我想计算第 2 列中的值的累积和dat.txt下面是第 1 列中的每个字符串。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);
这是一个使用函数的完全矢量化（尽管看起来有些混乱）的解决方案CUMSUM http://www.mathworks.com/access/helpdesk/help/techdoc/ref/cumsum.html and DIFF http://www.mathworks.com/access/helpdesk/help/techdoc/ref/diff.html随着逻辑索引 http://www.mathworks.com/access/helpdesk/help/techdoc/math/f1-85462.html#bq7egb6-1产生你想要的结果：
>> 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
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