blocks|key|3248373|text|itertools.chain()应该做到这一点。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3248374|它获取可迭代对象的列表，并逐个从每个可迭代对象中获取结果。|3248375|def+chain(*iterables):
++++for+it+in+iterables:
++++++++for+element+in+it:
++++++++++++yield+element|code-block|syntax|javascript|3248376|使用示例：|3248377|from+itertools+import+chain
generator+=+chain('ABC',+'DEF')

for+item+in+generator:
++++print(item)|3248378|输出：|3248379|A
B
C
D
E
F|3248380|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html#itertools.chain^0|0|H|0|H|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1A|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1C|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

<a href="http://docs.python.org/library/itertools.html#itertools.chain" rel="noreferrer"><code>itertools.chain()</code></a> should do it.
It takes a list of iterables and yields from each one by one
<pre class="lang-py prettyprint-override"><code>def chain(*iterables):
 for it in iterables:
 for element in it:
 yield element
</code></pre>
Usage example:
<pre><code>from itertools import chain
generator = chain('ABC', 'DEF')

for item in generator:
 print(item)
</code></pre>
Output:
<pre><code>A
B
C
D
E
F
</code></pre>

blocks|key|3248385|text|下面是一个代码示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3248386|from+itertools+import+chain

def+generator1():
++++for+item+in+'abcdef':
++++++++yield+item

def+generator2():
++++for+item+in+'123456':
++++++++yield+item

generator3+=+chain(generator1(),+generator2())
for+item+in+generator3:
++++print+item|code-block|syntax|javascript|3248387|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|$]]

A example of code:

<pre><code>from itertools import chain

def generator1():
 for item in 'abcdef':
 yield item

def generator2():
 for item in '123456':
 yield item

generator3 = chain(generator1(), generator2())
for item in generator3:
 print item
</code></pre>

blocks|key|3034239|text|使用itertools.chain.from_iterable，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3034240|def+genny(start):
++for+x+in+range(start,+start%2B3):
++++yield+x

y+=+[1,+2]
ab+=+[o+for+o+in+itertools.chain.from_iterable(genny(x)+for+x+in+y)]
print(ab)|code-block|syntax|javascript|3034241|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|$]]

With itertools.chain.from_iterable you can do things like:

<pre><code>def genny(start):
 for x in range(start, start+3):
 yield x

y = [1, 2]
ab = [o for o in itertools.chain.from_iterable(genny(x) for x in y)]
print(ab)
</code></pre>

blocks|key|3034246|text|简单的例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3034247|from+itertools+import+chain
x+=+iter([1,2,3])++++++#Create+Generator+Object+(listiterator)
y+=+iter([3,4,5])++++++#another+one
result+=+chain(x,+y)+++#Chained+x+and+y|code-block|syntax|javascript|3034248|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|$]]

<h1>Simple example:</h1>
<pre><code>from itertools import chain
x = iter([1,2,3]) #Create Generator Object (listiterator)
y = iter([3,4,5]) #another one
result = chain(x, y) #Chained x and y
</code></pre>

blocks|key|303713|text|如果你想保持生成器的分离，同时仍然迭代它们，你可以使用zip()：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|303714|注意:迭代在两个生成器中较短的一个处停止|303715|例如：|303716|for+(root1,+dir1,+files1),+(root2,+dir2,+files2)+in+zip(os.walk(path1),+os.walk(path2)):

++++for+file+in+files1:
++++++++#do+something+with+first+list+of+files

++++for+file+in+files2:
++++++++#do+something+with+second+list+of+files|code-block|syntax|javascript|303717|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you want to keep the generators separate but still iterate over them at the same time you can use zip():

NOTE: Iteration stops at the shorter of the two generators

For example:

<pre><code>for (root1, dir1, files1), (root2, dir2, files2) in zip(os.walk(path1), os.walk(path2)):

 for file in files1:
 #do something with first list of files

 for file in files2:
 #do something with second list of files
</code></pre>

blocks|key|3248484|text|假设我们必须生成生成器(gen1和gen+2)，并且我们想要执行一些额外的计算，需要这两个生成器的结果。我们可以通过map方法返回此类函数/计算的结果，而map方法又返回一个我们可以循环的生成器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3248485|在此场景中，函数/计算需要通过lambda函数实现。棘手的部分是我们打算在map及其lambda函数中做的事情。|3248486|建议解决方案的一般形式：|3248487|def+function(gen1,gen2):
++++++++for+item+in+map(lambda+x,+y:+do_somethin(x,y),+gen1,+gen2):
++++++++++++yield+item|code-block|syntax|javascript|3248488|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Lets say that we have to generators (gen1 and gen 2) and we want to perform some extra calculation that requires the outcome of both.
We can return the outcome of such function/calculation through the map method, which in turn returns a generator that we can loop upon. 

