blocks|key|1309716|text|>>>+t
[1,+3,+6]
>>>+[j-i+for+i,+j+in+zip(t[:-1],+t[1:])]++#+or+use+itertools.izip+in+py2k
[2,+3]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1309717|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&gt;&gt;&gt; t
[1, 3, 6]
&gt;&gt;&gt; [j-i for i, j in zip(t[:-1], t[1:])] # or use itertools.izip in py2k
[2, 3]
</code></pre>

blocks|key|1309772|text|其他答案是正确的，但是如果你在做数值工作，你可能想要考虑numpy。使用numpy，答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309773|v+=+numpy.diff(t)|code-block|syntax|javascript|1309774|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|$]]

The other answers are correct but if you're doing numerical work, you might want to consider numpy. Using numpy, the answer is:

<pre><code>v = numpy.diff(t)
</code></pre>

blocks|key|1094579|text|如果您不想使用numpy或zip，可以使用以下解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1094580|>>>+t+=+[1,+3,+6]
>>>+v+=+[t[i%2B1]-t[i]+for+i+in+range(len(t)-1)]
>>>+v
[2,+3]|code-block|syntax|javascript|1094581|entityMap^0|7|5|D|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

If you don't want to use <code>numpy</code> nor <code>zip</code>, you can use the following solution:

<pre><code>&gt;&gt;&gt; t = [1, 3, 6]
&gt;&gt;&gt; v = [t[i+1]-t[i] for i in range(len(t)-1)]
&gt;&gt;&gt; v
[2, 3]
</code></pre>

blocks|key|1310022|text|从Python+3.10开始，使用新的pairwise函数，可以在元素对之间滑动，从而在滚动对上映射：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1310023|from+itertools+import+pairwise

[y-x+for+(x,+y)+in+pairwise([1,+3,+6,+7])]
#+[2,+3,+1]|code-block|syntax|javascript|1310024|1310025|中间结果是：|1310026|pairwise([1,+3,+6,+7])
#+[(1,+3),+(3,+6),+(6,+7)]|1310027|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.10/library/itertools.html?highlight=pairwise#itertools.pairwise^0|1|B|J|8|J|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|17|8|@]|D|@]|E|$]]|$1|N|3|O|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

Starting in <code>Python 3.10</code>, with the new <a href="https://docs.python.org/3.10/library/itertools.html?highlight=pairwise#itertools.pairwise" rel="nofollow noreferrer"><code>pairwise</code></a> function it's possible to slide through pairs of elements and thus map on rolling pairs:
<pre><code>from itertools import pairwise

[y-x for (x, y) in pairwise([1, 3, 6, 7])]
# [2, 3, 1]
</code></pre>
<hr />
The intermediate result being:
<pre><code>pairwise([1, 3, 6, 7])
# [(1, 3), (3, 6), (6, 7)]
</code></pre>

blocks|key|1309909|text|我建议使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309910|v+=+np.diff(t)|code-block|syntax|javascript|1309911|这篇文章很简单，很容易阅读。|1309912|但是，如果您希望v具有与t相同的长度，那么|offset|length|style|CODE|1309913|v+=+np.diff([t[0]]+%2B+t)+#+for+python+3.x|1309914|或|1309915|v+=+np.diff(t+%2B+[t[-1]])|1309916|仅供参考:此功能仅适用于列表。|1309917|对于numpy数组|1309918|v+=+np.diff(np.append(t[0],+t))|1309919|entityMap^0|0|0|0|8|1|C|1|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|15|8|@$K|16|L|17|M|N]|$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|10|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|11|$]]

I would suggest using

<pre><code>v = np.diff(t)
</code></pre>

this is simple and easy to read. 

But if you want <code>v</code> to have the same length as <code>t</code> then

<pre><code>v = np.diff([t[0]] + t) # for python 3.x
</code></pre>

or

<pre><code>v = np.diff(t + [t[-1]])
</code></pre>

FYI: this will only work for lists.

for numpy arrays

<pre><code>v = np.diff(np.append(t[0], t))
</code></pre>

blocks|key|1094719|text|使用Python+3.8%2B中提供的:=+walrus运算符：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1094720|>>>+t+=+[1,+3,+6]
>>>+prev+=+t[0];+[-prev+%2B+(prev+:=+x)+for+x+in+t[1:]]
[2,+3]|code-block|syntax|javascript|1094721|entityMap^0|H|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Using the <code>:=</code> walrus operator available in Python 3.8+:

