从MATLAB数据采集的问题 [英] problems with data acquisition from MATLAB

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本文介绍了从MATLAB数据采集的问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

以前寻求帮助来读取文本数据C程序(一个Exe)生成的文件.

I previously asked for help in reading data from a text file generated by a C program (an Exe).

使用 @second 的解决方案,我解决了这个问题,但是昨天我发现输出文件比我预期的要复杂.

Using @second's solution, I solved the problem but yesterday I discovered that the output file is more complex than I had expected.

文件输出为:

V|0|0|0|t|0|1|1|4|11|T4|H13||||||||||||  
P|40|0.01|10|1|1|0|40|1|1|1||1|*||0|0|0  
*|A1|A1|A7|A16|F|F|F|F|F|F|||||||||||||  
*|codserv|area|codice|nome|tnom|tmin|tmax|pc|qc|susc|||||||
*|||||kV|kV|kV|MW|MVAR|S||||||||||||  
N|I|1|N01|N01|132|125.4|138.6|0|0||||||||
N|I|1|N02|N02|20|19|21|0|0|||||||||||||
N|I|1|N03|N03|20|19|21|1.013532234|0.49087611||||||||
N|I|1|N04|N04|20|19|21|0.390791617|0.189269056||||||||
N|I|1|N05|N05|20|19|21|0.180634542|0.121387171||||||||
N|I|1|N06|N06|20|19|21|0.709472564|0.343613323||||||||
N|I|1|N07|N07|20|19|21|0.103495727|0.069549543||||||||
N|I|1|N08|N08|20|19|21|0.351712456|0.170342158||||||||
N|I|1|N09|N09|20|19|21|0.097697904|0.06565339||||||||
N|I|1|N10|N10|20|19|21|0.162165157|0.078540184||||||||
N|I|1|N11|N11|20|19|21|0|0||||||||
*|A1|A8|A7|A7|F|F|F|F||F||F 
*|plev|area|codice|estr1|estr2|lung|imax|rsd|xsd|bsd1|bsd2|||
*|||||km|A|Ohm|Ohm||S||S    
L|I|D10203|N02|N03|1.884|360|0.41071|0.207886957|3.19E-08|3.19E-08|||||||||||||
L|I|D10304|N03|N04|1.62|360|0.35316|0.1787563|3.19E-08|3.19E-08|||||||||||||
L|I|D10405|N04|N05|0.532|360|0.11598|0.058702686|3.19E-08|3.19E-08|||||||||||||
L|I|D10506|N05|N06|1.284|360|0.27991|0.14168092|3.19E-08|3.19E-08|||||||||||||
L|I|D10607|N06|N07|1.618|280|0.53879|0.194766124|3.00E-08|3.00E-08|||||||||||||
L|I|D10708|N07|N08|0.532|280|0.17716|0.064039294|3.00E-08|3.00E-08|||||||||||||
L|I|D10809|N08|N09|2|360|0.436|0.220686791|3.19E-08|3.19E-08|||||||||||||
L|I|D10910|N09|N10|2.4|360|0.5232|0.264824149|3.19E-08|3.19E-08||||||||||||
*|A1|A8|A7|A7|F|F|A1|F|F|F|F|F|F||F||F||||||||||||||||||||||||| 
*|codserv|codice|estr1|estr2|vn1|vn2|nod1|varp|varm|np|Pb|rsd|xsd||bsd1||bsd2||||||||||||
*|||||kV|kV||%|%||MVA|%|%||%||%||||| 
%%%%%------%%%%%------%%%% **(read up to here)**
other unnecessary data

算法应:

  • 跳过前3行
  • 跳过第五行

  • skip the first 3 rows
  • skip fifth row

对于第四行*|codserv|area|codice|nome|tnom|tmin|tmax|pc|qc|susc|||||||,将每个字符串保存在向量为空的codeserv=[] area=[] codice=[] nome=[] tnom=[] tmin=[] tmax=[] pc=[] qc=[] susc=[]

For the fourth row *|codserv|area|codice|nome|tnom|tmin|tmax|pc|qc|susc|||||||, save each string in a vector empty codeserv=[] area=[] codice=[] nome=[] tnom=[] tmin=[] tmax=[] pc=[] qc=[] susc=[]

在向量的第四行之后填充向量和字符串的向量

Fill vectors with data and strings in the rows following the fourth

codeserv=[N N N N N N N N N N ....] 
area=[I I I I I I I ....] 
codice=[1 1 1 1 1 1 ...] 
nome=[N01 N02 N03 N04 N05 ] 
tnom=[N01 N02 N03 N04 N05] 
tmin=[132 20 20.....] 
tmax=[125.4 19 19 19 ....] 
pc=[138.6 21 21 21....] 
qc=[0 0 1.013532234 ....] 
susc=[0 0 0.49087611]

