blocks|key|1537611|text|import+itertools
itertools.flatten(+li+)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1537612|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>import itertools
itertools.flatten( li )
</code></pre>

blocks|key|1752638|text|有几个选择。首先，您可以只创建一个新列表，并将每个列表的内容添加到其中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1752639|li2+=+[]
for+sublist+in+li:
++++li2.extend(sublist)|code-block|syntax|javascript|1752640|或者，您可以使用itertools模块的chain函数，该函数生成一个包含多个迭代器中所有项的迭代器：|offset|length|style|CODE|1752641|import+itertools
li2+=+list(itertools.chain(*li))|1752642|如果采用这种方法，则可以在不创建中间列表的情况下生成字符串：|1752643|s+=+",".join(itertools.chain(*li))|1752644|entityMap^0|0|0|8|9|K|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

There's a couple choices. First, you can just create a new list and add the contents of each list to it:

<pre><code>li2 = []
for sublist in li:
 li2.extend(sublist)
</code></pre>

Alternately, you can use the <code>itertools</code> module's <code>chain</code> function, which produces an iterable containing all the items in multiple iterables:

<pre><code>import itertools
li2 = list(itertools.chain(*li))
</code></pre>

If you take this approach, you can produce the string without creating an intermediate list:

<pre><code>s = ",".join(itertools.chain(*li))
</code></pre>

blocks|key|1752730|text|对于第二个，有一个内置的string方法可以做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1752731|>>>+print+','.join(str(x)+for+x+in+li2)
"0,1,2,3,4,5,6,7,8"|code-block|syntax|javascript|1752732|对于第一个，您可以在理解列表中使用join：|1752733|>>>+print+",".join([",".join(str(x)+for+x+in+li])
"0,1,2,3,4,5,6,7,8"|1752734|但它更容易使用itertools.flatten：|1752735|>>>+import+itertools
>>>+print+itertools.flatten(li)
[0,1,2,3,4,5,6,7,8]
>>>+print+",".join(str(x)+for+x+in+itertools.flatten(li))
"0,1,2,3,4,5,6,7,8"|1752736|注意:+itertools是一个用迭代器来帮助你处理常见任务的模块，比如列表、元组或字符串……它很方便，因为它不存储正在处理的结构的副本，而是一个接一个地处理项目。|1752737|编辑:有趣的是，我正在学习很多方法来做到这一点。谁说只有一种好的方法呢？|1752738|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

For the second one, there is a built-in string method to do that :

<pre><code>&gt;&gt;&gt; print ','.join(str(x) for x in li2)
"0,1,2,3,4,5,6,7,8"
</code></pre>

For the first one, you can use join within a comprehension list :

<pre><code>&gt;&gt;&gt; print ",".join([",".join(str(x) for x in li])
"0,1,2,3,4,5,6,7,8"
</code></pre>

But it's easier to use itertools.flatten :

<pre><code>&gt;&gt;&gt; import itertools
&gt;&gt;&gt; print itertools.flatten(li)
[0,1,2,3,4,5,6,7,8]
&gt;&gt;&gt; print ",".join(str(x) for x in itertools.flatten(li))
"0,1,2,3,4,5,6,7,8"
</code></pre>

N.B : itertools is a module that help you to deal with common tasks with iterators such as list, tuples or string... It's handy because it does not store a copy of the structure you're working on but process the items one by one.

EDIT : funny, I am learning plenty of way to do it. Who said that there was only one good way to do it ?

blocks|key|1537868|text|我最喜欢的，也是最短的，是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1537869|li2+=+sum(li,+[])|code-block|syntax|javascript|1537870|和|1537871|s+=+','.join(li2)|1537872|编辑:使用sum而不是reduce+(感谢Thomas+Wouters!)|offset|length|style|CODE|1537873|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/17624/thomas-wouters^0|0|0|0|0|5|3|B|6|L|E|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]]|$1|K|3|L|5|6|7|12|8|@$M|13|N|14|O|P]|$M|15|N|16|O|P]]|9|@$M|17|N|18|1|19]]|A|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

My favorite, and the shortest one, is this:

<pre><code>li2 = sum(li, [])
</code></pre>

and

<pre><code>s = ','.join(li2)
</code></pre>

EDIT: use <code>sum</code> instead of <code>reduce</code>, (thanks <a href="https://stackoverflow.com/users/17624/thomas-wouters">Thomas Wouters</a>!)

blocks|key|1752891|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1752892|[+item+for+innerlist+in+outerlist+for+item+in+innerlist+]|code-block|syntax|javascript|1752893|将其直接转换为带分隔符的字符串：|1752894|','.join(str(item)+for+innerlist+in+outerlist+for+item+in+innerlist)|1752895|是的，'for+innerlist+in+outerlist‘和'for+item+in+innerlist’的顺序是正确的。即使循环的"body“位于listcomp的开头，嵌套循环(和'if‘子句)的顺序仍然与您将循环写出时相同：|1752896|for+innerlist+in+outerlist:
++++for+item+in+innerlist:
++++++++...|1752897|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|$]]