In this scenario, the function/calculation needs to be implemented via the lambda function.
The tricky part is what we aim to do inside the map and its lambda function.

General form of proposed solution:

<pre><code>def function(gen1,gen2):
 for item in map(lambda x, y: do_somethin(x,y), gen1, gen2):
 yield item
</code></pre>

blocks|key|303752|text|在Python+(3.5或更高版本)中，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|303753|def+concat(a,+b):
++++yield+from+a
++++yield+from+b|code-block|syntax|javascript|303754|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|$]]

In Python (3.5 or greater) you can do:

<pre><code>def concat(a, b):
 yield from a
 yield from b
</code></pre>

blocks|key|3034360|text|这里它使用了一个带有嵌套for的generator+expression：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3034361|a+=+range(3)
b+=+range(5)
ab+=+(i+for+it+in+(a,+b)+for+i+in+it)
assert+list(ab)+==+[0,+1,+2,+0,+1,+2,+3,+4]|code-block|syntax|javascript|3034362|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0289/#the-details^0|C|3|G|K|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]]]]

Here it is using a <a href="https://www.python.org/dev/peps/pep-0289/#the-details" rel="noreferrer">generator expression</a> with nested <code>for</code>s:

<pre><code>a = range(3)
b = range(5)
ab = (i for it in (a, b) for i in it)
assert list(ab) == [0, 1, 2, 0, 1, 2, 3, 4]
</code></pre>

blocks|key|305848|text|如果你只需要做一次并且不想再导入一个模块，有一个简单的解决方案……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|305849|只需这样做：|305850|for+dir+in+directory_1,+directory_2:
++++for+directory,+dirs,+files+in+os.walk(dir):
++++++++do_something()|code-block|syntax|javascript|305851|如果你真的想“连接”这两个生成器，那么你可以这样做：|305852|for+directory,+dirs,+files+in+(
++++++++x+for+osw+in+[os.walk(directory_1),+os.walk(directory_2)]+
+++++++++++++++for+x+in+osw
++++++++):
++++do_something()|305853|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you just need to do it once and do not wish to import one more module, there is a simple solutions...
just do:
<pre><code>for dir in directory_1, directory_2:
 for directory, dirs, files in os.walk(dir):
 do_something()
</code></pre>
If you really want to &quot;join&quot; both generators, then do :
<pre><code>for directory, dirs, files in (
 x for osw in [os.walk(directory_1), os.walk(directory_2)] 
 for x in osw
 ):
 do_something()
</code></pre>

blocks|key|300709|text|还可以使用解包运算符*|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|300710|concat+=+(*gen1(),+*gen2())|code-block|syntax|javascript|300711|注意:对于“非懒惰”的可迭代程序，它的工作效率最高。也可以用于不同类型的理解。生成器连接的首选方法是来自@Uduse的答案|300712|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/47592164/1774524^0|A|1|0|0|1G|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

One can also use unpack operator <code>*</code>:

<pre><code>concat = (*gen1(), *gen2())
</code></pre>

NOTE: Works most efficiently for 'non-lazy' iterables. Can also be used with different kind of comprehensions. Preferred way for generator concat would be from the answer from <a href="https://stackoverflow.com/a/47592164/1774524">@Uduse</a>

blocks|key|3034463|text|我想说，正如用户"wjandrea“在评论中所建议的那样，最好的解决方案是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3034464|def+concat_generators(*args):
++++for+gen+in+args:
++++++++yield+from+gen|code-block|syntax|javascript|3034465|它不会改变返回的类型，而且是真正的pythonic。|3034466|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would say that, as suggested in comments by user &quot;wjandrea&quot;, the best solution is
<pre><code>def concat_generators(*args):
 for gen in args:
 yield from gen
</code></pre>
It does not change the returned type and is really pythonic.

blocks|key|305895|text|python+2020更新:同时适用于3和python+2|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|305896|import+itertools

iterA+=+range(10,15)
iterB+=+range(15,20)
iterC+=+range(20,25)|code-block|syntax|javascript|305897|###+first+option

for+i+in+itertools.chain(iterA,+iterB,+iterC):
++++print(i)

#+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24|305898|###+alternative+option,+introduced+in+python+2.6

for+i+in+itertools.chain.from_iterable(+[iterA,+iterB,+iterC]+):
++++print(i)

#+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24|305899|itertools.chain()是基本的。|305900|如果你有一个迭代式的迭代器，itertools.chain.from_iterables是很方便的。例如，每个子目录的文件列表，如[+["src/server.py",+"src/readme.txt"],+["test/test.py"]+]。|CODE|305901|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/itertools.html#itertools.chain|1|https://docs.python.org/3/library/itertools.html#itertools.chain.from_iterable^0|J|A|0|0|0|0|0|H|0|0|1T|1L|E|U|1|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|Q|3|R|5|6|7|1D|8|@$9|1E|A|1F|B|S]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|T|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10]]|11|$5|W|X|Y|E|$Z|12]]]]

