有其他方法吗？
采纳答案by alecxe
You can use next():
您可以使用next()：
>>> a = [None, None, None, 1, 2, 3, 4, 5]
>>> next(item for item in a if item is not None)
1
If the list contains only Nones, it will throw StopIterationexception. If you want to have a default value in this case, do this:
如果列表只包含 Nones，它将抛出StopIteration异常。如果您想在这种情况下使用默认值，请执行以下操作：
>>> a = [None, None, None]
>>> next((item for item in a if item is not None), 'All are Nones')
All are Nones
回答by Jon Clements
Adapt from the following (you could one-liner it if you wanted):
改编自以下内容（如果你愿意，你可以单行它）：
values = (a, b, c, d)
not_None = (el for el in values if el is not None)
value = next(not_None, None)
This takes the first non Nonevalue, or returns Noneinstead.
这需要第一个非None值，或者返回None。
回答by pylang
first_trueis an itertoolsrecipe found in the Python 3 docs:
first_true是itertools在Python 3 文档中找到的配方：
def first_true(iterable, default=False, pred=None):
"""Returns the first true value in the iterable.
If no true value is found, returns *default*
If *pred* is not None, returns the first item
for which pred(item) is true.
"""
# first_true([a,b,c], x) --> a or b or c or x
# first_true([a,b], x, f) --> a if f(a) else b if f(b) else x
return next(filter(pred, iterable), default)
One may choose to implement the latter recipe or import more_itertools, a library that ships with itertoolsrecipes and more:
可以选择实现后一个 recipe 或 import more_itertools，一个带有itertools配方等的库：
> pip install more_itertools
Use:
用：
import more_itertools as mit
a = [None, None, None, 1, 2, 3, 4, 5]
mit.first_true(a, pred=lambda x: x is not None)
# 1
a = [None, None, None]
mit.first_true(a, default="All are None", pred=lambda x: x is not None)
# 'All are None'
Why use the predicate?
为什么要使用谓词？
"First non-None" item is not the same as "first True" item, e.g. [None, None, 0]where 0is the first non-None, but it is not the first Trueitem. The predicate allows first_trueto be useable, ensuring any first seen, non-None, falsey item in the iterable is still returned (e.g. 0, False) instead of the default.
“第一个非None”项目与“第一个True”项目不同，例如第一个非项目[None, None, 0]在哪里，但它不是第一个项目。谓词允许是可用的，确保迭代中的任何第一次看到的、非无的、虚假的项目仍然返回（例如，）而不是默认值。0NoneTruefirst_true0False
a = [None, None, None, False]
mit.first_true(a, default="All are None", pred=lambda x: x is not None)
# 'False'
回答by DannyMoshe
I think this is the simplest waywhen dealing with a small set of values (will work in a list comprehension as well):
我认为这是处理一小组值时最简单的方法（也适用于列表理解）：
firstVal = a or b or c or d
Will always return the first non "False" value which works in some cases (given you dont expect any values which could evaluate to false as @GrannyAching points out below)
将始终返回在某些情况下有效的第一个非“假”值（假设您不期望任何值可能评估为假，正如@GrannyAching 在下面指出的那样）
回答by Lars Blumberg
When the items in your list are expensive to calculate such as in
当列表中的项目计算成本很高时，例如
first_non_null = next((calculate(x) for x in my_list if calculate(x)), None)
# or, when receiving possibly None-values from a dictionary for each list item:
first_non_null = next((my_dict[x] for x in my_list if my_dict.get(x)), None)
then you might want to avoid the repetitive calculation and simplify to:
那么您可能希望避免重复计算并简化为：
first_non_null = next(filter(bool, (calculate(x) for x in my_list)), None)
# or:
first_non_null = next(filter(bool, (my_dict.get(x) for x in my_list)), None)
