2

我想为第 1 列dat.txt中的每个字符串计算下面第 2 列中的值的累积总和。所需的输出显示为dat2.txt：

dat.txt  dat2.txt
1 20     1 20  20   % 20 + 0
1 22     1 22  42   % 20 + 22
1 20     1 20  62   % 42 + 20
0 11     0 11  11
0 12     0 12  12
1 99     1 99  99   % 99 + 0   
1 20     1 20  119  % 20 + 99
1 50     1 50  169  % 50 + 119

这是我的初步尝试：

fid=fopen('dat.txt');
A  =textscan(fid,'%f%f');
in =cell2mat(A); 
fclose(fid);

i = find(in(2:end,1) == 1 & in(1:end-1,1)==1)+1;
out = in;
cumulative =in;
cumulative(i,2)=cumulative (i-1,2)+ cumulative(i,2);

fid = fopen('dat2.txt','wt');
format short g;
fprintf(fid,'%g\t%g\t%g\n',[out cumulative(:)]');
fclose(fid);
4

3 回答 3

3

不是完全矢量化的解决方案（它循环通过连续 1 的片段），但应该更快。它只为您的数据执行 2 个循环。使用 MATLAB 的CUMSUM函数。

istart = find(diff([0; d(:,1)])==1); %# start indices of sequential 1s
iend = find(diff([d(:,1); 0])==-1); %# end indices of sequential 1s

dcum = d(:,2);
for ind = 1:numel(istart)
    dcum(istart(ind):iend(ind)) = cumsum(dcum(istart(ind):iend(ind)));
end

dlmwrite('dat2.txt',[d dcum],'\t') %# write the tab-delimited file
于 2010-06-23T16:36:10.613 回答
3

这是一个完全矢量化（尽管看起来有些混乱）的解决方案，它使用函数CUMSUM和DIFF以及逻辑索引来产生您想要的结果：

>> data = [1 20;...  %# Initial data
           1 22;...
           1 20;...
           0 11;...
           0 12;...
           1 99;...
           1 20;...
           1 50];
>> data(:,3) = cumsum(data(:,2));       %# Add a third column containing the
                                        %#   cumulative sum of column 2
>> index = (diff([0; data(:,1)]) > 0);  %# Find a logical index showing where
                                        %#   continuous groups of ones start
>> offset = cumsum(index.*(data(:,3)-data(:,2)));  %# An adjustment required to
                                                   %#   zero the cumulative sum
                                                   %#   at the start of a group
                                                   %#   of ones
>> data(:,3) = data(:,3)-offset;        %# Apply the offset adjustment
>> index = (data(:,1) == 0);            %# Find a logical index showing where
                                        %#   the first column is zero
>> data(index,3) = data(index,2)        %# For each zero in column 1 set the
                                        %#   value in column 3 to be equal to
data =                                  %#   the value in column 2

     1    20    20
     1    22    42
     1    20    62
     0    11    11
     0    12    12
     1    99    99
     1    20   119
     1    50   169
于 2010-06-23T18:30:58.747 回答
0 
    d=[
1 20     
1 22     
1 20     
0 11     
0 12     
1 99     
1 20     
1 50
];
disp(d)

out=d;
%add a column
out(:,3)=0;

csum=0;
for(ind=1:length(d(:,2)))
    if(d(ind,1)==0)
        csum=0;           
        out(ind,3)=d(ind,2);    
    else
        csum=csum+d(ind,2);
        out(ind,3)=csum;    
    end

end

disp(out)
于 2010-06-23T14:48:08.040 回答