Returning Variables as Characters
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I have function such that
a= rand (1,10); % it generates random real numbers from 0 to 1
i=1:10;
if b(i)< 0.4;
b(i)= 'A';
else if
b(i) = 'B'
end
the purpose of this code is for example
b = 0.23 0.45 0.89 0.74 0.48 0.98 0.12 0.05 0.56 0.14
I want it to return as a result
b= A B B B B B A A B A
댓글 수: 2
Guillaume
2017년 12월 21일
@Dogukan,
We really don't appreciate when the original poster edit their question away once they've got an answer. This is not a private consulting service. The answers we post are not just for you, they're intended to help anybody with the same issue. If you edit the question away, the answers become meaningless.
It also makes us a lot less likely to answer your questions in the future.
Thankfully, MathWorks doesn't like it either and have a policy of restoring questions edited away to their original state, so it's also pointless.
답변 (2개)
Stephen23
2017년 12월 21일
편집: Stephen23
2017년 12월 21일
method one: character code:
>> b = char(66-(a<0.4))
b = ABBBBBAABA
method two: sprintf:
>> b = sprintf(' %c',66-(a<0.4))
b = A B B B B B A A B A
method three: indexing (char):
>> c = 'AB';
>> b = c(2-(a<0.4))
b = ABBBBBAABA
method four: indexing (cell):
>> c = {'A','B'}65_66
>> b = c(2-(a<0.4))
b = 'A' 'B' 'B' 'B' 'B' 'B' 'A' 'A' 'B' 'A'
method five: generate characters directly using randi:
>> char(randi([65,66],1,10))
ans = ABBBBBAABA
댓글 수: 0
Pawel Jastrzebski
2017년 12월 21일
편집: Pawel Jastrzebski
2017년 12월 21일
clear all;
clc;
data = rand (1,10)
%logical vector to establish when datapoint is 'A' and 'B'
% Condition 1: 'A' if 'data <= 0.5'
% Condition 2: 'B' if 'data > 0.5'
logicValA = data <= 0.5;
logicValB = data > 0.5;
% Use logical vectors to separate 'data' into 'A's' and 'B's':
% 'newData' as 'char vector' → all values concatenated
newData(logicValA) = 'A'
newData(logicValB) = 'B'
% 'newData1' as 'cell array' → all values separated
newData1 = {}
newData1(logicValA) = {'A'}
newData1(logicValB) = {'B'}
댓글 수: 2
Walter Roberson
2017년 12월 26일
A<B<C means to compare A and B first getting out 0 (false) or 1 (true), and then to compare that 0 or 1 to C. You need to test A<B&B<C
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