Counting number of certain letters

I'm trying to make a code that will output the number of a's and e's in my name. I have a basic understanding that I need to go through each element of the string and test if it's an 'a' or 'e' then add up each time. What am I doing wrong below?

myname = "jonathan"
  A = char(myname)
  numA=0;
  numE=0;
  for i=1:1:length(A)
      if i == 'a'
          numA = numA+1
      end    
      if i == 'e'
          numE = numE+1
      end    
  end    
  disp('The number of As: ');
  disp(numA);
  disp('The number of Es: ');
  disp(numE);

Answer 1

[per isakson]

"What am I doing wrong below?"

1. "to go through each element of the string" your for-loop is fine
2. "test if it's an 'a' or 'e' then add up each time" your tests don't test what you want them to test

The loop-counter, i, takes the numerical values 1,2,3,... You test whether this numerical value is equal to a character value. That smells. However, Matlab "thinks" that you want to compare the numerical value to the "ascii-value" of the character, e.g.

>> 97=='a'
ans =
     1

Here 1 stands for true

Replace

i == 'a'

by

A(i) == 'a'

to compare the i:th character of A to the character, 'a'

Answer 2

[Image Analyst]

Why not just take the histogram?

myname = 'jonathan'
% Convert to numbers:
myname = lower(myname) - '0' - 48
% Take the histogram.
edges = 1:26;
counts = histcounts(myname, edges)
% counts(1) is the number of "a"s in the word.
% counts(26) is the number of "z"s in the word.

[Shalini Lokanathan]

I am doing a task similar to this. Can you explain me what the last part of this piece means? I understand that the first parts makes the text into lowercase, but I don't understand the rest.

myname = lower(myname) - '0' - 48

[ABDALHADI ABU ZEYNEH]

hello

i am a phd student

i am using matlab, i meet a problem

how to find the occurrence of non consecutive letters

for example

the occurence of ab in aabbb=6 i mean that there are for choices to get a before "b"

in acacbbc we can find the word abc four times,i mean that there are for choices to get a before "b" and "b" before "c"

..

[Walter Roberson]

Clearer would be

myname = lower(myname) - 'a' + 1;