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牵手幸福

铜虫 (初入文坛)

[求助] 字符串处理 已有1人参与

n='0010000010010011000110101010011001011111100111111101011111011100000011101101101101001101100100100011001000110011'
我想要把以上字符串8位8位中间添加空格分开,也就是说输出格式为
n=‘00100000 10010011 00011010 10100110 01011111 10011111 11010111 11011100 00001110 11011011 01001101 01001010 00110010 00110011’
除了重新赋值以外,有没有简便的算法,求指教
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There is always time for Plan B .But don't begin with it.
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xiaomm09

金虫 (小有名气)

Table[StringTake[n, {i, i + 8}], {i, 1, StringLength[n]/8.}]
Mathematica版
追求完美!
7楼2014-05-05 00:10:01
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libralibra

至尊木虫 (著名写手)

骠骑将军

【答案】应助回帖

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感谢参与,应助指数 +1
牵手幸福: 金币+10, ★★★★★最佳答案 2014-05-03 09:02:46
如果是matlab的话
CODE:
n = '0010000010010011000110101010011001011111100111111101011111011100000011101101101101001101100100100011001000110011'
len = length(n);
row = len/8;
col = 8;
a = reshape(n,[col,row])';
b = repmat(' ',[row,1]);
c = [a,b];
d = reshape(c',[1,len+row])

结果:
CODE:
n =

0010000010010011000110101010011001011111100111111101011111011100000011101101101101001101100100100011001000110011


d =

00100000 10010011 00011010 10100110 01011111 10011111 11010111 11011100 00001110 11011011 01001101 10010010 00110010 00110011

>>  

matlab/VB/python/c++/Java写程序请发QQ邮件:790404545@qq.com
2楼2014-05-02 21:54:32
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牵手幸福

铜虫 (初入文坛)

引用回帖:
2楼: Originally posted by libralibra at 2014-05-02 21:54:32
如果是matlab的话
n = '0010000010010011000110101010011001011111100111111101011111011100000011101101101101001101100100100011001000110011'
len = length(n);
row = len/8;
col = 8;
a = reshape(n,)';
...

谢谢
There is always time for Plan B .But don't begin with it.
3楼2014-05-03 09:02:11
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colin_98

木虫 (正式写手)

还没想好

相同的变量,最后值变了,必须必是重新赋值了!
题中是字符串处理,必然涉及,因此无解。
http://hi.baidu.com/e_netcontrol
4楼2014-05-04 08:59:29
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