blocks|key|1556596|text|尝试使用range(100,-1,-1)，第三个参数是要使用的增量(有文档记录的here)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1556597|("range“选项，start，stop，step都记录在here中)|1556598|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/library/functions.html#range|1|https://docs.python.org/3/library/stdtypes.html?highlight=range#range^0|4|G|14|4|0|0|U|4|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|H|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

Try <code>range(100,-1,-1)</code>, the 3rd argument being the increment to use (documented <a href="https://docs.python.org/library/functions.html#range" rel="noreferrer">here</a>). 

("range" options, start, stop, step are documented <a href="https://docs.python.org/3/library/stdtypes.html?highlight=range#range" rel="noreferrer">here</a>)

blocks|key|1341141|text|在我看来，这是最具可读性的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341142|for+i+in+reversed(xrange(101)):
++++print+i,|code-block|syntax|javascript|1341143|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 my opinion, this is the most readable:

<pre><code>for i in reversed(xrange(101)):
 print i,
</code></pre>

blocks|key|1556648|text|for+i+in+range(100,+-1,+-1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1556649|和一些稍微更长(更慢)的解决方案：|unstyled|1556650|for+i+in+reversed(range(101))

for+i+in+range(101)[::-1]|1556651|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>for i in range(100, -1, -1)
</code></pre>

and some slightly longer (and slower) solution:

<pre><code>for i in reversed(range(101))

for i in range(101)[::-1]
</code></pre>

blocks|key|1341187|text|通常在Python中，您可以使用负索引从后面开始：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341188|numbers+=+[10,+20,+30,+40,+50]
for+i+in+xrange(len(numbers)):
++++print+numbers[-i+-+1]|code-block|syntax|javascript|1341189|结果：|1341190|50
40
30
20
10|1341191|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Generally in Python, you can use negative indices to start from the back:

<pre><code>numbers = [10, 20, 30, 40, 50]
for i in xrange(len(numbers)):
 print numbers[-i - 1]
</code></pre>

Result:

<pre><code>50
40
30
20
10
</code></pre>

blocks|key|1341253|text|另一种解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341254|z+=+10
for+x+in+range+(z):
+++y+=+z-x
+++print+y|code-block|syntax|javascript|1341255|结果：|1341256|10
9
8
7
6
5
4
3
2
1|1341257|提示:如果您使用此方法对列表中的索引进行反向计数，您将希望从'y‘值开始-1，因为您的列表索引将从0开始。|1341258|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Another solution:

<pre><code>z = 10
for x in range (z):
 y = z-x
 print y
</code></pre>

Result:

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

Tip:
If you are using this method to count back indices in a list, you will want to -1 from the 'y' value, as your list indices will begin at 0.

blocks|key|1341451|text|解决问题的简单答案可能是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341452|for+i+in+range(100):
++++k+=+100+-+i
++++print(k)|code-block|syntax|javascript|1341453|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 simple answer to solve your problem could be like this:

<pre><code>for i in range(100):
 k = 100 - i
 print(k)
</code></pre>

blocks|key|1341266|text|for+var+in+range(10,-1,-1)工作|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1341267|entityMap^0|0|Q|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>for var in range(10,-1,-1)</code> works

blocks|key|1341365|text|简短而甜蜜。这是我在上codeAcademy课程时的解决方案。按rev顺序打印字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341366|def+reverse(text):
++++string+=+""
++++for+i+in+range(len(text)-1,-1,-1):
++++++++string+%2B=+text[i]
++++return+string++++|code-block|syntax|javascript|1341367|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|$]]

Short and sweet. This was my solution when doing codeAcademy course. Prints a string in rev order. 

<pre><code>def reverse(text):
 string = ""
 for i in range(len(text)-1,-1,-1):
 string += text[i]
 return string 
</code></pre>

blocks|key|1341405|text|在你的例子100+-+i+i+in+range(+0,+101+)中，你总是可以从一个变量中增加范围和减去。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1341406|for+i+in+range(+0,+101+):
++++print+100+-+i|code-block|syntax|javascript|1341407|entityMap^0|5|7|D|K|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|$]]

You can always do increasing range and subtract from a variable in your case <code>100 - i</code> where <code>i in range( 0, 101 )</code>. 

<pre><code>for i in range( 0, 101 ):
 print 100 - i
</code></pre>

blocks|key|1341490|text|我想同时向后循环两个列表，所以我需要负索引。这是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341491|a=+[1,3,4,5,2]
for+i+in+range(-1,+-len(a),+-1):
++++print(i,+a[i])|code-block|syntax|javascript|1341492|结果：|1341493|-1+2
-2+5
-3+4
-4+3
-5+1|1341494|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I wanted to loop through a two lists backwards at the same time so I needed the negative index. This is my solution:

<pre><code>a= [1,3,4,5,2]
for i in range(-1, -len(a), -1):
 print(i, a[i])
</code></pre>

Result:

<pre><code>-1 2
-2 5
-3 4
-4 3
-5 1
</code></pre>

blocks|key|1341607|text|您还可以在python中创建自定义的反转机制。它可以在任何地方用来向后循环一个可迭代|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341608|class+Reverse:
++++"""Iterator+for+looping+over+a+sequence+backwards"""
++++def+__init__(self,+seq):
++++++++self.seq+=+seq
++++++++self.index+=+len(seq)

