entityMap|blocks|key|naqb|text|如果未指定，则范围从0索引开始。你想使用像这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b40fm|for+i+in+range(1,nI):
++++...|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Range starts from the 0 index if not otherwise specified. You want to use something like

<pre><code>for i in range(1,nI):
 ...
</code></pre>

entityMap|blocks|key|7kdo|text|如果您想遍历第二个项中的列表，只需使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|aae9q|(如果nI是列表的长度左右)。|cbjsf|for+i+in+range(1,+nI):
++++sum=0
++++for+v+in+range(nV):
++++++++for+j+in+range(nJ):
++++++++++++sum=sum%2Bx1[i][j][v]|code-block|syntax|javascript|5pclu|也许，你的函数的一部分在某个地方丢失了，但不管怎样，在一般的range()中是这样工作的：|7g112|range(start_from,+stop_at,+step_size)|cocju|即e。|6ao40|for+i+in+range(2,+7,+2):
++++print(i,+end='+')

Out:
2+4+6|7ctf1|编辑|3h07g|请记住:+python使用零索引，即第一个元素的索引为0，第二个元素的索引为-1，依此类推。|62tlo|默认情况下，|36b9b|从0开始，止于|15etp|传递的参数的值减一。|f3ao|如果有一个显式的|8ebqj|开始|f4cqs|，迭代从它的值开始。如果有一个|5qse0|步骤|80na4|，它继续，同时|2q9aj|返回值|5apgf|小于|acjj6|579ha|停止|3he4p|价值。|bkbg1|for+i+in+range(1,+7,+2):
++++print(i,+end='+')

Out:+
1+3+5++#+there's+no+7!|eelb4|详细描述|738l4|内置is|7j2h1|这里|bpjs8|..。^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|1U|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|1V|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|1W|9|@]|A|@]|B|$H|I]]|$2|J|4|K|6|7|8|1X|9|@]|A|@]|B|$]]|$2|L|4|M|6|G|8|1Y|9|@]|A|@]|B|$H|I]]|$2|N|4|O|6|7|8|1Z|9|@]|A|@]|B|$]]|$2|P|4|Q|6|G|8|20|9|@]|A|@]|B|$H|I]]|$2|R|4|S|6|7|8|21|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|22|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|23|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|24|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|25|9|@]|A|@]|B|$]]|$2|11|4|12|6|7|8|26|9|@]|A|@]|B|$]]|$2|13|4|14|6|7|8|27|9|@]|A|@]|B|$]]|$2|15|4|16|6|7|8|28|9|@]|A|@]|B|$]]|$2|17|4|18|6|7|8|29|9|@]|A|@]|B|$]]|$2|19|4|1A|6|7|8|2A|9|@]|A|@]|B|$]]|$2|1B|4|1C|6|7|8|2B|9|@]|A|@]|B|$]]|$2|1D|4|1E|6|7|8|2C|9|@]|A|@]|B|$]]|$2|1F|4|-4|6|7|8|2D|9|@]|A|@]|B|$]]|$2|1G|4|1H|6|7|8|2E|9|@]|A|@]|B|$]]|$2|1I|4|1J|6|7|8|2F|9|@]|A|@]|B|$]]|$2|1K|4|1L|6|G|8|2G|9|@]|A|@]|B|$H|I]]|$2|1M|4|1N|6|7|8|2H|9|@]|A|@]|B|$]]|$2|1O|4|1P|6|7|8|2I|9|@]|A|@]|B|$]]|$2|1Q|4|1R|6|7|8|2J|9|@]|A|@]|B|$]]|$2|1S|4|1T|6|7|8|2K|9|@]|A|@]|B|$]]]]

If you want to iterate through a list from a second item, just use <code>range(1, nI)</code> (if nI is the length of the list or so). 

<pre><code>for i in range(1, nI):
 sum=0
 for v in range(nV):
 for j in range(nJ):
 sum=sum+x1[i][j][v]
</code></pre>

Probaly, a part of your function just lost somewhere, but anyway, in in general range() works like this: 

<pre><code>range(start_from, stop_at, step_size)
</code></pre>

i. e. 

<pre><code>for i in range(2, 7, 2):
 print(i, end=' ')

Out:
2 4 6
</code></pre>

Edit

Please, remember: python uses zero indexing, i.e. first element has an index 0, the second - 1 etc. 

By default, <code>range</code> starts from 0 and stops at the value of the passed parameter minus one. If there's an explicit start, iteration starts from it's value. If there's a step, it continues while <code>range</code> returns values lesser than stop value.

