Question

help with matlab plzzzzz

Answer 1

%% 5.23
% The histc function can be used to determine the number of A's, B's,
% C's, D's, and F's in the grade data from problem 5.21, if the
% following grading scheme applies
% A > 90 to 100
% B > 80 to 90
% C > 70 to 80
% D > 60 to 70
% F > 0 to 60
% Use the bar function to create a histogram based on this scheme.
G = [68, 83, 61, 70, 75, 82, 57, 5, 76, 85, 62, 71, 96, 78, 76, 68, 72, 75, 83, 93]; % final test scores

newG = sort(G(:)); % final scores, starts with lowest then end with highest

[rrA, ccA] = find((newG < 100) & (newG > 90));
A = newG(rrA);

[rrB, ccB] = find((newG < 90) & (newG > 80));
B = newG(rrB);

[rrC, ccC] = find((newG < 80) & (newG > 70));
C = newG(rrC);

[rrD, ccD] = find((newG < 70) & (newG > 60));
D = newG(rrD);

[rrF, ccF] = find(newG < 60);
F = newG(rrF);

Answer 2

I am just trying to learn for myself how to make a 20x3 vector

Answer 3

I think I would need to use an if else statement, but how do I solve the problem?

Answer 4

@Loser66

Answer 5

%% 5.23
% The histc function can be used to determine the number of A's, B's,
% C's, D's, and F's in the grade data from problem 5.21, if the
% following grading scheme applies
% A > 90 to 100
% B > 80 to 90
% C > 70 to 80
% D > 60 to 70
% F > 0 to 60
% Use the bar function to create a histogram based on this scheme.
G = [68, 83, 61, 70, 75, 82, 57, 5, 76, 85, 62, 71, 96, 78, 76, 68, 72, 75, 83, 93]; % final test scores

newG = sort(G(:)); % final scores, starts with lowest then end with highest

A = find((newG < 100) & (newG > 90));
bar(A)

B = find((newG < 90) & (newG > 80));
bar(B)

C = find((newG < 80) & (newG > 70));
bar(C)

D = find((newG < 70) & (newG > 60));
bar(D)

F = find((newG < 60) & (newG > 50));
bar(F)

Answer 6

hahaha.. wrong person!! Sorry friend. I don't know.

Answer 7

@jim_thompson5910

Answer 8

remember the lab assignment you helped me with. I seem to have lost the program =( I have no clue where it is... ugh.

Answer 9

anyhow we can talk about that later. What about the question above?

Answer 10

which lab? and I'm still reading over and thinking of this code above

Answer 11

ahhhh i found it!!!!!

Answer 12

whew, I thought i lost it...

Answer 13

For the code above I was thinking about using an if else statement. However, I have not learned if else statements in my class. I don't think we learn that until chapter 6 or 7. We are currently in chapter 5.

Answer 14

one sec, trying something out

Answer 15

kk

Answer 16

ok I think I found what to do. You're on the right track. You just need to now extract the number of rows of each matrix then use those values to form a new matrix which will help form the bar graph

one sec while I write up that bit of code

Answer 17

wait hold on.

Answer 18

Basically I was trying to extract the indices or at least replace the indices with the correct letter grade. I was using [rrA, A] = find((newG > 90) & (newG < 100)); Why doesn't this spit out the grade score corresponding to the letter grade?

Answer 19

when you wrote

[rrA, ccA] = find((newG < 100) & (newG > 90));

you didn't need to have 2 outputs like that.

newG is a vector so you don't need to worry about the second dimension

----------------

you can write this

rrA = find((newG < 100) & (newG > 90));

and be just fine. The same applies to the other similar lines of code.

Answer 20

okay so then how to you get the letter grade to correspond with the score? Do you have to use another command to do this?

Answer 21

or do you take two commands and integrate them together as one?

Answer 22

`Basically I was trying to extract the indices or at least replace the indices with the correct letter grade. I was using [rrA, A] = find((newG > 90) & (newG < 100)); Why doesn't this spit out the grade score corresponding to the letter grade?`

the `find` function goes through a vector and it looks for values based on the conditions you gave it. In this case, you told matlab to look through newG and pull out values between 90 and 100 (exclude the endpoints). The find function simply returns the indices. Basically the addresses of where the values are located. It does NOT return the actual values. That's why you had to tack on `A = newG(rrA);` to actually pull out those values (in this case, 93 and 96)

Answer 23

yes I agree with the above statement. Go ahead and write the small bit of code up. We can then discuss it.