Like so:

<pre><code>[ item for innerlist in outerlist for item in innerlist ]
</code></pre>

Turning that directly into a string with separators:

<pre><code>','.join(str(item) for innerlist in outerlist for item in innerlist)
</code></pre>

Yes, the order of 'for innerlist in outerlist' and 'for item in innerlist' is correct. Even though the "body" of the loop is at the start of the listcomp, the order of nested loops (and 'if' clauses) is still the same as when you would write the loop out:

<pre><code>for innerlist in outerlist:
 for item in innerlist:
 ...
</code></pre>

blocks|key|1538058|text|试一试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538059|li=[[0,1,2],[3,4,5],[6,7,8]]
li2+=+[+y+for+x+in+li+for+y+in+x]|code-block|syntax|javascript|1538060|你可以这样读它：|1538061|给我每一笔钱的清单。|1538062|Y来自X。|1538063|X来自li。|1538064|要在字符串中映射它，请执行以下操作：|1538065|','.join(map(str,li2))|1538066|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Try that:

<pre><code>li=[[0,1,2],[3,4,5],[6,7,8]]
li2 = [ y for x in li for y in x]
</code></pre>

You can read it like this: 
Give me the list of every ys. 
The ys come from the xs. 
The xs come from li.

To map that in a string:

<pre><code>','.join(map(str,li2))
</code></pre>

blocks|key|1538141|text|有很多方法可以解决这个问题。我喜欢Numpy的工具，因为它通常已经在我做的任何事情中导入了，但是，如果你没有使用Numpy做其他事情，这可能不是一个好方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1538142|import+numpy
li+=+[[0,1,2],[3,4,5],[6,7,8]]
li2=li[0]+#first+element+of+array+to+merge
i=1+
while+i<len(li):
++++li2=numpy.concatenate((li2,li[i]))
++++i%2B=1
print+li2|code-block|syntax|javascript|1538143|这将打印0+1+2+3+4+5+6+7+8，然后您也可以将其转换为您的字符串。|1538144|entityMap|0|LINK|mutability|MUTABLE|url|http://numpy.scipy.org/^0|H|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

There are many ways to do this problem. I like <a href="http://numpy.scipy.org/" rel="nofollow noreferrer">Numpy</a>'s tools because it is normally already imported in everything I do. However, if you aren't using Numpy for anything else this probably isn't a good method.

<pre><code>import numpy
li = [[0,1,2],[3,4,5],[6,7,8]]
li2=li[0] #first element of array to merge
i=1 
while i&lt;len(li):
 li2=numpy.concatenate((li2,li[i]))
 i+=1
print li2
</code></pre>

This would print [0 1 2 3 4 5 6 7 8] and then you can convert this into your string too.

blocks|key|1538149|text|以下是一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538150|def+convert2DArrtostring(ndArr):
++++'''converts+2D+array+to+string'''
++++arr_str+=+"["
++++for+i+in+ndArr:
++++++++arr_str+%2B=+"["
++++++++for+j+in+i:
++++++++++++arr_str+%2B=+str(j)+%2B+"+"
++++++++arr_str+%2B=+"]\n"
++++arr_str+%2B=+"]"
++++return+arr_str|code-block|syntax|javascript|1538151|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|$]]

Here is a way:
<pre><code>def convert2DArrtostring(ndArr):
 '''converts 2D array to string'''
 arr_str = &quot;[&quot;
 for i in ndArr:
 arr_str += &quot;[&quot;
 for j in i:
 arr_str += str(j) + &quot; &quot;
 arr_str += &quot;]\n&quot;
 arr_str += &quot;]&quot;
 return arr_str
</code></pre>

List Comprehension for me seems to be like the opaque block of granite that regular expressions are for me. I need pointers.

Say, I have a 2D list:

<pre><code>li = [[0,1,2],[3,4,5],[6,7,8]]
</code></pre>

I would like to merge this either into one long list

<pre><code>li2 = [0,1,2,3,4,5,6,7,8]
</code></pre>

or into a string with separators:

<pre><code>s = "0,1,2,3,4,5,6,7,8"
</code></pre>

Really, I'd like to know how to do both.

How do I merge a 2D array in Python into one string with List Comprehension?

列表理解对我来说就像是一块不透明的花岗岩，正则表达式就是我的。我需要指点。比方说，我有一个2D列表：li = [[0,1,2],[3,4,5],[6,7,8]]我想把它合并成一个长长的列表li2 = [0,1,2,3,4,5,6,7,8]或转换为带分隔符的字符串：s = "0,1,2,3,4,5,6,7,8"真的，我想...

问如何将Python中的二维数组合并为一个具有列表理解功能的字符串？
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