How would I replace a string of numbers with more numbers using for and if statements?

3 ビュー (過去 30 日間)
Here is my code so far. I'm trying to make it so that if the number is "1" in the string, it replaces it with five zeros. If the number is "2" in the string it replaces that with 5 ones, and if the number is "3" in the string, it replaces it with 5 twos and make that all a different string of numbers called d. The end should look like this: d = [0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,2,2,2,2,2]. I really need help, I am awful with matlab.
if true
t = [1,2,1,2,1,3];
% For Loop
for n = 1:length(t)
if t(n) == 1
d(n) = 0,0,0,0,0;
elseif t(n) == 2
d(n) = 1,1,1,1,1;
else t(n) == 3
d = 2,2,2,2,2;
end
end
end

採用された回答

madhan ravi
madhan ravi 2018 年 11 月 3 日
編集済み: madhan ravi 2018 年 11 月 3 日
t = string([1,2,1,2,1,3]);
d=cell(1,numel(t)); %pre-allocation
for i = 1:numel(t)
if t(i)=='1'
d{i}=[ones(1,5)*0];
elseif t(i)=='2'
d{i}=[ones(1,5)*1];
elseif t(i)=='3'
d{i}=[ones(1,5)*2];
end
end
d = horzcat(d{:})
command window:
>> COMMUNITY
d =
Columns 1 through 13
0 0 0 0 0 1 1 1 1 1 0 0 0
Columns 14 through 26
0 0 1 1 1 1 1 0 0 0 0 0 2
Columns 27 through 30
2 2 2 2
>>
  2 件のコメント
Jacob Roach
Jacob Roach 2018 年 11 月 3 日
編集済み: Jacob Roach 2018 年 11 月 3 日
Wow, I didn't even know about the horzcat command, do you think you would be able to explain your thought process or something? I know this is a simple matlab code for you probably, but it's so frustrating not being able to understand why you're doing what you do. Thank you so much!
madhan ravi
madhan ravi 2018 年 11 月 3 日
1)just created 1 by 5 cell array;
2) made a loop until the number of cell array;
3)made if and else if conditions to store elements in each cell when the condition is satisfied;
4)finally concatenated them horizontally.
What's the advantage?
Cells are huge containers they can compress how much ever you feed inside them unlike the normal arrays

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その他の回答 (1 件)

Tyler Johns
Tyler Johns 2018 年 11 月 3 日
Keep in mind when programming there is usually more than one way to accomplish your goal. Here is an additional solution. This may not be the most efficient way since the size of vector d changes every loop iteration but it is still a solution nonetheless.
t = [1 2 1 2 1 3];
d=[]; % create empty vector d
for n = 1:length(t)
if t(n)==1
d = [d 0 0 0 0 0]; %add new numbers to end of vector d
elseif t(n)==2
d = [d 1 1 1 1 1];
elseif t(n)==3
d = [d 2 2 2 2 2];
end
end
disp(d)
  4 件のコメント
Jacob Roach
Jacob Roach 2018 年 11 月 3 日
Alright, so basically I need to make a "random walker" that walks in either direction using a for loop. It can walk forward one, back one back one forward one, it just has to be random for a thousand times. Then I need to use comet to simulate the walk, and im just not sure how to set up this code. So far I have this. I need to plot the walk as a function of time. I don't know what I'm doing wrong.
%%Part 1
% Simulation of Walker
t = linspace(0,1001,50000);
x = 0;
d = randn(0,2);
for n = 1:1001
if d == 0
x = x;
elseif d == 1
x = x + 1;
else d == 2
x = x - 1;
end
end
comet(x,t);
Stephen23
Stephen23 2018 年 11 月 3 日
Follow madhan ravi's answer: it shows good practice: a preallocated cell array to store the intermediate arrays.

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