<pre><code>for i in range(1, 7, 2):
 print(i, end=' ')

Out: 
1 3 5 # there's no 7!
</code></pre>

Detailed description of <code>range</code> build-in is <a href="https://docs.python.org/3/library/functions.html#func-range" rel="nofollow noreferrer">here</a>.

entityMap|blocks|key|b581p|text|首先要记住，python使用零索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|c3mca|除了使用range函数来获取所需项目的索引或切片来获取元素之外，您可以遍历列表。|7g4li|我认为这里变得令人困惑的是，在您的示例中，值和索引是相同的，因此为了澄清，我将使用以下列表作为示例：|bdtmo|I+=+['a',+'b',+'c',+'d',+'e']
nI+=+len(I)+#+5|code-block|syntax|javascript|7bteu|range函数将允许您遍历索引：|kp8l|for+i+in+range(1,+nI):
++++print(i)
#+Prints:
#+1
#+2
#+3
#+4|7dbum|如果您想使用range函数访问这些值，您应该这样做：|e0b3u|for+index+in+range(1,+nI):
++++i+=+I[index]
++++print(i)
#+Prints:
#+b
#+c
#+d
#+e|687ep|您也可以使用数组切片来实现这一点，而您甚至不需要|otoc|..。数组切片会返回一个包含切片的新数组。|2br3d|切片已完成|5kcqh|其中所有三个参数都是可选的，并且：|1jmbf|ehr9g|是要包含在切片中的第一个索引。|c35us|dqhs9|是要从切片中排除的第一个元素的索引。|b4uum|9oq72|是每个下一个索引的步数，从(正如预期的)|629ar|(如果|467q4|是2，并从1开始，它得到每个奇数索引)。|4usav|示例：|188t8|for+i+in+I[1:]:
++++print(i)
#+Prints:
#+b
#+c
#+d
#+e^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|1I|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|1J|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|1K|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|1L|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|1M|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|1N|9|@]|A|@]|B|$J|K]]|$2|P|4|Q|6|7|8|1O|9|@]|A|@]|B|$]]|$2|R|4|S|6|I|8|1P|9|@]|A|@]|B|$J|K]]|$2|T|4|U|6|7|8|1Q|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|1R|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|1S|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|1T|9|@]|A|@]|B|$]]|$2|11|4|-4|6|7|8|1U|9|@]|A|@]|B|$]]|$2|12|4|13|6|7|8|1V|9|@]|A|@]|B|$]]|$2|14|4|-4|6|7|8|1W|9|@]|A|@]|B|$]]|$2|15|4|16|6|7|8|1X|9|@]|A|@]|B|$]]|$2|17|4|-4|6|7|8|1Y|9|@]|A|@]|B|$]]|$2|18|4|19|6|7|8|1Z|9|@]|A|@]|B|$]]|$2|1A|4|1B|6|7|8|20|9|@]|A|@]|B|$]]|$2|1C|4|1D|6|7|8|21|9|@]|A|@]|B|$]]|$2|1E|4|1F|6|7|8|22|9|@]|A|@]|B|$]]|$2|1G|4|1H|6|I|8|23|9|@]|A|@]|B|$J|K]]]]

First thing is to remember that python uses zero indexing.

You can iterate throught the list except using the range function to get the indexes of the items you want or slices to get the elements.

What I think is becoming confusing here is that in your example, the values and the indexes are the same so to clarify I'll use this list as example:

<pre><code>I = ['a', 'b', 'c', 'd', 'e']
nI = len(I) # 5
</code></pre>

The range function will allow you to iterate through the indexes:

<pre><code>for i in range(1, nI):
 print(i)
# Prints:
# 1
# 2
# 3
# 4
</code></pre>

If you want to access the values using the range function you should do it like this:

<pre><code>for index in range(1, nI):
 i = I[index]
 print(i)
# Prints:
# b
# c
# d
# e
</code></pre>

You can also use array slicing to do that and you don't even need <code>nI</code>. Array slicing returns a new array with your slice. 
The slice is done with <code>the_list_reference[start:end:steps]</code> where all three parameters are optional and: 
<code>start</code> is the index of the first to be included in the slice 
<code>end</code> is the index of the first element to be excluded from the slice 
<code>steps</code> is how many steps for each next index starting from (as expected) the <code>start</code> (if <code>steps</code> is 2 and start with 1 it gets every odd index). 
Example:

