Common Commands in Low-Level File I/O

Transcription

1 Common Commands in Low-Level File I/O feof(fid), which refers to end-of-file, returns 1 if a previous operation set the end-of-file indicator for the specified file. tline = fgetl(fid) returns the next line of the specified file, removing the newline characters. tline is a text string unless the line contains only the end-of-file marker. In this case, tline is the numeric value clear all; 2 clc 3 % main 4 fid=fopen('fgetl.m'); % open a file 5 tline=fgetl(fid); 6 while ischar(tline) % ischar tells if tline is a char 7 disp(tline) 8 tline=fgetl(fid); 9 end 10 fclose(fid); 1 / 33

2 Close Files fclose(fid) closes an open file. fclose( all ) closes all open files. You can also use fclose all. fclose returns a status of 0 when the close operation is successful. Otherwise, it returns 1. 2 / 33

3 fprintf Recall that fprintf(format, A 1,..., A n ) formats data and displays the results on the screen 1. fprintf(fid, format, A 1,..., A n ) applies the format to all elements of arrays A 1,..., A n in column order, and writes the data to a text file. format: Format of the output fields, specified as a string. A1,..., A n : arrays. Try fprintf(1, format, A 1,..., A n ). 1 Recall disp. 3 / 33

4 Recall that MATLAB reserves fids 1 and 2 for standard output on the screen, and standard error, respectively. fprintf(1, This is standard output!\n ); fprintf(2, This is standard error!\n ); sprintf(formatspec, A 1,..., A n ), similar to fprintf but returns the results as a string. 4 / 33

5 Example fprintf can print multiple numeric values and literal text to the screen. 1 A=[ ; ]; 2 format='x is %4.2f meters or %8.3f mm\n'; 3 fprintf(format,a) % print on the screen 1 >> 2 3 X is 9.90 meters or mm 4 X is 8.80 meters or mm 5 X is 7.70 meters or mm %4.2f specifies that the first value in each line of output is a floating-point number with a field width of four digits, including two digits after the decimal point. Can you explain %8.3f? 5 / 33

6 fprintf Escape Characters %%: Percent character \\: Backslash \b: Backspace \n: New line \t: Horizontal tab 6 / 33

7 fprintf format Integer, signed: %d or %i Integer, unsigned %u: Base 10 %o: Base 8 %x: Base 16 Floating-point number 2 %f: Fixed-point notation %e: Exponential notation, such as e+00 Characters %c: Single character %s: String 2 See IEEE / 33

8 Example: grep in UNIX/Linux-like Systems findstr(s1, S2) returns the starting indices of any occurrences of the shorter of the two strings in the longer. 1 function grep(filename, pattern) 2 fid=fopen(filename, 'r'); 3 line number=1; 4 while feof(fid)==0 5 line=fgetl(fid); 6 matched=findstr(line, pattern); 7 if ~isempty (matched) % tell if matched is... empty array 8 fprintf('%d: %s \n', line number, line); 9 end 10 line number=line number + 1; 11 end 12 fclose(fid); 8 / 33

9 1 >> grep('grep.m','matched') 2 3 6: matched=findstr(line, pattern); 4 7: if ~isempty (matched) % tell if matched is... empty array 9 / 33

10 Exercise Write a short table of the exponential function to a text file called exp.txt, which looks like: 1 x exp(x) / 33

11 1 clear all; 2 clc; 3 % main 4 x=0:.1:1; 5 A=[x', (exp(x))']; % recall that ' is a... transposition operator. 6 fid=fopen('exp.txt', 'w'); 7 fprintf(fid, '%6s %12s\n', 'x', 'exp(x)'); 8 fprintf(fid, '%6.2f %12.8f\n', A); 9 fclose(fid); 10 dos('start exp.txt'); % Open exp.txt dos( command ), for Windows systems, calls upon the shell to execute the given command. Try dos( start ). 11 / 33

12 fscanf A=fscanf(fid, format, size) reads data from the file specified by file identifier fid, converts it according to the specified format string, and returns it in matrix A. fscanf populates A in column order. fscanf can be used to skip specific characters in a sample file, and return only numeric data. 12 / 33