Answer 24

ok this should work. See the attached m file

Answer 25

oops I have it in reverse order. For some reason I told myself to reverse m1 through m5. My bad. I fixed it

Answer 26

oh nice, all I needed was to use the size() command? I don't need the cc in each set of brackets?


rrA = find((newG < 100) & (newG > 90));
A = newG(rrA)

rrB = find((newG < 90) & (newG > 80));
B = newG(rrB)

rrC = find((newG < 80) & (newG > 70));
C = newG(rrC)

rrD= find((newG < 70) & (newG > 60));
D = newG(rrD)

rrF = find((newG < 60) & (newG > 0));
F = newG(rrF)

Answer 27

`I don't need the cc in each set of brackets?`

no because matrix newG is really a vector. When you use the find function on a vector, it will spit out just 1 output value: the index or indices where the values are located

Answer 28

it spits out M = the value and I = indice? This would be two output technically?

Answer 29

I guess if you really want to be technical, newG is a 20x1 matrix

when you say
[rrA, ccA] = find( ... blah blah stuff dealing with newG .... )
it will return different values for rrA, but ccA will always be equal to 1 because there's only one column here. So ccA isn't needed

Answer 30

^ okay thanks for that detailed statement. It makes sense. Let me finish putting the rest of the code in the script and playing around with it a bit.

Answer 31

http://www.mathworks.com/help/matlab/ref/find.html

if you want the values, then you have to say

`[row,col,v] = find(___) `

that description given on the link says...

`"also returns vector v, which contains the nonzero elements of X."`

though when I test it, it doesn't seem to work. So I'd just do it the way we're already doing it

Answer 32

newG =

5
57
61
62
68
68
70
71
72
75
75
76
76
78
82
83
83
85
93
96

Answer 33

here is newG so when I run [m1, n1] = size(A) it outputs

m1 =

2


n1 =

1

m1 = row2 and n1 = col1 of newG ?

Answer 34

no, matrix is being used here

A =

93
96

matrix A has 2 rows, 1 column

Answer 35

ahhhhh okay I see

Answer 36

matrix A is what you set up: the matrix of scores that correspond to A grades

Answer 37

now I noticed that you don't have any "or equal to" as part of any of your inequality signs. So values like 70 are not part of any bars at all. This score is completely excluded which isn't what you want

Answer 38

I guess the only issue is the score 70 so that's the good news

Answer 39

So the size() command puts the numbers in ascending order.
[m1,m2,m3,m4] = size(X) returns m1 = 2, m2 = 3, m3 = 4, m4 = 1
[m,n] = size(X) returns m = 2, n = 12

Answer 40

I read your statement and thinking.

Answer 41

let me think of over `[m1,m2,m3,m4] = size(X)`

Answer 42

uh oh, now you have overlap

Answer 43

basically what you are saying is if you don't include the equal to or less than sign you will not output 70 because it doesn't include 70 if we only use <

Answer 44

you are right overlap. let me fix this

Answer 45

consider this bit of code


rrA = find((newG <= 100) & (newG >= 90));
A = newG(rrA);

rrB = find((newG <= 90) & (newG >= 80));
B = newG(rrB);


notice how 90 is going to be in matrix AND in matrix B

Answer 46

G = [68, 83, 61, 70, 75, 82, 57, 5, 76, 85, 62, 71, 96, 78, 76, 68, 72, 75, 83, 93]; % final test scores

newG = sort(G(:)); % final scores, starts with lowest then end with highest

rrA = find((newG <= 100) & (newG > 90));
A = newG(rrA);

rrB = find((newG <= 90) & (newG > 80));
B = newG(rrB);

rrC = find((newG <= 80) & (newG > 70));
C = newG(rrC);

rrD= find((newG <= 70) & (newG > 60));
D = newG(rrD);

rrF = find(newG <= 60);
F = newG(rrF);

% get the sizes of each matrix
[m1, n1] = size(A)
[m2, n2] = size(B)
[m3, n3] = size(C)
[m4, n4] = size(D)
[m5, n5] = size(F)

Answer 47

much better, then you'll end up with this histogram

Answer 48

before, it was perfectly symmetrical. Now it has 1 more score added to the D bar (going from height of 4 to 5)

Answer 49

oh btw, but don't forget to put `clc; clear all;` at the top. This should be done at the top of any script so you always start over fresh each time you run the script.

