blocks|key|2809323|text|以下是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2809324|a+=+1010101010010100
k+=+[]
m+=+[int(i)+for+i+in+str(a)]

for+j+in+range(0,+len(m),+4):
++++k.append(m[j:j%2B4])|code-block|syntax|javascript|2809325|输出：|2809326|[[1,+0,+1,+0],+[1,+0,+1,+0],+[1,+0,+0,+1],+[0,+1,+0,+0]]|2809327|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|$]]

Here is my solution:

<pre><code>a = 1010101010010100
k = []
m = [int(i) for i in str(a)]

for j in range(0, len(m), 4):
 k.append(m[j:j+4])
</code></pre>

output:

<pre><code>[[1, 0, 1, 0], [1, 0, 1, 0], [1, 0, 0, 1], [0, 1, 0, 0]]
</code></pre>

blocks|key|840990|text|你的代码中有很多问题-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|840991|我认为你想做的是-+b.index(x)%253+==+0而不是b.index(x)!=0%253+(这没有多大意义，除非b.index(x)是0，否则总是假的。|offset|length|style|CODE|840992|你行中有一个冒号-+tmp_list.append(x):，这是一个语法错误(你可能应该删除python中的逐位|unordered-list-item|840993|
|840994|运算符，我认为你应该使用b.index(x)%253+==+0运算符而不是and运算符&给出元素存在的第一个索引，因为你的字符串只包含1和0，这永远不会超过索引2。相反，您应该考虑使用enumerate函数，该函数将为您提供元素的索引和值。|840995|840996|一个建议-|840997|840998|您不应该有if+b.index(x)==0:条件，您可以只在循环外部定义temp_list。|ordered-list-item|840999|使用if/elif+(而不是所有的if)。|841000|841001|841002|示例代码：|841003|b+=+str(a)
list1+=+[]
tmp_list+=+[]
for+i,x+in+enumerate(b):
++++if+i%254!=0+or+i==0:
++++++++tmp_list.append(x)
++++else:
++++++++list1.append(tmp_list)
++++++++tmp_list=[]
++++++++tmp_list.append(x)|code-block|syntax|javascript|841004|输出-|841005|[['1',+'0',+'1',+'0'],+['1',+'0',+'1',+'0'],+['1',+'0',+'0',+'1']]|841006|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/functions.html#enumerate^0|0|A|H|U|F|0|A|J|0|0|C|H|Z|3|15|1|2I|9|2I|9|0|0|0|0|0|5|H|1|2|7|H|2|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1J|8|@$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1O|8|@$D|1P|E|1Q|F|G]]|9|@]|A|$]]|$1|K|3|L|5|6|7|1R|8|@]|9|@]|A|$]]|$1|M|3|N|5|J|7|1S|8|@$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]|$D|1Z|E|20|F|G]]|9|@$D|21|E|22|1|23]]|A|$]]|$1|O|3|-4|5|6|7|24|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|25|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|26|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|27|8|@$D|28|E|29|F|G]]|9|@]|A|$]]|$1|V|3|W|5|U|7|2A|8|@$D|2B|E|2C|F|G]|$D|2D|E|2E|F|G]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|2F|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|2G|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|2H|8|@]|9|@]|A|$]]|$1|11|3|12|5|13|7|2I|8|@]|9|@]|A|$14|15]]|$1|16|3|17|5|6|7|2J|8|@]|9|@]|A|$]]|$1|18|3|19|5|13|7|2K|8|@]|9|@]|A|$14|15]]|$1|1A|3|-4|5|6|7|2L|8|@]|9|@]|A|$]]]|1B|$1C|$5|1D|1E|1F|A|$1G|1H]]]]

You have lots of issues in your code -

<ol>
<li>I think you want to do - <code>b.index(x)%3 == 0</code> and not <code>b.index(x)!=0%3</code> (which does not make much sense, and is always false unless b.index(x) is 0.</li>
<li>You have a colon in the line - <code>tmp_list.append(x):</code> , this is a syntax error (you should probably remove it).</li>
<li><code>&amp;</code> is the bitwise and operator in python, I believe you should be using <code>and</code> operator instead of <code>&amp;</code> .</li>
<li><code>list.index(&lt;element&gt;)</code> gives the first index where the element exists, since your string only consists of 1s and 0s , this would never go past index 2. Instead you should think about using <a href="https://docs.python.org/2/library/functions.html#enumerate" rel="nofollow"><code>enumerate</code></a> function , that would provide you the index as well as the value of the element.</li>
</ol>

A suggestion -

<ol>
<li>You should not have the <code>if b.index(x)==0:</code> condition, you can just define the temp_list outside the for loop.</li>
<li>Use <code>if/elif</code> (instead of all <code>if</code>s) .</li>
</ol>

Example code -

<pre><code>b = str(a)
list1 = []
tmp_list = []
for i,x in enumerate(b):
 if i%4!=0 or i==0:
 tmp_list.append(x)
 else:
 list1.append(tmp_list)
 tmp_list=[]
 tmp_list.append(x)
</code></pre>

output -

<pre><code>[['1', '0', '1', '0'], ['1', '0', '1', '0'], ['1', '0', '0', '1']]
</code></pre>

blocks|key|2809379|text|其中之一应该可以(如果您的数字长度%254+!=+0，则会提供不同的结果)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2809380|num+=+1010101010010100
[str(num)[i:i%2B4]+for+i+in+range(0,+len(str(num)),+4)]
[str(num)[i*4:i*4%2B4]+for+i+in+range(0,+len(str(num))//4)]|code-block|syntax|javascript|2809381|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|$]]

One of these should do (will provide different results in case your number length % 4 != 0):

