我在4195X1中包含的数据被称为z1。我想提取120块数据，并使用matlab将其标记为z1_1_120，z1_120_240，z1_240_360等。我既要提取它们，也要以这种方式在循环中标记它们。这是我到目前为止所做的，但是不确定如何进行：
load(z1)
for i = 1:4195
q=z1(i);
q1(i,:)=q;
q2=q1(1:120:end);
end
z1=q2(1:end);
正如丹尼尔评论，不要这样做！
如果需要描述的标签类型，则可以使用Table或Struct数据类型。
解决方案很简单：不要这样做。这是你膝盖上的一枪
以下解决方案创建一个表和一个结构。该表需要一些填充，因为所有列的长度必须相同（并且4195不是120的倍数）。
代码有点复杂，我试图在不使用for循环的情况下解决它（以使解决方案更有效[并且更有趣]）。我希望您希望遵循该代码并从中学习... ...>
这里是一个代码示例（注释中有解释：）>
% Fill z1 with sequential numbers (just for testing) z1 = (1:4195)'; z1_len = length(z1); % Compute length remainder from 120 len_mod120 = mod(z1_len, 120); pad_len = mod(120 - len_mod120, 120); % If length of z1 is not a multiple of 120, add zeros padding at the end. pad_z1 = padarray(z1, pad_len, 0, 'post'); % After padding, length of z1 is 4120 % Reshape pad_z1 into a matrix where each row is 120 elements % Z1(:, 1) gets z1(1:120), Z1(:, 2) gets z1(121:240)... Z1 = reshape(pad_z1, [], length(pad_z1)/120); % Build naming indices name_idx = zeros(1, 2*length(pad_z1)/120); name_idx(1:2:end) = 1:120:length(pad_z1); %First naming index: 1 121 241 361 ... name_idx(2:2:end) = name_idx(1:2:end) + 120-1; %Second naming index: 120 240 360 480 % String of elements names separated by space str_names = sprintf('z1_%d_%d ', name_idx); % 'z1_1_120 z1_121_240 z1_241_360 z1_361_480 ... % Build cell array of names var_names = split(str_names(1:end-1)); %{'z1_1_120'}, {'z1_121_240'}, {'z1_121_240'} % Build table, where each column is 120 elements, and column names are 'z1_1_120' 'z1_121_240' 'z1_121_240' % A table is useful, if you don't care about the zero padding we added at the beginning % https://www.mathworks.com/matlabcentral/answers/376985-how-to-convert-string-to-variable-name T = array2table(Z1, 'VariableNames', var_names); % Convert table to struct: % S.z1_1_120 holds first 120 elements, S.z1_121_240 holds next 120 elements... S = table2struct(T, 'ToScalar', true); % Fix the last field in the stract - should contain only 115 elements (not 120) S = rmfield(S, var_names{end}); % Remove the last field from the struct % last_field_name = 'z1_4081_4195' last_field_name = sprintf('z1_%d_%d', name_idx(end-1), z1_len); % Add the last field (only 195 elemtns) S.(last_field_name) = z1(end-len_mod120+1:end);表T：
T = 120×35 table z1_1_120 z1_121_240 z1_241_360 z1_361_480 ... ________ __________ __________ __________ 1 121 241 361 2 122 242 362 3 123 243 363 4 124 244 364 5 125 245 365 6 126 246 366 ... ... ... ...访问第一个元素的示例：
T.z1_1_120(1)S Struct S：
S = struct with fields: z1_1_120: [120×1 double] z1_121_240: [120×1 double] z1_241_360: [120×1 double] z1_361_480: [120×1 double] z1_481_600: [120×1 double] z1_601_720: [120×1 double] z1_721_840: [120×1 double] ... z1_4081_4195: [115×1 double]访问第一个元素的示例：
S.z1_1_120(1)