blocks|key|2934566|text|你可能想改变这一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2934567|if+len(v1)+>+len(v2):|code-block|syntax|javascript|2934568|至：|2934569|if+len(v1)+<+len(v2):|2934570|这样，当v1较短时，您将迭代到v1中的元素数量，这将防止您越过边界。|2934571|请注意，这也会抛出一个错误，因为newVector是一个长度为0的列表，而您访问的内容超出了它的范围。你就得改变|offset|length|style|CODE|2934572|newVector[index]+%2B=+v1[index]+%2B+v2[index]|2934573|至|2934574|newVector.append(v1[index]+%2B+v2[index])|2934575|但是，请注意，这可以更简单地完成，如下所示：|2934576|def+addVectors(v1,+v2):
++++return+map(sum,+zip(v1,+v2))|2934577|ETA:要用零填充列表，请执行以下操作：|2934578|import+itertools
def+addVectors(v1,+v2):
++++return+map(sum,+itertools.izip_longest(v1,+v2,+fillvalue=0))|2934579|例如：|2934580|addVectors([1,+2,+3,+4,+5],+[1,+2])
#+[2,+4,+3,+4,+5]|2934581|entityMap^0|0|0|0|0|0|G|9|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1E|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1G|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1H|8|@$O|1I|P|1J|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1L|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1M|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1O|8|@]|9|@]|A|$E|F]]|$1|12|3|13|5|6|7|1P|8|@]|9|@]|A|$]]|$1|14|3|15|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|16|3|17|5|6|7|1R|8|@]|9|@]|A|$]]|$1|18|3|19|5|D|7|1S|8|@]|9|@]|A|$E|F]]|$1|1A|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|1B|$]]

You probably meant to change the line:

<pre><code>if len(v1) &gt; len(v2):
</code></pre>

to:

<pre><code>if len(v1) &lt; len(v2):
</code></pre>

That way, you are iterating to the number of elements in v1 when v1 is shorter, which prevents you from going over the edge.

Note that this would also throw an error because <code>newVector</code> is a list of length 0 and you are accessing outside of its range. You'd have to change

<pre><code>newVector[index] += v1[index] + v2[index]
</code></pre>

to

<pre><code>newVector.append(v1[index] + v2[index])
</code></pre>

However, note that this can be done much more simply as:

<pre><code>def addVectors(v1, v2):
 return map(sum, zip(v1, v2))
</code></pre>

ETA: To pad the list with zeros, do:

<pre><code>import itertools
def addVectors(v1, v2):
 return map(sum, itertools.izip_longest(v1, v2, fillvalue=0))
</code></pre>

For example:

<pre><code>addVectors([1, 2, 3, 4, 5], [1, 2])
# [2, 4, 3, 4, 5]
</code></pre>

blocks|key|2524306|text|为什么不直接用这个呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2524307|def+sum_lists(a,+b):
++++return+[x[0]+%2B+x[1]+for+x+in+zip(a,+b)]

sum_lists([1,+2,+3],+[4,+5,+6])++++#+->+[5,+7,+9]|code-block|syntax|javascript|2524308|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Why not just use this?

<pre><code>def sum_lists(a, b):
 return [x[0] + x[1] for x in zip(a, b)]

sum_lists([1, 2, 3], [4, 5, 6]) # -&gt; [5, 7, 9]
</code></pre>

blocks|key|4572008|text|您可以比较列表的长度，这是正确的。但随后您需要更改所需的操作。也就是说，当list1比list2长的时候，你应该只遍历list2长度的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4572009|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You compare the lengths of the list, which is correct. But then you change the required operations. I.e. when list1 is longer than list2, you should only loop over the elements for the length of list2.

blocks|key|4662185|text|你的问题出在这里：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4662186|if+len(v1)+>+len(v2):
++++for+index+in+range(len(v1)):
++++++++print(index)
++++++++newVector[index]+%2B=+v1[index]+%2B+v2[index]|code-block|syntax|javascript|4662187|您要确保len(v1)+>+len(v2)，然后在range(len(v1))上迭代。|offset|length|style|CODE|4662188|在您的示例中，您试图访问并不存在的v2[2]。|4662189|更新：|4662190|为了响应您的编辑，您可以使用类似以下内容：|4662191|def+addVectors(v1,+v2):
++++if+len(v1)+>+len(v2):
++++++++map(sum,+zip(v1,+v2)).extend(v1[len(v2):])
++++else:
++++++++map(sum,+zip(v1,+v2)).extend(v2[len(v1):])|4662192|entityMap^0|0|0|4|H|P|E|0|H|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|V|$]]

Your problem lies here:

<pre><code>if len(v1) &gt; len(v2):
 for index in range(len(v1)):
 print(index)
 newVector[index] += v1[index] + v2[index]
</code></pre>

You ensure that <code>len(v1) &gt; len(v2)</code>, but then iterate over <code>range(len(v1))</code>. 