<pre><code>num = 1010101010010100
[str(num)[i:i+4] for i in range(0, len(str(num)), 4)]
[str(num)[i*4:i*4+4] for i in range(0, len(str(num))//4)]
</code></pre>

blocks|key|2809383|text|b+=+str(a)
n+=+4
[list([b[i:i%2Bn])+for+i+in+range(0,len(b),n)]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2809384|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>b = str(a)
n = 4
[list([b[i:i+n]) for i in range(0,len(b),n)]
</code></pre>

blocks|key|847508|text|如果你想把你的序列分成4个一组，你可以简单地这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|847509|>>>+b='1010101010010100'

>>>+[b[i:i%2B4]+for+i+in+range(0,+len(b),+4)]
['1010',+'1010',+'1001',+'0100']|code-block|syntax|javascript|847510|或者，如果您恰好需要一个列表列表，则在list(..)中包装该项：|offset|length|style|CODE|847511|>>>+[list(b[i:i%2B4])+for+i+in+range(0,+len(b),+4)]
[['1',+'0',+'1',+'0'],+['1',+'0',+'1',+'0'],+['1',+'0',+'0',+'1'],+['0',+'1',+'0',+'0']]|847512|此外，如果子列表中的元素必须为int|847513|>>>+[list(int(c)+for+c+in+b[i:i%2B4])+for+i+in+range(0,+len(b),+4)]
[[1,+0,+1,+0],+[1,+0,+1,+0],+[1,+0,+0,+1],+[0,+1,+0,+0]]|847514|entityMap^0|0|0|J|8|0|0|F|3|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

If you want to split your sequence in groups of 4, you can simply do:

<pre><code>&gt;&gt;&gt; b='1010101010010100'

&gt;&gt;&gt; [b[i:i+4] for i in range(0, len(b), 4)]
['1010', '1010', '1001', '0100']
</code></pre>

or wrap the item in <code>list(..)</code> if you need exactly a list of lists:

<pre><code>&gt;&gt;&gt; [list(b[i:i+4]) for i in range(0, len(b), 4)]
[['1', '0', '1', '0'], ['1', '0', '1', '0'], ['1', '0', '0', '1'], ['0', '1', '0', '0']]
</code></pre>

moreover, if the elements in sublists have to be <code>int</code>:

<pre><code>&gt;&gt;&gt; [list(int(c) for c in b[i:i+4]) for i in range(0, len(b), 4)]
[[1, 0, 1, 0], [1, 0, 1, 0], [1, 0, 0, 1], [0, 1, 0, 0]]
</code></pre>

blocks|key|2809403|text|列表理解是你的朋友(就像其他很多时候一样)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2809404|a+=+1010101010010100
[list(str(a))[i:i%2B4]+for+i+in+range(0,len(str(a)),4)]|code-block|syntax|javascript|2809405|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|$]]

List comprehension is your friend (as so many other times):

<pre><code>a = 1010101010010100
[list(str(a))[i:i+4] for i in range(0,len(str(a)),4)]
</code></pre>

blocks|key|2809441|text|一种有效的解决方案可能是这样的，来自itertools+receipes|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2809442|它返回迭代器，所以它应该很快。|2809443|from+itertools+import+izip_longest

def+grouper(iterable,+n,+fillvalue=None):
++++args+=+[iter(iterable)]+*+n
++++return+izip_longest(fillvalue=fillvalue,+*args)

numbers+=+str(1010101010010100)
nth+=+3
result+=+grouper(numbers,+nth)|code-block|syntax|javascript|2809444|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/itertools.html#recipes^0|I|I|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Kind of effective solution could be this, from the <a href="https://docs.python.org/2/library/itertools.html#recipes" rel="nofollow">itertools receipes</a>

It returns iterator, so it should be pretty fast.

<pre><code>from itertools import izip_longest

def grouper(iterable, n, fillvalue=None):
 args = [iter(iterable)] * n
 return izip_longest(fillvalue=fillvalue, *args)

numbers = str(1010101010010100)
nth = 3
result = grouper(numbers, nth)
</code></pre>

blocks|key|4275393|text|import+math

a+=+1010101010010100
res+=+[]

long+=+math.ceil(math.log10(a))
digits+=+[0]+*+(4+-+long+%25+4)+if+long+%25+4+else+[]
digits.extend([int(i)+for+i+in+str(a)])

for+i+in+xrange(0,+len(digits),+4):
++++res.append(digits[i:i%2B4])

print(res)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4275394|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>import math

a = 1010101010010100
res = []

long = math.ceil(math.log10(a))
digits = [0] * (4 - long % 4) if long % 4 else []
digits.extend([int(i) for i in str(a)])

for i in xrange(0, len(digits), 4):
 res.append(digits[i:i+4])

print(res)
</code></pre>

I have a integer that looks like this

<pre><code>a = 1010101010010100
</code></pre>

that I would like to turn into something like this

<pre><code>list = [[1,0,1,0],[1,0,1,0],[1,0,0,1],[0,1,0,0]]
</code></pre>

What I've tried to do is something like

<pre><code>b = str(a)
list1 = [];
for x in b:
 if b.index(x)==0:
 tmp_list = [];
 if b.index(x)!=0%3:
 tmp_list.append(x):
 if b.index(x)==0%3 &amp; b.index(x)!=0:
 list1.append(tmp_list)
 tmp_list=[]; 
</code></pre>

I don't get result as expected.

integer manipulation with 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。

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我有一个整数，看起来像这样a = 1010101010010100我想变成这样的东西list = [[1,0,1,0],[1,0,1,0],[1,0,0,1],[0,1,0,0]]我尝试做的事情是这样的b = str(a)list1 = [];for x in b:    if b.index(x)==0:      ...

问使用python进行整数操作
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问使用python进行整数操作EN