13 Example 1 clear all; 2 clc 3 str=[ 4 '78' char(176) 'C '; 5 '72' char(176) 'C '; 6 '64' char(176) 'C '; 7 '66' char(176) 'C '; 8 '49' char(176) 'C '; 9 ]; 10 % char(176) is the symbol of degree 11 fid=fopen('temperature.txt', 'w'); 12 fprintf(fid, '%s\n', str'); 13 fclose(fid); 14 % main 15 fid=fopen('temperature.txt', 'r'); 16 [A, count]=fscanf(fid, ['%d', char(176), 'C']) 17 fclose(fid); 13 / 33

14 1 >> 2 3 A = count = / 33

15 Binary Files fread(fid, size, precision) interprets values in the file according to the form and size described by precision. Valid entries for size are: N: read N elements into a column vector. inf : read to the end of the file. [M, N]: read elements to fill an M-by-N matrix, in column order. Valid entries for precision are: uchar : unsigned integer, 8 bits. int64 : integer, 64 bits. uint64 : unsigned integer, 64 bits. float64 : floating point, 64 bits. Note that 64 can be replaced by 8, 16, and 32. fwrite(fid, A, precision) translates the values of A according to the form and size described by precision. 15 / 33

16 Example Create a binary file containing a 3-by-3 magic square, whose element is stored as 4-byte integers. 1 clear all; 2 clc 3 % main 4 A=magic(3) 5 fid = fopen('magic3.txt', 'w'); 6 fwrite(fid, A, 'int32'); 7 fclose(fid); 8 fid = fopen('magic3.txt', 'r'); 9 fread(fid,100, 'int32') 16 / 33

17 Access to Internet urlread(url, Name, Value) returns the contents of a URL as a string. 1 contents = urlread(' 2 ~d /matlab.html'); 3 fid=fopen('matlab.html','w'); 4 fprintf(fid,'%s', contents); 5 fclose all; 6 dos('start matlab.html') Try sendmail, ftp. 17 / 33

18 Exercise 假設三元一次聯立方程式儲存在 txt 檔案內 請寫一個程式, 讀取 txt 檔內聯立方程式的係數矩陣 A 和常數陣列 b, 格式如下 : 利用 function 的方式呼叫 Gauss Elimination 解 A x = b 的問題 以 The solution is (x, y, z) =... 的格式, 將結果加回本來的檔案內 整個流程希望可以只用一個指令就可以執行完畢, 所以要把以上的動作都包裝成 function 輸入 gausssolver(myfile) 即可完成三元一次聯立方程式求解 18 / 33

19 1 function gausssolver(str) 2 % main 3 fid=fopen(str,'r'); 4 k=1; 5 temp(3,4)=0; 6 while k<4 7 temp(k,:)=str2num(fgetl(fid)); 8 k=k+1; 9 end 10 fclose all; 11 A=temp(:,1:3); b=temp(:,4); % init A and b 12 [y s]=gauss(a,b); 13 if s==1 % a flag to tell if succeeded 14 fid=fopen(str,'a'); 15 fprintf(fid,'\nthe solution is (%f,%f,%f).',y); 16 fclose all; 17 show=['start ', str]; 18 dos(show); 19 disp('done.') 20 else 19 / 33

20 21 disp('failed in Gauss Elimination.') 22 end 1 function [y s]=gauss(a,b) 2 % s is a flag to indicate the success of Gauss... elimination / 33

21 Exercise 讀取 txt 檔內的聯立方程式, 並將分離出係數矩陣 A 和常數矩陣 b 即,txt 檔案內可能存 : 3x + 2y z = 1 2x 2y + 4z = 2 x + 0.5y z = 0 所以需要把 x, y, z 與 = 和係數分離開來 21 / 33

22 But... gausssolver1 cannot deal with the strings like 3x + 2y 1z = 1. So, let s find a way out. 1 function gausssolver2(str) 2 % main 3 fid=fopen(str,'r'); 4 k=1; 5 temp=cell(3,1); 6 while k<4 7 temp{k}=fgetl(fid); 8 k=k+1; 9 end 10 fclose all; 11 [A,b]=init matrix(temp) % deal with the strings 12 [y s]=gauss(a,b); 13 if s==1 % a flag to tell if succeeded 22 / 33

23 Title 14 fid=fopen(str,'a'); 15 fprintf(fid,'\nthe solution is (%f,%f,%f).',y); 16 fclose all; 17 show=['start ', str]; 18 dos(show); 19 disp('done.') 20 else 21 disp('failed in Gauss Elimination.') 22 end 23 / 33

