Python 使用 a.any() 或 a.all()
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Use a.any() or a.all()
提问by Moe Steen
x = np.arange(0,2,0.5)
valeur = 2*x
if valeur <= 0.6:
print ("this works")
else:
print ("valeur is too high")
here is the error I get:
这是我得到的错误：
if valeur <= 0.6:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
I have read several posts about a.any() or a.all() but still can't find a way that really clearly explain how to fix the problem. I see why Python does not like what I wrote but I am not sure how to fix it.
我已经阅读了几篇关于 a.any() 或 a.all() 的帖子，但仍然找不到真正清楚地解释如何解决问题的方法。我明白为什么 Python 不喜欢我写的东西，但我不知道如何修复它。
回答by poke
If you take a look at the result of valeur <= 0.6, you can see what's causing this ambiguity:
如果您查看 的结果valeur <= 0.6，您可以看到造成这种歧义的原因：
>>> valeur <= 0.6
array([ True, False, False, False], dtype=bool)
So the result is another array that has in this case 4 boolean values. Now what should the result be? Should the condition be true when one value is true? Should the condition be true only when all values are true?
所以结果是另一个数组，在这种情况下有 4 个布尔值。现在结果应该是什么？当一个值为真时，条件是否应该为真？只有当所有值都为真时，条件才应该为真吗？
That's exactly what numpy.anyand numpy.alldo. The former requires at least one true value, the latter requires that all values are true:
这正是numpy.any和numpy.all做的。前者要求至少有一个真值，后者要求所有值都为真：
>>> np.any(valeur <= 0.6)
True
>>> np.all(valeur <= 0.6)
False
回答by hpaulj
You comment:
你评论：
valeur is a vector equal to [ 0. 1. 2. 3.] I am interested in each single term. For the part below 0.6, then return "this works"....
valeur 是一个等于 [ 0. 1. 2. 3.] 的向量，我对每一项都感兴趣。对于低于 0.6 的部分，则返回“this works”....
If you are interested in each term, then write the code so it deals with each. For example.
如果您对每个术语都感兴趣，那么编写代码以便处理每个术语。例如。
for b in valeur<=0.6:
if b:
print ("this works")
else:
print ("valeur is too high")
This will write 2 lines.
这将写入 2 行。
The error is produced by numpycode when you try to use it a context that expects a single, scalar, value. if b:...can only do one thing. It does not, by itself, iterate through the elements of bdoing a different thing for each.
numpy当您尝试在需要单个标量值的上下文中使用它时，代码会产生该错误。 if b:...只能做一件事。它本身不会遍历b为每个元素做不同事情的元素。
You could also cast that iteration as list comprehension, e.g.
您还可以将该迭代转换为列表理解，例如
['yes' if b else 'no' for b in np.array([True, False, True])]
回答by Gursel Karacor
This should also work and is a closer answer to what is asked in the question:
这也应该有效，并且是对问题中所问内容的更接近答案：
for i in range(len(x)):
if valeur.item(i) <= 0.6:
print ("this works")
else:
print ("valeur is too high")