In your example, you're trying to access <code>v2[2]</code>, which doesn't exist.

UPDATE:

In response to your edit, you could use something like this:

<pre><code>def addVectors(v1, v2):
 if len(v1) &gt; len(v2):
 map(sum, zip(v1, v2)).extend(v1[len(v2):])
 else:
 map(sum, zip(v1, v2)).extend(v2[len(v1):])
</code></pre>

blocks|key|2934712|text|要填充长度不均匀的列表，可以使用itertools|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2934713|>>>+import+itertools
>>>+map(sum,+itertools.izip_longest([1,2,3],+[1,2],+fillvalue+=+0))
[2,+4,+3]|code-block|syntax|javascript|2934714|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html#itertools.izip_longest^0|G|9|G|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

To fill the lists of uneven length, you can use <a href="http://docs.python.org/library/itertools.html#itertools.izip_longest" rel="nofollow"><code>itertools</code></a>:

<pre><code>&gt;&gt;&gt; import itertools
&gt;&gt;&gt; map(sum, itertools.izip_longest([1,2,3], [1,2], fillvalue = 0))
[2, 4, 3]
</code></pre>

blocks|key|2934741|text|您的IndexError是因为您试图在newVector为空列表时写入newVectorindex。您需要将其初始化为一堆零，或者使用append。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2934742|>>>+first+=+[1,2,3]
>>>+second+=+[1,2]
>>>+output+=+[]
>>>+for+i,+item+in+enumerate(first):
...+++additional+=+second[i]+if+i+<+len(second)+else+0
...+++output.append(item+%2B+additional)
...
>>>+output
[2,+4,+3]|code-block|syntax|javascript|2934743|为了确保len(first)+>+len(second)，您可以这样做：|2934744|first,+second+=+first,+second+if+len(first)+>+len(second)+else+second,+first|2934745|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Your IndexError is because you're trying to write to newVector[index] when newVector is an empty list. You need to either initialize it to a bunch of zeros, or use append instead.

<pre><code>&gt;&gt;&gt; first = [1,2,3]
&gt;&gt;&gt; second = [1,2]
&gt;&gt;&gt; output = []
&gt;&gt;&gt; for i, item in enumerate(first):
... additional = second[i] if i &lt; len(second) else 0
... output.append(item + additional)
...
&gt;&gt;&gt; output
[2, 4, 3]
</code></pre>

And to ensure that len(first) > len(second), you can do something like this:

<pre><code>first, second = first, second if len(first) &gt; len(second) else second, first
</code></pre>

blocks|key|2524360|text|或者你也可以试试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2524361|def+add_vector(vector1,+vector2):
index+=+len(vector1)+-+1
new+=+[]
while+index+>=+0:
++++result+=+vector1[index]+%2B+vector2[index]
++++new.append(result)
++++index+-=1
new.reverse()
return+new|code-block|syntax|javascript|2524362|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Or you can try 

<pre><code>def add_vector(vector1, vector2):
index = len(vector1) - 1
new = []
while index &gt;= 0:
 result = vector1[index] + vector2[index]
 new.append(result)
 index -=1
new.reverse()
return new
</code></pre>

EDIT: Thank you for the responses, which are helpful, but I'm not sure they're getting at the core of my problem. In some cases it would be nice to simply select the lower of the lists and then add the values accordingly, but in THIS case I actually want to treat uneven lists as though the shorter list had zero values for the values it was missing. So I want [1, 2, 3] + [1, 2] to function as [1, 2, 3] + [1, 2, 0]. I don't think zip, or reversing my operator, will accomplish this. 

I'm trying to create a function that adds the corresponding values of two lists and returns a new list with the sums of each of the indices of the two original lists:

<pre><code>def addVectors(v1, v2):
 print(v1[0], v1[1], v2[0])
 newVector = []
 if len(v1) &gt; len(v2):
 for index in range(len(v1)):
 print(index)
 newVector[index] += v1[index] + v2[index]
 else:
 for index in range(len(v2)):
 print(index)
 newVector[index] += v2[index] + v1[index]
 return newVector


addVectors([1, 2, 3], [1, 2])
</code></pre>

Yet I'm getting an error stating that the list index is out of range? Not sure what I'm doing wrong in this seemingly simple program....

Function Adding Two Lists

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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编辑:感谢你的回复，这些回复很有帮助，但我不确定它们是否解决了我的问题的核心。在某些情况下，简单地选择列表中较低的一个，然后相应地添加值会很好，但在这种情况下，我实际上希望处理不均匀的列表，就好像较短的列表缺少的值为零一样。所以我想让1，2，3+ 1，2作为函数。我不认为zip或颠倒我的运算符可以实现这一点。我正在尝试...

问函数添加两个列表
EN

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问函数添加两个列表EN