2020 update: Work in both python 3 and python 2
<pre><code>import itertools

iterA = range(10,15)
iterB = range(15,20)
iterC = range(20,25)
</code></pre>
<pre><code>### first option

for i in itertools.chain(iterA, iterB, iterC):
 print(i)

# 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
</code></pre>
<pre><code>### alternative option, introduced in python 2.6

for i in itertools.chain.from_iterable( [iterA, iterB, iterC] ):
 print(i)

# 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
</code></pre>
<a href="https://docs.python.org/3/library/itertools.html#itertools.chain" rel="nofollow noreferrer">itertools.chain()</a> is the basic.
<a href="https://docs.python.org/3/library/itertools.html#itertools.chain.from_iterable" rel="nofollow noreferrer">itertools.chain.from_iterables</a> is handy if you have an iterable of iterables. For example a list of files per subdirectory like <code>[ [&quot;src/server.py&quot;, &quot;src/readme.txt&quot;], [&quot;test/test.py&quot;] ]</code>.

blocks|key|303069|text|(免责声明:仅限Python+3！)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|303070|语法类似于您想要的是使用splat运算符来展开两个生成器：|303071|for+directory,+dirs,+files+in+(*os.walk(directory_1),+*os.walk(directory_2)):
++++do_something()|code-block|syntax|javascript|303072|解释：|303073|这有效地将两个生成器单级展平为N元组的3元组(来自os.walk)，如下所示：|offset|length|style|CODE|303074|((directory1,+dirs1,+files1),+(directory2,+dirs2,+files2),+...)|303075|然后，您的for循环将遍历这个N元组。|303076|当然，通过简单地将外括号替换为括号，您可以获得3元组的列表，而不是3元组的N元组：|303077|for+directory,+dirs,+files+in+[*os.walk(directory_1),+*os.walk(directory_2)]:
++++do_something()|303078|这会产生类似如下的结果：|303079|[(directory1,+dirs1,+files1),+(directory2,+dirs2,+files2),+...]|303080|优点：|303081|这种方法的优点是不需要导入任何东西，代码也不多。|303082|缺点：|303083|缺点是您将两个生成器转储到一个集合中，然后迭代该集合，从而有效地执行两次遍历，并潜在地使用大量内存。|303084|entityMap^0|0|0|0|0|P|7|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1F|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1G|8|@$M|1H|N|1I|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1J|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1K|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1L|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1M|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|F|7|1O|8|@]|9|@]|A|$G|H]]|$1|12|3|13|5|6|7|1P|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|1R|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|1S|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|1B|$]]

(Disclaimer: Python 3 only!)
Something with syntax similar to what you want is to use the splat operator to expand the two generators:
<pre class="lang-py prettyprint-override"><code>for directory, dirs, files in (*os.walk(directory_1), *os.walk(directory_2)):
 do_something()
</code></pre>
<h3>Explanation:</h3>
This effectively performs a single-level flattening of the two generators into an N-tuple of 3-tuples (from <code>os.walk</code>) that looks like:
<pre class="lang-py prettyprint-override"><code>((directory1, dirs1, files1), (directory2, dirs2, files2), ...)
</code></pre>
Your for-loop then iterates over this N-tuple.
Of course, by simply replacing the outer parentheses with brackets, you can get a list of 3-tuples instead of an N-tuple of 3-tuples:
<pre class="lang-py prettyprint-override"><code>for directory, dirs, files in [*os.walk(directory_1), *os.walk(directory_2)]:
 do_something()
</code></pre>
This yields something like:
<pre class="lang-py prettyprint-override"><code>[(directory1, dirs1, files1), (directory2, dirs2, files2), ...]
</code></pre>
<h3>Pro:</h3>
The upside to this approach is that you don't have to import anything and it's not a lot of code.
<h3>Con:</h3>
The downside is that you dump two generators into a collection and then iterate over that collection, effectively doing two passes and potentially using a lot of memory.

I want to change the following code

<pre><code>for directory, dirs, files in os.walk(directory_1):
 do_something()

for directory, dirs, files in os.walk(directory_2):
 do_something()
</code></pre>

to this code:

<pre><code>for directory, dirs, files in os.walk(directory_1) + os.walk(directory_2):
 do_something()
</code></pre>

I get the error:

<blockquote>
 unsupported operand type(s) for +: 'generator' and 'generator'
</blockquote>

How to join two generators in Python?

How to join two generators in Python?

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋 

腾讯云代码助手

CODING DevOps

Cloud Studio

SDK中心

API中心

命令行工具

我想要更改以下代码for directory, dirs, files in os.walk(directory_1):    do_something()for directory, dirs, files in os.walk(directory_2):    do_something()这段代码：for dire...

问如何在Python中连接两个生成器？
EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Python中连接两个生成器？EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Python中连接两个生成器？
EN