MATLAB中ismember函数处理nx1字符数组查找空格异常问题

Let's break down why your current approach isn't working and how to fix it:

The Root Cause #

When you convert a string array (like your String2) using convertStringsToChars, you don't get a 2D character array—you get a nx1 cell array, where each cell contains a separate character array.

Your original working example used a single character array, so ismember worked as expected. But for the cell array:

• In Plan A, ismember(Char2, StringSearchedSpaces) compares entire cell elements (each being a character array) to a single space character, which will always return false (0).
• In Plan B, Char2(z,:) accesses the cell container instead of the actual character array inside it—you need curly braces {} to pull out the content of each cell.

Corrected Solutions #

Option 1: Fixed For Loop #

Update your Plan B to use curly braces to access each cell's character array:

StringSearchedSpaces = ' '; % White space
String2 = ["Random info in middle one, "+ ...
"Random info still continues. ", "Random info in middle two. "+ ...
"Random info still continues. "];
Char2 = convertStringsToChars(String2);

% Fixed Plan B
PlanBString2Spaceidx = cell(size(Char2)); % Initialize cell array to store results
for z = 1:length(Char2)
    PlanBString2Spaceidx{z} = ismember(Char2{z}, StringSearchedSpaces);
end

This stores a logical array in each cell of PlanBString2Spaceidx, matching your working example: each array has 1s where spaces are present, 0s elsewhere.

Option 2: Simplify with cellfun #

You can skip the manual loop using cellfun, which applies a function to every element of a cell array:

% Clean alternative with cellfun
Spaceidx = cellfun(@(x) ismember(x, StringSearchedSpaces), Char2, 'UniformOutput', false);

The 'UniformOutput', false argument tells MATLAB to return a cell array of results (since each character array might have a different length).

Verifying the Result #

To confirm it's working, display one of the results:

% Show logical array for the first string
disp(Spaceidx{1})

You'll see 1s in positions where spaces exist, identical to the behavior of your 1x1 example.

内容的提问来源于stack exchange，提问作者Div