blocks|key|366423|text|请注意，可以在片中使用变量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366424|l+=+['a','+b','+c','+d','+e']
c_index+=+l.index("c")
l2+=+l[:c_index]|code-block|syntax|javascript|366425|这将在l2中放置l的前两个条目|366426|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|$]]

Note that you can use a variable in a slice:

<pre><code>l = ['a',' b',' c',' d',' e']
c_index = l.index("c")
l2 = l[:c_index]
</code></pre>

This would put the first two entries of l in l2

blocks|key|366405|text|在蟒蛇里，这叫做切片。下面是python's+slice+notation的一个示例|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|366406|>>>+list1+=+['a','b','c','d','e','f','g','h',+'i',+'j',+'k',+'l']
>>>+print+list1[:5]
['a',+'b',+'c',+'d',+'e']
>>>+print+list1[-7:]
['f',+'g',+'h',+'i',+'j',+'k',+'l']|code-block|syntax|javascript|366407|请注意，您如何可以正面或负面地分割。当你使用负数时，这意味着我们从右向左切分。|366408|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/509211/the-python-slice-notation^0|E|N|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]]]]

In python, it's called slicing. Here is an example of <a href="https://stackoverflow.com/questions/509211/the-python-slice-notation">python's slice notation</a>:

<pre><code>&gt;&gt;&gt; list1 = ['a','b','c','d','e','f','g','h', 'i', 'j', 'k', 'l']
&gt;&gt;&gt; print list1[:5]
['a', 'b', 'c', 'd', 'e']
&gt;&gt;&gt; print list1[-7:]
['f', 'g', 'h', 'i', 'j', 'k', 'l']
</code></pre>

Note how you can slice either positively or negatively. When you use a negative number, it means we slice from right to left.

blocks|key|2308466|text|如果您已经知道了指数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2308467|list1+=+['x','y','z','a','b','c','d','e','f','g']
indices+=+[(0,+4),+(5,+9)]
print+[list1[s:e%2B1]+for+s,e+in+indices]|code-block|syntax|javascript|2308468|请注意，我们将%2B1添加到末尾，以使范围包括在内.|2308469|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|$]]

If you already know the indices:

<pre><code>list1 = ['x','y','z','a','b','c','d','e','f','g']
indices = [(0, 4), (5, 9)]
print [list1[s:e+1] for s,e in indices]
</code></pre>

Note that we're adding +1 to the end to make the range inclusive...

blocks|key|2308417|text|list1a=list[:5]
list1b=list[5:]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2308418|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|$]]

blocks|key|1166363|text|如果您有多个索引，或者知道需要得到的索引范围，那么可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1166364|split_points+-将拆分字符串或列表的点|1166365|K-您需要拆分的范围，例如=3|1166366|split_points+=+[i+for+i+in+range(0,+len(string),+k)]

parts+=+[string[ind:ind+%2B+k]+for+ind+in+split_points]|code-block|syntax|javascript|1166367|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|$]]

One of the ways to do it if you have multiple indexes or know the range of indexes you need to get:
split_points - points where you will split your string or list
k - the range you need to split, example = 3
<pre><code>split_points = [i for i in range(0, len(string), k)]

parts = [string[ind:ind + k] for ind in split_points]
</code></pre>

blocks|key|2308445|text|list1=['x','y','z','a','b','c','d','e','f','g']
find=raw_input("Enter+string+to+be+found")
l=list1.index(find)
list1a=[:l]
list1b=[l:]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2308446|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>list1=['x','y','z','a','b','c','d','e','f','g']
find=raw_input("Enter string to be found")
l=list1.index(find)
list1a=[:l]
list1b=[l:]
</code></pre>

blocks|key|366057|text|考虑以下示例的核心伪码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366058|def+slice_it(list_2be_sliced,+indices):
++++"""Slices+a+list+at+specific+indices+into+constituent+lists.
++++"""
++++indices.append(len(list_2be_sliced))
++++return+[list_2be_sliced[indices[i]:indices[i%2B1]]+for+i+in+range(len(indices)-1)]|code-block|syntax|javascript|366059|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|$]]

Consider the core pesudocode of the following example:
<pre><code>def slice_it(list_2be_sliced, indices):
 &quot;&quot;&quot;Slices a list at specific indices into constituent lists.
 &quot;&quot;&quot;
 indices.append(len(list_2be_sliced))
 return [list_2be_sliced[indices[i]:indices[i+1]] for i in range(len(indices)-1)]
</code></pre>

blocks|key|2308581|text|如果输入是要拆分数组的索引列表，那么我就是这样做的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2308582|#input
list1+=+['x','y','z','a','b','c','d','e','f','g']
split_points+=+[2,5,8]+

#split+array+on+indices:
s+=+split_points%2B[len(list1)]++#must+contain+index+beyond+last+element,+alternatively+use+directly+split_points.append(len(list1))
print([list1[i1:i2]+for+i1,i2+in+zip([0]%2Bs[:-1],s)])

>>>+[['x',+'y'],+['z',+'a',+'b'],+['c',+'d',+'e'],+['f',+'g']]|code-block|syntax|javascript|2308583|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|$]]

This is the way I do it, if the input is a list of indices on which to split an array:
<pre><code>#input
list1 = ['x','y','z','a','b','c','d','e','f','g']
split_points = [2,5,8] 

#split array on indices:
s = split_points+[len(list1)] #must contain index beyond last element, alternatively use directly split_points.append(len(list1))
print([list1[i1:i2] for i1,i2 in zip([0]+s[:-1],s)])

&gt;&gt;&gt; [['x', 'y'], ['z', 'a', 'b'], ['c', 'd', 'e'], ['f', 'g']]
</code></pre>

How do I split a list into sub-lists based on index ranges?
e.g. original list:
<pre><code>list1 = [x,y,z,a,b,c,d,e,f,g]
</code></pre>
using index ranges 0–4:
<pre><code>list1a = [x,y,z,a,b]
</code></pre>
using index ranges 5–9:
<pre><code>list1b = [c,d,e,f,g]
</code></pre>
I already known the (variable) indices of list elements which contain certain string and want to split the list based on these index values.
Also need to split into variable number of sub-lists, i.e.:
<pre><code>list1a
list1b
.
.
list1[x]
</code></pre>

Split a list into sub-lists based on index ranges

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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如何根据索引范围将列表拆分为子列表？例如，原始清单：list1 = [x,y,z,a,b,c,d,e,f,g]使用索引范围0-4：list1a = [x,y,z,a,b]使用索引范围5-9：list1b = [c,d,e,f,g]我已经知道了列表元素的(变量)索引，这些元素包含某些字符串，并希望根据这些索引值拆分列表。还需要划分成可变数量的子列表，即：list1alist1b..list1[x]

问根据索引范围将列表拆分为子列表。
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回答 8

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问根据索引范围将列表拆分为子列表。EN

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问根据索引范围将列表拆分为子列表。
EN