<pre><code>&gt;&gt;&gt; t = [1, 3, 6]
&gt;&gt;&gt; prev = t[0]; [-prev + (prev := x) for x in t[1:]]
[2, 3]
</code></pre>

blocks|key|1309734|text|好的。我想我找到了正确的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309735|v+=+[x[0]-x[1]+for+x+in+zip(t[1:],t[:-1])]|code-block|syntax|javascript|1309736|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|$]]

Ok. I think I found the proper solution:
<pre><code>v = [x[0]-x[1] for x in zip(t[1:],t[:-1])]
</code></pre>

blocks|key|1094651|text|函数式方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094652|>>>+import+operator
>>>+a+=+[1,3,5,7,11,13,17,21]
>>>+map(operator.sub,+a[1:],+a[:-1])
[2,+2,+2,+4,+2,+4,+4]|code-block|syntax|javascript|1094653|使用生成器：|1094654|>>>+import+operator,+itertools
>>>+g1,g2+=+itertools.tee((x*x+for+x+in+xrange(5)),2)
>>>+list(itertools.imap(operator.sub,+itertools.islice(g1,1,None),+g2))
[1,+3,+5,+7]|1094655|使用索引：|1094656|>>>+[a[i%2B1]-a[i]+for+i+in+xrange(len(a)-1)]
[2,+2,+2,+4,+2,+4,+4]|1094657|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

A functional approach:

<pre><code>&gt;&gt;&gt; import operator
&gt;&gt;&gt; a = [1,3,5,7,11,13,17,21]
&gt;&gt;&gt; map(operator.sub, a[1:], a[:-1])
[2, 2, 2, 4, 2, 4, 4]
</code></pre>

Using generator:

<pre><code>&gt;&gt;&gt; import operator, itertools
&gt;&gt;&gt; g1,g2 = itertools.tee((x*x for x in xrange(5)),2)
&gt;&gt;&gt; list(itertools.imap(operator.sub, itertools.islice(g1,1,None), g2))
[1, 3, 5, 7]
</code></pre>

Using indices:

<pre><code>&gt;&gt;&gt; [a[i+1]-a[i] for i in xrange(len(a)-1)]
[2, 2, 2, 4, 2, 4, 4]
</code></pre>

blocks|key|1094904|text|还可以使用以下命令将差异转换为易于阅读的转换矩阵|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094905|v+=+t.reshape((c,r)).T+-+t.T|code-block|syntax|javascript|1094906|其中c+=列表中的项数，r+=1，因为列表基本上是一个向量或一维数组。|offset|length|style|CODE|1094907|entityMap^0|0|0|2|1|C|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

You can also convert the difference into an easily readable transition matrix using
<pre><code>v = t.reshape((c,r)).T - t.T
</code></pre>
Where <code>c</code> = number of items in the list and <code>r</code> = 1 since a list is basically a vector or a 1d array.

blocks|key|1309838|text|我的方式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309839|>>>v+=+[1,2,3,4,5]
>>>[v[i]+-+v[i-1]+for+i,+value+in+enumerate(v[1:],+1)]
[1,+1,+1,+1]|code-block|syntax|javascript|1309840|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|$]]

My way

<pre><code>&gt;&gt;&gt;v = [1,2,3,4,5]
&gt;&gt;&gt;[v[i] - v[i-1] for i, value in enumerate(v[1:], 1)]
[1, 1, 1, 1]
</code></pre>

Given a list of numbers, how does one find differences between every (<code>i</code>)-th elements and its (<code>i+1</code>)-th?

Is it better to use a <code>lambda</code> expression or maybe a list comprehension?

For example: 

Given a list <code>t=[1,3,6,...]</code>, the goal is to find a list <code>v=[2,3,...]</code> because <code>3-1=2</code>, <code>6-3=3</code>, etc.

Python: Finding differences between elements of a list

Python

给定一个数字列表，如何找到每个(i)-th元素和它的(i+1)-th之间的差异？使用lambda表达式还是使用列表理解更好？例如：给定一个列表t=[1,3,6,...]，目标是找到一个列表v=[2,3,...]，因为3-1=2、6-3=3等。

问Python:查找列表元素之间的差异
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回答 10

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