2015-10-06 150 views
1

我想繪製包含x,y,z,t的數千個數據點的圖形。的數據是在一個.txt文件和如下所示:Matlab 4dplot(x,y,z,t)

  • [X,Y,Z,時間]
  • [50.9160,12.2937,-44.9963,0.0]
  • [50.9160,12.2937, - 44.9963,0.8]
  • [50.9160,12.2937,-44.9963,1.8]
  • [50.9160,12.2937,-44.9963,2.8]
  • [50.9160,12.2937,-44.9963,3.8]
  • [50.9160,12.2937 ,-44.9963,4.9]
  • [50.9160,12.2937,-44.9963,8.8]
  • [50.9160,12.2937,-44.9963,11.1]
  • [50.9160,12.2937,-44.9963,11.7]
  • [50.9160,12.2937,-44.9963,12.8 ]
  • [50.8989,12.3248,-45.0376,13.7]
  • [50.8989,12.3248,-45.0376,14.9]
  • [50.8989,12.3248,-45.0376,15.7]
  • [50.8989,12.3248,-45.0376 ,17.2]
  • [50。 8989,12.3248,-45.0376,17.7]
  • 等(超過一千個數據點)

我一起打開文本文件的行思,創建一個循環,將搶X,Y, z,t並使用scatter3進行繪圖。如果有人能讓我開始瞭解MATLAB代碼應該是什麼樣子,那很好。

+1

不能Matlab文件讓你開始?嘗試一些併發布一些代碼,然後再詢問:http://www.mathworks.com/help/matlab/ref/csvread.html和http://www.mathworks.com/help/matlab/ref/scatter3.html首先 – Dan

+0

正如丹說,你應該閱讀這些文件,並嘗試其中的一些例子。我能看到的最大問題是如何產生時間。對於這個問題,我認爲有時間顏色編碼將是最好的,但在看到結果之前很難確定。此外,有時間作爲顏色代碼是有點直觀的。但是,這將有助於指示線路(因爲數據被採樣) – patrik

+1

嘿。看看4D繪圖的不同可能性:https://stackoverflow.com/questions/27659632/reconstructing-three-dimensions-image-matlab/27660039#27660039 https://stackoverflow.com/questions/29229988/visualize-a -three-dimensional-array-like-cubic-lattice-using-matlab/29233108#29233108或https://stackoverflow.com/questions/31828064/constructing-voxels-of-a-3d-cube-in-matlab/31829681 #31829681 –

回答

1

我真的不明白你想如何在4D中進行繪圖,但我猜你希望以3D繪圖,隨着時間的推移動態改變繪圖。

但不管怎麼說,有關文件:

首先,文本閱讀,真是簡單。

  1. 你只需要創建一個變量,即file_text_rd,然後使用這個變量來打開文件:

    file_text_rd = fopen('data.txt','r');

    的第一個參數是.txt文件的名稱(請注意,您需要將目錄設置爲.txt文件所在的文件夾)。第二個參數表明您希望從文本文件中讀取。然後,您可以從文件中讀取數據並將其放入具有不同功能的變量中(取決於您最適合的變量)。例如,

    var=fgetl(file_text_rd);

    會把文件內容的第一行中的變量。

    var=fscanf(file_text_rd,%c);

    將把.txt文件的全部內容在變量等其他閱讀功能freadfgets。所以,根據功能的不同,您可能想要使用一些循環功能來填充您的內容var

  2. 當你與你的文件讀取完成後,需要關閉該文件用:

    fclose(file_text_rd), clear file_text_rd;

接下來的部分是可能更多麻煩一點點。這是你將你閱讀的字符轉換爲整數的部分。我爲你寫了一段代碼,來說明實現這一點的一種方法。在這個解決方案中,我使用了fscanf函數。

