Question

Matlab help

Answer 1

@jim_thompson5910

Answer 2

clc ; clear % clears command and workspace window
load('weather_data'); % loads data
month = 1; % defined variable
m = weatherdata(month,:); % extract months data

m_avg = mean(m) % average rain fall per day in inches
m_max = max(m) % max. rain fall per day in inches
d_max = find(m == m_max) % days in which max. rain fell in inches
d_zero = find(m == 0) % days in which zero inches fell per day
nd_zero = length(d_zero) % number of days in which zero inches fell per day
d_nonzero = find(m>0) % days in which at least some rain fell
nd_nonzero = length(d_nonzero) % number of days in which at least some rain fell
m_avg_nonzero = mean(m(d_nonzero)) % average inches per day of only days in which some rain fell

Answer 3

@jim_thompson5910 same assignment. I am on the last section.

Answer 4

how does the script run so far?

Answer 5

Test all of your values for the month of January manually to make sure they are correct.
Now, improve your script to handle the non-day data points. That is, set the month variable
to 2 (February), which has multiple slots with value -99999, and make your script ignore any such
values. Note, this is not the same as setting -99999 to zero, as was done in the example. This would
invalidate any averages computed for such a month.
Run your improved script for the February data and check the results against manual calcula-
tions for that month.
You may wish to test your script against other months' data. When you are convinced your results
and variable names are correct, go through and use the semicolon to supress all screen output, then
submit your lab02.m le via D2L to the Lab 2 assignment folder.

Answer 6

it runs good

Answer 7

ok great

Answer 8

`Test all of your values for the month of January manually to make sure they are correct` so they want you to compute the mean, median, max, etc without using matlab. So use another calculator or you can do it by hand if you really want to take the long route

Answer 9

the good news is that you'll only focus on row 1

Answer 10

give me a sec. I will use the stat function in my 89

Answer 11

The reason why I'm having you use another calculator is because when they say "test to make sure it's correct", they're implying that matlab may be incorrect. Ideally matlab is error free, but it doesn't hurt to check with another calculator.

Answer 12

so the mean matched.

Answer 13

that's good

Answer 14

m_avg =

27.3548


m_max =

272


d_max =

3


d_zero =

Columns 1 through 16

1 2 4 5 6 7 11 12 13 16 17 20 21 25 26 27

Columns 17 through 18

28 29


nd_zero =

18


d_nonzero =

3 8 9 10 14 15 18 19 22 23 24 30 31


nd_nonzero =

13


m_avg_nonzero =

65.2308

Answer 15

however the max. don't match
calculator is 224 and matlab is 272

Answer 16

matlab has decimal points after the whole number but in my calculator it only shows 27

Answer 17

and I only input the values on first row in weatherdata matrix

Answer 18

hmm let me think

Answer 19

nd_nonzero = length(d_nonzero) % number of days in which at least some rain fell
m_avg_nonzero = mean(m(d_nonzero)) % average inches per day of only days in which some rain fell

Answer 20

these two lines of code are correct?

Answer 21

272 is definitely in row 1, column 3 so that's definitely larger than 224

Answer 22

`d_nonzero = find(m>0)` will look though matrix m and will return the indices where the elements are nonzero

so that's why the first element of d_nonzero is 3 because element 3 is the first nonzero element in the first row

Answer 23

length(d_nonzero) will basically count how many elements are in the matrix d_nonzero

so assigning that to nd_nonzero is valid

Answer 24

`m(d_nonzero)` will spit out the elements in m that are nonzero. What is happening is that you input the nonzero indices (stored in d_nonzero matrix) into matrix m and the output is what you see if you kicked out the nonzero elements in row 1

taking the mean of that will give the avg of the nonzero elements

so that line looks valid too

Answer 25

the error is somewhere in your calculator (probably in the input) because your TI89 said that the max was 224 when 272 is actually the max

Answer 26

kk let me look at it again

Answer 27

Okay i must have missed a few numbers. I get a max of 272 and an average of 27.258

Answer 28

hmm matlab is saying 27.3548

Answer 29

whoa yeah calculator is saying 27.258

Answer 30

excel is saying 27.35483871

Answer 31

btw horizontal scrolling is a pain so I prefer the original 31x12 matrix to avoid horizontal scrolling. Plus it's easier to access all the data

Answer 32

Now, improve your script to handle the non-day data points. That is, set the month variable
to 2 (February), which has multiple slots with value -99999, and make your script ignore any such
values. Note, this is not the same as setting -99999 to zero, as was done in the example. This would
invalidate any averages computed for such a month.

Answer 33

in the example in the textbook it said to change -99999 to 0's which I did. Do I need to go back and chaange these values to -99999

Answer 34

if it's already done, there's no sense in redoing it. It seems like tedious busy work. But your teacher may want you to show how to do it anyway. I'm not sure how s/he operates

Answer 35

so now I need to set month = 2;

Answer 36

I guess you would design the script to handle cases IF there were -9999 there
so I'd just put it in but I wouldn't change the values in the spreadsheet to -9999. That seems like overkill in my opinion

of course, you could easily do a simple search/replace to save time

Answer 37

I'd do it like this

`d_negative = find(m == -9999) `
`m(d_negative) = 0`

that should do two things
1) the first line in the code above will go through vector m and find all the -9999 values. The locations are stored in vector d_negative

2) the next line will assign the elements that are -9999 to 0. I haven't tested it though so this method may not work. If not, then you'll have to use a for loop to go through each index one by one

Answer 38

why is it m() the m on the outside of the parenthesis? Is that a command?