<pre><code>for i in I[1:]:
 print(i)
# Prints:
# b
# c
# d
# e
</code></pre>

entityMap|blocks|key|6rcsm|text|您可以简单地使用切片：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b13k7|for+item+in+I[1:]:
++++print(item)|code-block|syntax|javascript|fpsmu|如果你想要索引，使用pythonic风格|1aujs|START+=+1
for+index,+item+in+enumerate(I[START:],+START):
++++print(item,+index)^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|O|9|@]|A|@]|B|$F|G]]]]

You can simply use slicing:
<pre><code>for item in I[1:]:
 print(item)
</code></pre>
And if you want indexing, use pythonic-style <code>enumerate</code>:
<pre><code>START = 1
for index, item in enumerate(I[START:], START):
 print(item, index)
</code></pre>

entityMap|blocks|key|9sctg|text|您应该避免使用内置的名称。如果确实需要，可以在末尾添加一个下划线，例如sum|type|unstyled|depth|inlineStyleRanges|entityRanges|data|eoc8i|_|857hj|，但这可能是一个提示，您应该使用内置的。|d0qpp|对于迭代变量，建议使用|apjt8|除非你有理由给它起别的名字。|7c87f|如果使用numpy数组，则可以使用for+value遍历所有值|72uih|如果你把|ag4bb|在循环内部，那么它每次都会被重置到那个位置。|cue0h|你可以使用itertools.product来“去嵌套”for循环，例如|aukd8|sum(x[i][j][v]+for+i,+j,+v+in+itertools.product(range(nI),+range(nJ),+range(nV))) |code-block|syntax|javascript|5i24e|为了减少多个for循环，Itertools有一个乘积函数，它提供了一个生成器，允许您将多个迭代器展平为一个可迭代的元组，这实际上与嵌套的for循环相同。唯一的区别是值的排序。|88tf6|from+itertools+import+product

for+x,+y+in+product(range(3),+range(4,+7)):
++++print(x,+y)|fgqsa|0+4
0+5
0+6
1+4
1+5
1+6
2+4
2+5
2+6

[Program+finished]^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|13|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|14|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|15|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|16|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|17|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|18|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|19|9|@]|A|@]|B|$]]|$2|O|4|P|6|7|8|1A|9|@]|A|@]|B|$]]|$2|Q|4|R|6|7|8|1B|9|@]|A|@]|B|$]]|$2|S|4|T|6|U|8|1C|9|@]|A|@]|B|$V|W]]|$2|X|4|Y|6|7|8|1D|9|@]|A|@]|B|$]]|$2|Z|4|10|6|U|8|1E|9|@]|A|@]|B|$V|W]]|$2|11|4|12|6|U|8|1F|9|@]|A|@]|B|$V|W]]]]

<ul>
<li>You should avoid using names of built-ins. If you really must, add an underscore at the end e.g. sum_, but it's probably a hint that you should be using the built-in instead.</li>
<li>For iteration variables, it's recommended you use<code>i, j, k, m, n</code>unless you have a reason to name it something else.</li>
<li>If you are using numpy arrays, you can loop over all values using for value in <code> np.nditer(...)</code></li>
<li>If you put <code>sum = 0</code> inside of the loop, then it'll get reset to that every time.</li>
<li>You can &quot;de-nest&quot; for loops using itertools.product e.g.</li>
</ul>
<pre><code>sum(x[i][j][v] for i, j, v in itertools.product(range(nI), range(nJ), range(nV))) 
</code></pre>
<ul>
<li>To reduce multiple for loops, Itertools has a product function that gives you a generator that will allow you to flatten multiple iterables into one iterable of tuples that effectively is the same as nested for loops. The only difference is ordering of the values.</li>
</ul>
<pre><code>from itertools import product

for x, y in product(range(3), range(4, 7)):
 print(x, y)
</code></pre>
<pre><code>0 4
0 5
0 6
1 4
1 5
1 6
2 4
2 5
2 6

[Program finished]
</code></pre>

I have this for-loop. I want <code>i in range(nI)</code> to start from the second number in the <code>I</code> list. Could you guide me?

<pre><code>I=[0,1,2,3,4,5,6]
nI=len(I)
for i in range(nI):
 sum=0
 for v in range(nV):
 for j in range(nJ):
 sum=sum+x1[i][j][v]
return sum
</code></pre>

How to start from second index for for-loop

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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 我有这个for-loop。我想 中的第二个数字开始 列表。你能给我指路吗？ I=[0,1,2,3,4,5,6]nI=len(I)for i in range(nI):    sum=0    for v in range(nV):        for j in range(nJ):            sum=s...

问如何从for循环的第二个索引开始
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问如何从for循环的第二个索引开始EN