blocks|key|895012|text|diffs+=+[lst[x+%2B+1]+-+lst[x]+for+x+in+range(len(lst)+-+1)]
for+x+in+diffs:
++++print+x+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|895013|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>diffs = [lst[x + 1] - lst[x] for x in range(len(lst) - 1)]
for x in diffs:
 print x 
</code></pre>

blocks|key|2513364|text|it+=+iter(lst)
it.next()
print+[(x,+y,+y+-+x)+for+(x,+y)+in+itertools.izip(lst,+it)]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2513365|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>it = iter(lst)
it.next()
print [(x, y, y - x) for (x, y) in itertools.izip(lst, it)]
</code></pre>

blocks|key|4587477|text|使用zip，并用它自己压缩列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4587478|l+=+[1,+2,+4,+7]
[b+-+a+for+a,+b+in+zip(l,+l[1:])]

#+[1,+2,+3]|code-block|syntax|javascript|4587479|entityMap^0|2|3|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|$]]

Use <code>zip</code>, and zip the list with itself.

<pre><code>l = [1, 2, 4, 7]
[b - a for a, b in zip(l, l[1:])]

# [1, 2, 3]
</code></pre>

blocks|key|4587522|text|如果您正在操作数值数据，请考虑使用numpy|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4587523|import+numpy+as+np

lst+=+[3.18,10.57,14.95]
arr+=+np.array(lst)

diff+=+np.diff(arr)

>>>+diff
array([+7.39,++4.38])|code-block|syntax|javascript|4587524|如果有必要，您可以将其转换回list：|4587525|diff_list+=+list(diff)|4587526|否则，您可以迭代它，就像迭代列表一样：|4587527|for+item+in+diff:+
++++print(item)

7.39
4.38|4587528|编辑：我计时的五个解决方案非常接近，所以选择一个更容易阅读的解决方案|offset|length|style|BOLD|4587529|t+=+timeit.Timer("[b+-+a+for+a,+b+in+zip(l,+l[1:])]",+"l+=+range(int(1e6))")
print(t.timeit(1))
>>>+0.523894071579

t+=+timeit.Timer("list(np.diff(np.array(l)))",+"import+numpy+as+np;+l+=+range(int(1e6))")
print(t.timeit(1))
>>>+0.484916915894

t+=+timeit.Timer("diffs+=+[l[x+%2B+1]+-+l[x]+for+x+in+range(len(l)+-+1)]",+"l+=+range(int(1e6))")
print(t.timeit(1))
>>>+0.363043069839

t+=+timeit.Timer("[(x,+y,+y+-+x)+for+(x,+y)+in+itertools.izip(l,+it)]",+"l+=+range(int(1e6));+it+=+iter(l);+it.next()")
print(t.timeit(1))
>>>+0.54354596138

#+pairwise+solution
t+=+timeit.Timer("a,+b+=+itertools.tee(l);+next(b,+None);+[(x,+y)+for+x,+y+in+itertools.izip(a,+b)]",+"l+=+range(int(1e6));")
print(t.timeit(1))
>>>+0.477301120758|4587530|entityMap^0|0|0|0|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

If you are manipulating numerical data, consider using numpy

<pre><code>import numpy as np

lst = [3.18,10.57,14.95]
arr = np.array(lst)

diff = np.diff(arr)

&gt;&gt;&gt; diff
array([ 7.39, 4.38])
</code></pre>

You can convert it back to list if you have to:

<pre><code>diff_list = list(diff)
</code></pre>

Otherwise you can iterate over it just like you iterate over a list:

<pre><code>for item in diff: 
 print(item)

7.39
4.38
</code></pre>

EDIT: the five solutions I timed were pretty close to each other, so choose the one that's easier to read

<pre><code>t = timeit.Timer("[b - a for a, b in zip(l, l[1:])]", "l = range(int(1e6))")
print(t.timeit(1))
&gt;&gt;&gt; 0.523894071579

t = timeit.Timer("list(np.diff(np.array(l)))", "import numpy as np; l = range(int(1e6))")
print(t.timeit(1))
&gt;&gt;&gt; 0.484916915894

t = timeit.Timer("diffs = [l[x + 1] - l[x] for x in range(len(l) - 1)]", "l = range(int(1e6))")
print(t.timeit(1))
&gt;&gt;&gt; 0.363043069839

t = timeit.Timer("[(x, y, y - x) for (x, y) in itertools.izip(l, it)]", "l = range(int(1e6)); it = iter(l); it.next()")
print(t.timeit(1))
&gt;&gt;&gt; 0.54354596138

# pairwise solution
t = timeit.Timer("a, b = itertools.tee(l); next(b, None); [(x, y) for x, y in itertools.izip(a, b)]", "l = range(int(1e6));")
print(t.timeit(1))
&gt;&gt;&gt; 0.477301120758
</code></pre>

blocks|key|2513462|text|你需要来自itertools的pairwise()+recipe，很多Python的优点都来自于它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2513463|>>>+for+x,y+in+pairwise(lst):+
...+++++print(y,+"+-+",+x,+"+=+",+y+-+x)
...+
10.57++-++3.18++=++7.390000000000001
14.95++-++10.57++=++4.379999999999999|code-block|syntax|javascript|2513464|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html|1|http://docs.python.org/library/itertools.html#recipes^0|5|9|F|A|5|9|0|Q|6|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]|$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

You need the <code>pairwise()</code> <a href="http://docs.python.org/library/itertools.html#recipes" rel="nofollow">recipe</a> from <a href="http://docs.python.org/library/itertools.html" rel="nofollow"><code>itertools</code></a>, from where lots of Python goodness comes.

<pre><code>&gt;&gt;&gt; for x,y in pairwise(lst): 
... print(y, " - ", x, " = ", y - x)
... 
10.57 - 3.18 = 7.390000000000001
14.95 - 10.57 = 4.379999999999999
</code></pre>

How can I compare one element with the next one in a sorted list, and print out their differences. Any help would be appreciated.

<pre><code>Eg: 
lst = [3.18,10.57,14.95,...]
10.57 - 3.18 = 7.39
14.95 - 10.57 = 4.38
...
</code></pre>

compare list elements

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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如何将一个元素与排序列表中的下一个元素进行比较，并打印出它们之间的差异。任何帮助都将不胜感激。Eg: lst = [3.18,10.57,14.95,...]10.57 - 3.18 =  7.3914.95 - 10.57 = 4.38...

问比较列表元素
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