blocks|key|590647|text|你在找string+slicing.|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|590648|>>>+x+=+"1.012345e0070.123414e-004-0.1234567891.21423"
>>>+x[2:10]
'012345e0'|code-block|syntax|javascript|590649|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/introduction.html#strings^0|3|F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You're looking for <a href="http://docs.python.org/tutorial/introduction.html#strings" rel="noreferrer">string slicing.</a>

<pre><code>&gt;&gt;&gt; x = "1.012345e0070.123414e-004-0.1234567891.21423"
&gt;&gt;&gt; x[2:10]
'012345e0'
</code></pre>

blocks|key|590659|text|尝试此函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590660|x+=+"1.012345e0070.123414e-004-0.1234567891.21423"
while+len(x)>0:
++v+=+x[:12]
++print+v
++x+=+x[12:]|code-block|syntax|javascript|590661|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|$]]

Try this function:

<pre><code>x = "1.012345e0070.123414e-004-0.1234567891.21423"
while len(x)&gt;0:
 v = x[:12]
 print v
 x = x[12:]
</code></pre>

blocks|key|590665|text|from+itertools+import+izip_longest

def+grouper(n,+iterable,+padvalue=None):
++++return+izip_longest(*[iter(iterable)]*n,+fillvalue=padvalue)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|590666|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>from itertools import izip_longest

def grouper(n, iterable, padvalue=None):
 return izip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
</code></pre>

blocks|key|806418|text|由于您希望以一种不寻常的方式进行迭代，因此生成器是一个很好的抽象方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806419|def+chunks(s,+n):
++++"""Produce+`n`-character+chunks+from+`s`."""
++++for+start+in+range(0,+len(s),+n):
++++++++yield+s[start:start%2Bn]

nums+=+"1.012345e0070.123414e-004-0.1234567891.21423"
for+chunk+in+chunks(nums,+12):
++++print+chunk|code-block|syntax|javascript|806420|产生：|806421|1.012345e007
0.123414e-00
4-0.12345678
91.21423|806422|(看起来不太对劲，但这些是12个字符的块)|806423|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|$]]

Since you want to iterate in an unusual way, a generator is a good way to abstract that:

<pre><code>def chunks(s, n):
 """Produce `n`-character chunks from `s`."""
 for start in range(0, len(s), n):
 yield s[start:start+n]

nums = "1.012345e0070.123414e-004-0.1234567891.21423"
for chunk in chunks(nums, 12):
 print chunk
</code></pre>

produces:

<pre><code>1.012345e007
0.123414e-00
4-0.12345678
91.21423
</code></pre>

(which doesn't look right, but those are the 12-char chunks)

blocks|key|590772|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590773|step+=+12
for+i+in+range(0,+len(string),+12):
++++slice+=+string[i:step]
++++step+%2B=+12|code-block|syntax|javascript|590774|这样，在每次迭代中，您将获得一个包含14个字符的切片。|590775|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|$]]

you can do it like this:

<pre><code>step = 12
for i in range(0, len(string), 12):
 slice = string[i:step]
 step += 12
</code></pre>

in this way on each iteration you will get one slice of 14 characters.

blocks|key|806524|text|我一直在想，既然字符串加法运算是可以通过一个简单的逻辑实现的，可能除法应该是这样的。当除以一个数字时，它应该被该长度拆分。所以这可能就是你要找的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806525|class+MyString:
++++def+__init__(self,+string):
++++++++self.string+=+string
++++def+__div__(self,+div):
++++++++l+=+[]
++++++++for+i+in+range(0,+len(self.string),+div):
++++++++++++l.append(self.string[i:i%2Bdiv])
++++++++return+l

>>>+m+=+MyString(s)
>>>+m/3
['abc',+'bdb',+'fbf',+'bfb']


>>>+m+=+MyString('abcd')
>>>+m/3
['abc',+'d']|code-block|syntax|javascript|806526|如果您不想创建一个全新的类，只需使用此函数重新包装上述代码的核心，|806527|>>>+def+string_divide(string,+div):
+++++++l+=+[]
+++++++for+i+in+range(0,+len(string),+div):
+++++++++++l.append(string[i:i%2Bdiv])
+++++++return+l

>>>+string_divide('abcdefghijklmnopqrstuvwxyz',+15)
['abcdefghijklmno',+'pqrstuvwxyz']|806528|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 always thought, since string addition operation is possible by a simple logic, may be division should be like this. When divided by a number, it should split by that length. So may be this is what you are looking for. 

<pre><code>class MyString:
 def __init__(self, string):
 self.string = string
 def __div__(self, div):
 l = []
 for i in range(0, len(self.string), div):
 l.append(self.string[i:i+div])
 return l

&gt;&gt;&gt; m = MyString(s)
&gt;&gt;&gt; m/3
['abc', 'bdb', 'fbf', 'bfb']


&gt;&gt;&gt; m = MyString('abcd')
&gt;&gt;&gt; m/3
['abc', 'd']
</code></pre>

If you don't want to create an entirely new class, simply use this function that re-wraps the core of the above code,

<pre><code>&gt;&gt;&gt; def string_divide(string, div):
 l = []
 for i in range(0, len(string), div):
 l.append(string[i:i+div])
 return l

&gt;&gt;&gt; string_divide('abcdefghijklmnopqrstuvwxyz', 15)
['abcdefghijklmno', 'pqrstuvwxyz']
</code></pre>

I can't figure out how to do this with string methods:

In my file I have something like 1.012345e0070.123414e-004-0.1234567891.21423... which means there is no delimiter between the numbers. 

Now if I read a line from this file I get a string like above which I want to split after e.g. 12 characters. 
There is no way to do this with something like <code>str.split()</code> or any other string method as far as I've seen but maybe I'm overlooking something?

Thx

Split string by count of characters

我不知道如何使用字符串方法做到这一点：在我的文件中，有类似于1.012345e0070.123414e-004-0.1234567891.21423...的内容这意味着数字之间没有分隔符。现在，如果我从这个文件中读取一行，我会得到一个类似上面的字符串，我想在例如12个字符之后拆分它。就我所见，没有办法用str.spli...

问按字符数拆分字符串
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