Answer 50

y = [m1, m2, m3, m4, m5]


y =

2 4 7 5 2

Answer 51

looks good, those y values are the heights of the bars

Answer 52

yes, but does the 2 represents the size(A)

Answer 53

Size(A) = 2x1 matrix because we only had two scores in between 90 and 100

Answer 54

it represents the number of rows in A, so yes in a way

Answer 55

btw I realized that you could do

m1 = size(A,1)

to get the number of rows

Answer 56

size(A,1) returns the number of rows in A
size(A,2) returns the number of columns in A

Answer 57

`Size(A) = 2x1 matrix because we only had two scores in between 90 and 100`
yes correct

Answer 58

awesome and my code works. I am just glad I understand what is going on. Thanks for your help. I need to work on my lab. If I have anymore questions I will make a new thread.

Answer 59

actually can you take a look at my last problem? I finished it just want to make sure my logic is correct

Answer 60

%% 5.25
% Use the randn function to create 1000 values in a normal (Gaussian)
% distribution of numbers with a mean of 70 and a standard deviation
% of 3.5. Create a histogram of the data set you calculated.

% given information
n = 1000;
mu = 70;
S = 3.5;

R = normrnd(70, 3.5, 1000, 1); % normal Gaussian
nbins = 15; % number of bins
hist(R, nbins);

Answer 61

why did you set up n,mu,S when you didn't use them?

Answer 62

you used hard coded values instead

Answer 63

because I wasn't thinking...

Answer 64

I also notice it says to use the randn function but you wrote normrnd

is there a difference? hmm let me research

Answer 65

I found a example using normrnd() instead of randn()

Answer 66

https://answers.yahoo.com/question/index?qid=20100608095023AAaafrT
the page says

`"Documentation states that normrnd uses the randn function"`

strange how that works lol

Answer 67

I searched "normal Gaussian" and found on mathworks site normrnd. So it used it because it worked.

Answer 68

ok normrnd it is

Answer 69

My instructor has us putting cell blocks on one script (%%). I have 8 to 10 problems on one script separated in cells. Is there a way to only run one cell at a time? What I have been doing is suppressing each cell before running the code of interest.

Answer 70

I suppress by using '%' on each line of code

Answer 71

you can run the whole file or you can run one particular cell only

Answer 72

you don't have to suppress with comments

Answer 73

I meant suppress the code not the comments

Answer 74

how are you running the script? are you pushing a button or hitting a key on the keyboard?

Answer 75

I click the little green arrow on the editor tab called 'run'

Answer 76

http://www.mathworks.com/help/matlab/matlab_prog/run-sections-of-programs.html

go to the part `Run the code in the current section.`

Answer 77

by 'section' they mean 'cell' (I think anyway)

Answer 78

yep you are right it works. I just have to click in the cell. it turns yellow showing me I am in the cell.

Answer 79

yeah its wherever the blinking text cursor is

Answer 80

I was probably confusing myself because I left old output in the command window which might have made me think it was running more than one mini program.

Answer 81

yeah which is why clc is a good idea. But I'm not sure whether to put it at the top of each cell or put it at the top of the entire script file

Answer 82

it happens to me a lot too

Answer 83

so should I put clc; clear in each cell? Make this a habit? Instructor never advised us to put these commands in each tab

Answer 84

cell* not tab

Answer 85

Again I'm not sure. If one cell depends on the other, then 'clear all' is a bad idea to put at the top of each cell. If you delete data set up in cell A, and cell B needs it, then you'll run into errors or get weird results

but if each cell is independent and separate from one another, then `clc;clear all` at the top of each cell is probably not a bad idea.

Answer 86

let me try it

Answer 87

I guess clc isn't as bad because yes the command window is cleared out, but at least the data variables are kept intact

Answer 88

If cell B depends on cell A, I would save the data needed for cell B in a separate MAT file. That way you can safely clear all the variables out and then load in whatever data you need for cell B

If you do that then you can have `clc; clear all;` at the top of each cell and have no issues if certain cells depend on other cells

Answer 89

clear all clears what again? 'clear' clears the command window.

Answer 90

So running clc; clear all will erase whatever output is located in the command window. It won't clear what is the workspace

Answer 91

http://www.mathworks.com/help/matlab/ref/clc.html

clc clears the command window

`clc clears all input and output from the Command Window display, giving you a "clean screen."`

Answer 92

maybe not use clear all but clear instead?

Answer 93

The `clear` function removes variables from the workspace
`clear all` will get rid of EVERYTHING in the workspace regardless of what type of data type it is

apparently, and I'm just learning this, saying `clear` is the same as `clear all` though I haven't fully tested this

http://www.mathworks.com/help/matlab/ref/clear.html

Answer 94

it won't clear anything in the work space when using clear all

Answer 95

`clear removes all variables from the current workspace, releasing them from system memory.`

Answer 96

clear all should empty the entire workspace