Using an if/else statement inside of a for loop

Hi! I am trying to write an if else statement inside of a for loop in order to determine how many people surveyed had a specific response. I posted my code below. Every time I run it instead of generating the numbers, it generates my fprintf statement that amount of time. I am just really confused because I followed the template my professor gave and it just isn't working for me. Thank you,
y=load('Class19_survey.txt');
ag=0;
ne=0;
dis=0;
for k=1:length(y)
if y(k)>=4
ag=ag+1;
fprintf('\nThe number of people who agree is\n',ag)
elseif y(k)==3
ne=ne+1
fprintf('\nThe number of neutral responses is\n',ne)
else y(k)<=2
dis=dis+1
fprintf('\nThe number of disagree responses is\n', dis)
end
end

댓글 수: 1

Sorry here is an easier version to read
y=load('Class19_survey.txt');
ag=0;
ne=0;
dis=0;
for k=1:length(y)
if y(k)>=4
ag=ag+1;
fprintf('\nThe number of people who agree is\n',ag)
elseif y(k)==3
ne=ne+1
fprintf('\nThe number of neutral responses is\n',ne)
else y(k)<=2
dis=dis+1
fprintf('\nThe number of disagree responses is\n', dis)
end
end

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 채택된 답변

KL
KL 2017년 11월 1일

0 개 추천

Perhaps you want something like this,
y=load('Class19_survey.txt');
ag=0;
ne=0;
dis=0;
for k=1:length(y)
if y(k)>=4
ag=ag+1;
elseif y(k)==3
ne=ne+1;
else y(k)<=2
dis=dis+1;
end
end
fprintf('\nThe number of people who agree is\n',ag)
fprintf('\nThe number of neutral responses is\n',ne)
fprintf('\nThe number of disagree responses is\n', dis)
but you could also do it without the loop and if statements,
ag = length(y(y>=4));
ne = length(y(y==3));
dis = length(y(y<=2));

추가 답변 (1개)

Walter Roberson
Walter Roberson 2017년 11월 1일

0 개 추천

y=load('Class19_survey.txt');
ag=0;
ne=0;
dis=0;
for k=1:length(y)
if y(k)>=4
ag=ag+1;
elseif y(k)==3
ne=ne+1;
else y(k)<=2
dis=dis+1;
end
end
fprintf('\nThe number of people who agree is\n',ag);
fprintf('\nThe number of neutral responses is\n',ne);
fprintf('\nThe number of disagree responses is\n', dis);

카테고리

도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

질문:

2017년 11월 1일

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2020년 6월 29일

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