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poster · 2019-12-10, 20:41

no time scores 1 10 123 2 11 22 3 12 22 4 50 55 5 60 22 6 70 66 . . . . . . nnn 以上是我的txt文件的内容（一千行）。

1st column - number of samples 2nd column - time (from beginning to end ->accumulated) 3rd column - scores 我想创建一个新文件，该文件将是分数的每三个样本的总和除以同一样本的时间差。

eg (123+22+22)/ (12-10) = 167/2 = 83.5 (55+22+66)/(70-50) = 143/20 = 7.15 新的txt文件

83.5 7.15 . . . n 到目前为止，我有此代码：

fid=fopen('data.txt') data = textscan(fid,'%*d %d %d') time = (data{1}) score= (data{2}) for sample=1:length(score) ..... // I'm stucked here .. end ....

回答：

%# Easier to load with importdata data = importdata('data.txt',' ',1); %# Get the number of rows n = size(data,1); %# Column IDs time = 2;score = 3; %# The interval size (3 in your example) interval = 3; %# Pre-allocate space new_data = zeros(numel(interval:interval:n),1); %# For each new element in the new data index = 1; %# This will ignore elements past the closest (floor) multiple of 3 as requested for i = interval:interval:n %# First and last elements in a batch a = i-interval+1; b = i; %# Compute the new data new_data(index) = sum( data(a:b,score) )/(data(b,time)-data(a,time)); %# Increment index = index+1; end

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2019-12-10, 20:41	#1
poster 高级会员注册日期: 2019-11-21 帖子: 3,006 声望力: 66	循环MATLAB问题 no time scores 1 10 123 2 11 22 3 12 22 4 50 55 5 60 22 6 70 66 . . . . . . nnn 以上是我的txt文件的内容（一千行）。 1st column - number of samples 2nd column - time (from beginning to end ->accumulated) 3rd column - scores 我想创建一个新文件，该文件将是分数的每三个样本的总和除以同一样本的时间差。 eg (123+22+22)/ (12-10) = 167/2 = 83.5 (55+22+66)/(70-50) = 143/20 = 7.15 新的txt文件 83.5 7.15 . . . n 到目前为止，我有此代码： fid=fopen('data.txt') data = textscan(fid,'%d %d %d') time = (data{1}) score= (data{2}) for sample=1:length(score) ..... // I'm stucked here .. end .... 回答：* %# Easier to load with importdata data = importdata('data.txt',' ',1); %# Get the number of rows n = size(data,1); %# Column IDs time = 2;score = 3; %# The interval size (3 in your example) interval = 3; %# Pre-allocate space new_data = zeros(numel(interval:interval:n),1); %# For each new element in the new data index = 1; %# This will ignore elements past the closest (floor) multiple of 3 as requested for i = interval:interval:n %# First and last elements in a batch a = i-interval+1; b = i; %# Compute the new data new_data(index) = sum( data(a:b,score) )/(data(b,time)-data(a,time)); %# Increment index = index+1; end 更多&回答...