%Open file for reading 
file_text_rd=fopen('data.txt','r'); 
%Read the content (%c means you are reading characters) 
var=fscanf(file_text_rd,'%c'); 
%Converse the characters to double. Have in mind the ascii values of the 
%chars, so you can get the actual number value of the numbers in the string 
%by subtracting the 48 of the original value, since the zero in ascii is 
%numbered as 48 (in decimal system). 
conv_var=double(var)-48; 
%Define the initial value of your variable (all zeros) 
final_var=zeros(1,4); 
%Row counter 
count_r=1; 
%Column counter 
count_c=1; 
%Divider 
times=10; 
%Dot flag 
dot=0; 
%Negative sign flag 
negative_sign=0; 
%This for loop is for testing every single character from the first to the 
%last 
for i=1:size(conv_var,2)-1  
    %This if condition is for: 
    %1. Checking if the character is a number between 0 and 9 
    %2. Checking if the character is a dot 
    %3. Checking if the character is a minus sign 
    %4. Checking if the character is a comma 
    %All other characters are not of interest. 
    if (conv_var(i)>=0 && conv_var(i) <=9) || conv_var(i) == -2 || conv_var(i) == -3 || conv_var(i) == -4 

     %If it's not a comma (that means we are still on the last number we 
     %were working on) we go in this section 
     if conv_var(i)~= -4 
      %If it's not a minus sign we go in this section 
      if conv_var(i) ~= -3 
       %If the dot flag hasn't been set to 1 yet (no dot in the 
       %string has yet been found) we don't enter this section 
       if dot==1 
        %If the flag HAS been set, then the number just found 
        %on the sequence is divided by 10 and then added to the 
        %old versison, since if we are reading the number 
        %'50.9160', the 9 has to be divided by 10 and then 
        %added to 50 
        final_var(count_r,count_c)=final_var(count_r,count_c)+conv_var(i)/times; 
        %The divider now rizes because the next found number 
        %would be 10 times smaller than the one found just now. 
        %For example, in '50.9160', 1 is 10 times less than 9 
        times=times*10; 
       else 
        %This condition is needed so we don't add the ascii 
        %number equivalent to the dot to the number we are 
        %working on. 
        if conv_var(i)~=-2 
         %We multiply the old version of the number we are 
         %working on, since if we are reading the number 
         %'50.9160', first we will read 5, then we will read 0, 
         %so we will need to multiply 5 by 10 and then add the 0 
         %and so on... 
         final_var(count_r,count_c)=final_var(count_r,count_c)*10+conv_var(i); 
        else 
         %If the character found IS the dot, then we just 
         %set the flag 
         dot=1; 
        end 
       end 
      else 
       %If the character found IS the negative_sign, then we set 
       %the flag for the negative_sign, so we can multiply the 
       %number we are working on atm with -1. 
       negative_sign=1; 
      end    
     else 
      %We get in this section if we found a comma character (or the 
      %ascii equvalent of the comma sign, more accurately) 
      if negative_sign==1 
       %This is the part where we multiply the number by -1 if 
       %we've found a minus sign before we found the comma 
       final_var(count_r,count_c)=-final_var(count_r,count_c); 
      end 

      %Here we add 1 to the column counter, since we are ready to 
      %work with the next number 
      count_c=count_c+1; 

      %We reset all the flags and the divider 
      dot=0; 
      times=10; 
      negative_sign=0; 
     end 

    end 

    %The number -38 in ascii is the equivalent of NL, or the end of the 
    %line sign (which we can't see), which actually means there was an "Enter" pressed at this point 
    if conv_var(i)==-38 
     %We set the column counter to one since, we will work now with the 
     %first number of the next four parameters 
     count_c=1; 

     %We increment the row counter so we can start saving the new values 
     %in the second row of our matrix 
     count_r=count_r+1; 

     %We set the next row initially to be all-zeros 
     final_var(count_r,:)=zeros(1,4); 

     %We reset the flags 
     dot=0; 
     times=10; 
     negative_sign=0; 
    end 
end 

%We close the file, since our work is done (you can put this line after the 
%fscanf if you like) 
fclose(file_text_rd), clear file_text_rd; 

我相信,你對如何繪製你的4D數據的一些想法,即使我真的不能趕上那部分了。

我希望我幫忙, 博揚

+0

謝謝你的幫助!我有一個進一步的問題。在我的數據中,括號前有負號。例如: - [50.9160,12.2937,-44.9963,0.0] - [50.9160,12.2937,-44.9963,0.8]。你能否修改你的代碼,以便在開始時不考慮負數? – Sagistic

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