24 Assume that we are solving a system of linear equations with 3 unknowns and 3 constraints. 1 function [A b]=init matrix(temp) 2 3 delete char=['x','y','z','=']; 4 A=zeros(3,4); 5 loc=zeros(1,4); % indicators for deleting characters 6 for i=1:3 7 temp conversion=temp{i}; % Notice that temp... is a cell array. 8 j=1; 9 k=1; 10 while j<length(temp conversion) && k<5 11 if temp conversion(j)==delete char(k) 12 loc(k)=j; 13 k=k+1; 14 end 15 j=j+1; 24 / 33

25 16 end 17 A(i,:)=[str2num(temp conversion(1:loc(1)-1)), str2num(temp conversion(loc(1)+1:loc(2)-1)), str2num(temp conversion(loc(2)+1:loc(3)-1)), str2num(temp conversion(loc(4)+1: length(temp conversion)))] 22 end 23 b=a(:,4); 24 A=A(:,1:3); In the exercise, parsing is a very tedious task. You can imagine more examples in daily life. 25 / 33

26 Code Style A more clean source code could be like this: 1 function [y s]=gausssolver3(str,var) 2 % main 3 temp=read(str,var); 4 [A,b]=init matrix(temp) 5 [y s]=gauss(a,b); 6 write(str,y,s); 7 end 8 9 function temp=read(str,var) 10 varnum=length(var); 11 fid=fopen(str,'r'); 12 k=1; 13 temp=cell(varnum,1); 14 % while feof(fid)==0 15 while k<varnum+1 16 temp{k}=fgetl(fid); 26 / 33

27 Title 17 k=k+1; 18 end 19 fclose all; 20 end function write(str,y,s) 23 if s==1 24 fid=fopen(str,'a'); 25 fprintf(fid,'\nthe solution is ('); 26 for i=1:length(y)-1 27 fprintf(fid,'%f,',y(i)); 28 end 29 fprintf(fid,'%f).',y(length(y))); 30 fclose all; 31 show=['start ', str]; 32 dos(show); 33 disp('done.') 34 else 35 disp('failed in Gauss Elimination.') 27 / 33

28 Title 36 end 37 end 28 / 33

29 Exercise (Upgraded) Features: 1. 解任意由 n 條方程式組成的 n 元一次聯立方程式 2. 處理諸如 x + y z = 之類的方程式 即, 讓程式可以讀出 x 的係數是 1, z 的係數是 1 等 (Try.) 1 function [y s]=gausssolver4(str,var) 2 % main [A,b]=init matrix 1(temp,var) / 33

30 1 function [A b]=init matrix 1(temp,var) 2 delete char=[var,'=']; 3 varnum=length(delete char)-1; 4 A=zeros(varnum,varnum+1);loc(1:varnum+1)=0; 5 for i=1:varnum 6 temp conversion=temp{i}; 7 j=1; 8 k=1; 9 while j<length(temp conversion) && k<varnum+2 10 if temp conversion(j)==delete char(k) 11 loc(k)=j; 12 k=k+1; 13 end 14 j=j+1; 15 end 16 %% WARNING 17 for j=1:varnum+1 18 if j==1 19 A(i,j)=str2num(temp conversion(1:loc(j)-1)); 20 else if j==varnum+1 30 / 33

31 21 A(i,j)=str2num(temp conversion(loc(j) :length(temp conversion))); 23 else 24 A(i,j)=str2num(temp conversion(loc(j-1) :loc(j)-1)); 26 end 27 end 28 end 29 %% WARNING 30 end 31 b=a(:,varnum+1); 32 A=A(:,1:varnum); 31 / 33

32 Check List Before the end of this short class, you should check whether or not you have been familiar with: 1. (Problem Formulation) Make a clear definition of problem. Inputs Outputs 2. (Algorithm) Organize the resource and steps to reach the goal. 3. (Implementation) Implement the algorithm using a programming language, such as MATLAB. 4. (Debugging) There are always bugs in the program. Ready to fight? 5. (Complexity Analysis) Improve the program if the computation cost is unacceptably high. 6. (Generalization) Make your program more flexible. 32 / 33

33 Practice Makes Better Data Structures and Algorithms in C++, Michael T. Goodrich, Roberto Tamassia, and David M. Mount, 2/e, 2011 Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein, 3/e, 2009 高中生程式解題系統 33 / 33