++++def+__iter__(self):
++++++++return+self

++++def+__next__(self):
++++++++if+self.index+==+0:
++++++++++++raise+StopIteration
++++++++self.index+-=+1
++++++++return+self.seq[self.index]


>>>+d+=+[1,2,3,4,5]
>>>+for+i+in+Reverse(d):
...+++print(i)
...+
5
4
3
2
1|code-block|syntax|javascript|1341609|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|$]]

You can also create a custom reverse mechanism in python. Which can be use anywhere for looping an iterable backwards

<pre><code>class Reverse:
 """Iterator for looping over a sequence backwards"""
 def __init__(self, seq):
 self.seq = seq
 self.index = len(seq)

 def __iter__(self):
 return self

 def __next__(self):
 if self.index == 0:
 raise StopIteration
 self.index -= 1
 return self.seq[self.index]


&gt;&gt;&gt; d = [1,2,3,4,5]
&gt;&gt;&gt; for i in Reverse(d):
... print(i)
... 
5
4
3
2
1
</code></pre>

blocks|key|1341339|text|我在一个codeacademy练习中尝试过(在字符串中反转字符，而不使用反转的nor+:：-1)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341340|def+reverse(text):
++++chars=+[]
++++l+=+len(text)
++++last+=+l-1
++++for+i+in+range+(l):
++++++++chars.append(text[last])
++++++++last-=1

++++result=+""+++
++++for+c+in+chars:
++++++++result+%2B=+c
++++return+result
print+reverse('hola')|code-block|syntax|javascript|1341341|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|$]]

I tried this in one of the codeacademy exercises (reversing chars in a string without using reversed nor :: -1)

<pre><code>def reverse(text):
 chars= []
 l = len(text)
 last = l-1
 for i in range (l):
 chars.append(text[last])
 last-=1

 result= "" 
 for c in chars:
 result += c
 return result
print reverse('hola')
</code></pre>

blocks|key|1341568|text|哦，好的，读错了问题，我猜它是关于在数组中向后返回？如果是这样的话，我有这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341569|array+=+["ty",+"rogers",+"smith",+"davis",+"tony",+"jack",+"john",+"jill",+"harry",+"tom",+"jane",+"hilary",+"jackson",+"andrew",+"george",+"rachel"]


counter+=+0+++

for+loop+in+range(len(array)):
++++if+loop+<=+len(array):
++++++++counter+=+-1
++++++++reverseEngineering+=+loop+%2B+counter
++++++++print(array[reverseEngineering])|code-block|syntax|javascript|1341570|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|$]]

Oh okay read the question wrong, I guess it's about going backward in an array? if so, I have this:

<pre><code>array = ["ty", "rogers", "smith", "davis", "tony", "jack", "john", "jill", "harry", "tom", "jane", "hilary", "jackson", "andrew", "george", "rachel"]


counter = 0 

for loop in range(len(array)):
 if loop &lt;= len(array):
 counter = -1
 reverseEngineering = loop + counter
 print(array[reverseEngineering])
</code></pre>

blocks|key|1557181|text|您可能希望在python中使用reversed函数。在我们开始编写代码之前，我们必须记住，range函数总是返回一个列表(或者一个我不知道的元组)，所以range(5)将返回[0,+1,+2,+3,+4]。reversed函数反转列表或元组，因此reversed(range(5))将为[4,+3,+2,+1,+0]，因此您的解决方案可能是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1557182|for+i+in+reversed(range(100)):
++++print(i)|code-block|syntax|javascript|1557183|entityMap^0|F|8|19|5|24|8|2F|F|2V|8|3F|I|3Z|F|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|L|$]]

You might want to use the <code>reversed</code> function in python.
Before we jump in to the code we must remember that the <code>range</code>
function always returns a list (or a tuple I don't know) so <code>range(5)</code> will return <code>[0, 1, 2, 3, 4]</code>. The <code>reversed</code> function reverses a list or a tuple so <code>reversed(range(5))</code> will be <code>[4, 3, 2, 1, 0]</code> so your solution might be:
<pre class="lang-py prettyprint-override"><code>for i in reversed(range(100)):
 print(i)
</code></pre>

blocks|key|1341693|text|它在我身上运行得很好|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1341694|for+i+in+range(5)[::-1]:
++++print(i,end='+')|code-block|syntax|javascript|1341695|输出:4+3+2+1+0|1341696|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|$]]

It works well with me
<pre><code>for i in range(5)[::-1]:
 print(i,end=' ')
</code></pre>
output : 4 3 2 1 0

blocks|key|1556912|text|a+=+10
for+i+in+sorted(range(a),+reverse=True):
++++print+i|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1556913|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>a = 10
for i in sorted(range(a), reverse=True):
 print i
</code></pre>

I am trying to loop from 100 to 0. How do I do this in Python?

<code>for i in range (100,0)</code> doesn't work.

Loop backwards using indices in Python?

Python

我正在尝试从100循环到0。我如何在Python中做到这一点？for i in range (100,0)不工作。

问在Python中使用索引进行反向循环？
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回答 16

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