Answer 39

it's used to access values in the matrix/vector

simple uses are....
m(1) returns the first value in vector m
m(2) returns the second value in vector m
and so on...


more complicated uses are...
m([1 3 5]) will return the 1st, 3rd, 5th value in vector m
m([2 25 312]) will return the 2nd, the 25th, and the 312th element in matrix m


so when we say
m(d_negative)
keep in mind that d_negative is a matrix. It's a matrix of indices

Answer 40

month = 2; % defined variable
d_negative = find(m == -99999);
m(d_negative) = 0

m =

Columns 1 through 16

0 0 272 0 0 0 0 9 26 1 0 0 0 50 85 0

Columns 17 through 31

0 90 9 0 0 1 15 224 0 0 0 0 0 33 33

Answer 41

let's say

d_negative = [1 5 8 12]

saying
m(d_negative)
will return the 1st, 5th, 8th and 12th elements in vector m

Answer 42

I want you to try something

try assigning all the 0 values to something else, like 9999 just to see it in action

Answer 43

m_avg =

27.3548


m_max =

272


d_max =

3


d_zero =

Empty matrix: 1-by-0


nd_zero =

0


d_nonzero =

Columns 1 through 16

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Columns 17 through 31

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31


nd_nonzero =

31


m_avg_nonzero =

27.3548


m =

Columns 1 through 16

0 0 272 0 0 0 0 9 26 1 0 0 0 50 85 0

Columns 17 through 31

0 90 9 0 0 1 15 224 0 0 0 0 0 33 33

>> m =

Columns 1 through 16

0 0 272 0 0 0 0 9 26 1 0 0 0 50 85 0

Columns 17 through 31

0 90 9 0 0 1 15 224 0 0 0 0 0 33 33

Answer 44

try this bit of code and show me what the output is

`d_negative = find(m == 0) `
`m(d_negative) = 9999`

you can delete this section later

Answer 45

m =

Columns 1 through 8

-99999 -99999 272 -99999 -99999 -99999 -99999 9

Columns 9 through 16

26 1 -99999 -99999 -99999 50 85 -99999

Columns 17 through 24

-99999 90 9 -99999 -99999 1 15 224

Columns 25 through 31

-99999 -99999 -99999 -99999 -99999 33 33

Answer 46

ok so you used -9999 instead of 9999 ? it looks like that bit of code works. You can delete it now. It was just to test it out

Answer 47

clc ; clear % clears command and workspace window
load('weather_data'); % loads data
month = 1; % defined variable
m = weatherdata(month,:); % extract months data

m_avg = mean(m) % average rain fall per day in inches
m_max = max(m) % max. rain fall per day in inches
d_max = find(m == m_max) % days in which max. rain fell in inches
d_zero = find(m == 0) % days in which zero inches fell per day
nd_zero = length(d_zero) % number of days in which zero inches fell per day
d_nonzero = find(m > 0) % days in which at least some rain fell
nd_nonzero = length(d_nonzero) % number of days in which at least some rain fell
m_avg_nonzero = mean(m(d_nonzero)) % average inches per day of only days in which some rain fell

month = 2; % defined variable
d_negative = find(m == -9999);
m(d_negative) = 0

Answer 48

so I have to suppress all screen outputs. Why though?

Answer 49

clc ; clear % clears command and workspace window
load('weather_data'); % loads data
month = 1; % defined variable
m = weatherdata(month,:); % extract months data

m_avg = mean(m); % average rain fall per day in inches
m_max = max(m); % max. rain fall per day in inches
d_max = find(m == m_max); % days in which max. rain fell in inches
d_zero = find(m == 0); % days in which zero inches fell per day
nd_zero = length(d_zero); % number of days in which zero inches fell per day
d_nonzero = find(m > 0); % days in which at least some rain fell
nd_nonzero = length(d_nonzero); % number of days in which at least some rain fell
m_avg_nonzero = mean(m(d_nonzero)); % average inches per day of only days in which some rain fell

month = 2; % defined variable
d_negative = find(m == -9999);
m(d_negative) = 0;

Answer 50

because the command window gets cluttered up with extra data if you don't suppress output. Usually the person reading the report doesn't care how the data was computed. They usually want the final result(s)

Answer 51

if you're dealing with huge matrices, it's a good idea to suppress output otherwise it gets very cluttered

Answer 52

okay sounds reasonable. I got docked points for not doing that on the last lab assignment

Answer 53

your code looks good. I'm assuming at this point you've fully tested everything a few times. If so, then it looks good enough to be submitted to your teacher

Answer 54

yeah I will probably save it and turn it in tomorrow. I may just look at once more before submitting it. Its not due until Thursday.

Answer 55

I need to get some rest though. Thanks for your help and have a goodnight!

Answer 56

you're welcome

Answer 57

wrong code. I need to make a new thread