blocks|key|1783938|text|for+n+in+range(6,0,-1):
++++print+n
#+prints+[6,+5,+4,+3,+2,+1]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1783939|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>for n in range(6,0,-1):
 print n
# prints [6, 5, 4, 3, 2, 1]
</code></pre>

blocks|key|1783991|text|这已经很晚了，但我只想补充一下，还有一种更好的方法:使用reversed|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1783992|for+i+in+reversed(range(10)):
++++print+i|code-block|syntax|javascript|1783993|提供：|1783994|4
3
2
1
0|1783995|entityMap^0|S|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

This is very late, but I just wanted to add that there is a more elegant way: using <code>reversed</code> 

<pre><code>for i in reversed(range(10)):
 print i
</code></pre>

gives:

<pre><code>4
3
2
1
0
</code></pre>

blocks|key|1784013|text|for+n+in+range(6,0,-1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1784014|这将为您提供6,5,4,3,2,1|unstyled|offset|length|style|CODE|1784015|至于|1784016|for+n+in+reversed(range(0,6))|1784017|会给你5,4,3,2,1,0|1784018|entityMap^0|0|6|B|0|0|0|3|B|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@$G|U|H|V|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|Y|8|@$G|Z|H|10|I|J]]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>for n in range(6,0,-1)
</code></pre>

This would give you <code>6,5,4,3,2,1</code>

As for

<pre><code>for n in reversed(range(0,6))
</code></pre>

would give you <code>5,4,3,2,1,0</code>

blocks|key|1997696|text|for+n+in+range(6,0,-1):
++++print+n|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1997697|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>for n in range(6,0,-1):
 print n
</code></pre>

blocks|key|1997704|text|>>>+range(6,+0,+-1)
[6,+5,+4,+3,+2,+1]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1997705|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; range(6, 0, -1)
[6, 5, 4, 3, 2, 1]
</code></pre>

blocks|key|1784036|text|0是条件值当此条件为真时，循环将继续执行。10是初始值。1是修饰符，其中可以是简单的减量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1784037|for+number+in+reversed(range(0,10,1)):
print+number;|code-block|syntax|javascript|1784038|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|$]]

0 is conditional value when this condition is true, loop will keep executing.10 is the initial value. 1 is the modifier where may be simple decrement.

<pre><code>for number in reversed(range(0,10,1)):
print number;
</code></pre>

blocks|key|1784067|text|虽然晚了一步，但对于任何要创建自己的函数或想看看它是如何工作的人来说，这里有一个额外的功能，可以根据所需的增量重新排列开始-停止值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1784068|def+RANGE(start,+stop=None,+increment=1):
++++if+stop+is+None:
++++++++stop+=+start
++++++++start+=+1

++++value_list+=+sorted([start,+stop])

++++if+increment+==+0:
++++++++print('Error!+Please+enter+nonzero+increment+value!')
++++else:
++++++++value_list+=+sorted([start,+stop])
++++++++if+increment+<+0:
++++++++++++start+=+value_list[1]
++++++++++++stop+=+value_list[0]
++++++++++++while+start+>=+stop:
++++++++++++++++worker+=+start
++++++++++++++++start+%2B=+increment
++++++++++++++++yield+worker
++++++++else:
++++++++++++start+=+value_list[0]
++++++++++++stop+=+value_list[1]
++++++++++++while+start+<+stop:
++++++++++++++++worker+=+start
++++++++++++++++start+%2B=+increment
++++++++++++++++yield+worker|code-block|syntax|javascript|1784069|负增量：|1784070|for+i+in+RANGE(1,+10,+-1):
++++print(i)|1784071|或者，反转开始-停止：|1784072|for+i+in+RANGE(10,+1,+-1):
++++print(i)|1784073|输出：|1784074|10
9
8
7
6
5
4
3
2
1|1784075|常规增量：|1784076|for+i+in+RANGE(1,+10):
++++print(i)|1784077|1784078|1
2
3
4
5
6
7
8
9|1784079|零增量：|1784080|for+i+in+RANGE(1,+10,+0):
++++print(i)|1784081|1784082|'Error!+Please+enter+nonzero+increment+value!'|1784083|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|W|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|Z|3|10|5|6|7|1K|8|@]|9|@]|A|$]]|$1|11|3|12|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|13|3|P|5|6|7|1M|8|@]|9|@]|A|$]]|$1|14|3|15|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|16|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|17|$]]

Late to the party, but for anyone tasked with creating their own or wants to see how this would work, here's the function with an added bonus of rearranging the start-stop values based on the desired increment:

<pre><code>def RANGE(start, stop=None, increment=1):
 if stop is None:
 stop = start
 start = 1

 value_list = sorted([start, stop])

 if increment == 0:
 print('Error! Please enter nonzero increment value!')
 else:
 value_list = sorted([start, stop])
 if increment &lt; 0:
 start = value_list[1]
 stop = value_list[0]
 while start &gt;= stop:
 worker = start
 start += increment
 yield worker
 else:
 start = value_list[0]
 stop = value_list[1]
 while start &lt; stop:
 worker = start
 start += increment
 yield worker
</code></pre>

Negative increment:

<pre><code>for i in RANGE(1, 10, -1):
 print(i)
</code></pre>

Or, with start-stop reversed:

<pre><code>for i in RANGE(10, 1, -1):
 print(i)
</code></pre>

Output:

<pre><code>10
9
8
7
6
5
4
3
2
1
</code></pre>

Regular increment:

<pre><code>for i in RANGE(1, 10):
 print(i)
</code></pre>

Output:

<pre><code>1
2
3
4
5
6
7
8
9
</code></pre>

Zero increment:

<pre><code>for i in RANGE(1, 10, 0):
 print(i)
</code></pre>

Output:

<pre><code>'Error! Please enter nonzero increment value!'
</code></pre>

blocks|key|1784098|text|For+python3，其中-1表示在每个步骤for+n+in+range(6,0,-1):+++++print(n)中要递减的值|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1784099|entityMap^0|E|2|N|10|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

For python3 where <code>-1</code> indicate the value that to be decremented in each step
<code>
for n in range(6,0,-1):
 print(n)
</code>

I could be wrong (just let me know and I'll delete the question) but it seems python won't respond to

<pre><code>for n in range(6,0):
 print n
</code></pre>

I tried using xrange and it didn't work either. How can I implement that?

Decreasing for loops in Python impossible?

我可能是错的(只要让我知道，我会删除这个问题)，但python似乎不会响应for n in range(6,0):    print n我试过使用xrange，它也不起作用。我该如何实现它呢？

问在Python中减少for循环是不可能的？
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