  • 对以字母L开头的数据执行相同的操作.阅读此行codice|estr1|estr2|lung|imax|rsd|xsd||bsd1||bsd2,并用L

  • Do the same with the data starting with the letter L. Read this line codice|estr1|estr2|lung|imax|rsd|xsd||bsd1||bsd2 and fill the vectors with values in the lines beginning with L

    plev=[L L L L L L L ....] 
    area=[I I I I I I I ....]
    codice=[D10203 D10304 ...] 
    estr1=[N02 N03 N04 N05  ...] 
    estr2=[N03 N04 N05...] 
    lung=[1.884 1,662 ....] 
    imax=[360 360 .....] 
    rsd=[number....] 
    xsd=[number....] 
    bsd1=[number ....] 
    bsd2=[number....]
    

  • 我试图改写上一个问题中的代码,但是鉴于以NL开头的行不知道我需要多少行才能读取第一个字符串并计算数字L是.

    I tried to adapt the code from the previous question, but given that the lines that start with N and L do not know how many I need to know how to read the first string and count the number N and L are.

    read
    [vp***NNNNNNNNNNNNNNNNNNNNNNNNLLLLLLLLLLLLLLLLLLLLL***] 
    
    length N 
    
    length L 
    
    skip 1 2 3 line
    
    read 4 line, create vector
    
    codeserv=[N N N N N N N N N N ....] 
    area=[I I I I I I I ....] 
    codice=[1 1 1 1 1 1 ...] 
    nome=[N01 N02 N03 N04 N05 ] 
    tnom=[N01 N02 N03 N04 N05] 
    tmin=[132 20 20.....] 
    tmax=[125.4 19 19 19 ....] 
    pc=[138.6 21 21 21....] 
    qc=[0 0 1.013532234 ....] 
    susc=[0 0 0.49087611]
    
    skip length N +1 line
    
    read *|plev|area|codice|estr1|estr2|lung|imax|rsd|xsd|bsd1|bsd2|||
    
    skip length N +3 line
    
    create
    
    plev=[L L L L L L L ....] 
    area=[I I I I I I I ....]
    codice=[D10203 D10304 ...] 
    estr1=[N02 N03 N04 N05  ...] 
    estr2=[N03 N04 N05...] 
    lung=[1.884 1,662 ....] 
    imax=[360 360 .....] 
    rsd=[number....] 
    xsd=[number....] 
    bsd1=[number ....] 
    bsd2=[number....]
    
    close the cycle 
    

    我希望这是可以理解的.我最大的问题是计算文本中的NL.

    I hope it is understandable. My biggest problem is counting the N and L in the text.

    推荐答案

    function readtest2()
    
    fid = fopen('test2.txt'); 
    
    skipLines(3)
    names1 = getNames;
    skipLines(1);
    nEntries1 = countPrefix('N');
    data1 = textscan(fid,'%s %s %d %s %s %d %d %f %f %f %[| ]', nEntries1, 'delimiter','|');
    
    
    skipLines(2)
    names2 = getNames;
    skipLines(1);
    nEntries2 = countPrefix('L');
    data2 = textscan(fid,'%s %s %s %s %s %f %d %f %f %f %f %[| ]', nEntries2, 'delimiter','|');
    
    fclose(fid);
    
    getData(data1, names1);
    getData(data2, names2);
    
    
    
    function names = getNames()
      names = fgetl(fid);
      names = textscan(names,'%s','delimiter','|');
    end
    
    function getData(data, names)
    
      for i = 1:size(data,2)-1
        values = ( data{i}(1:end));
        if(iscell(values))
          values = cell2mat(values);
        end
    
        name = names{1}{i+1};
    
        % very basic error checking
        if(~strcmp(name, ''))
    
          %save the value in the calling work space
          assignin('base', name, values)
        end
      end
    
    end
    
    
    function skipLines(n)
      while(n > 0)
        fgetl(fid);
        n = n - 1;
      end
    end
    
    function n = countPrefix(prefix)
      pos = ftell(fid);
    
      n = 0;
      currLine = fgetl(fid);
      while(currLine(1) == prefix)
        currLine = fgetl(fid);
        n = n + 1;
      end
      fseek(fid, pos, 'bof');
    end
    
    
    end
    

    这篇关于从MATLAB数据采集的问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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