Question

matlab

Answer 1

? @raffle_snaffle

Answer 2

@jim_thompson5910

Answer 3

what's your question @raffle_snaffle ?

Answer 4

how do you extract certain years out of a large matrix using the find command?

Answer 5

which page are you on?

Answer 6

page 3?

Answer 7

page 2 of 4 bottom of the page where I need to find a precip_2013 = find()

Answer 8

wait hold on think i figured it out.

Answer 9

ok post what you get

Answer 10

precip_2013 = find((gcdata2 < 2014) & (gcdata2 < 2015) & (gcdata2 < 2016))

Answer 11

okay maybe not

Answer 12

you just want 2013, so why not say

`indices = find(year == 2013)`

recall that you stored all the year values in the `year` vector. The stored indices will house the addresses of the rows where the precipitation data corresponds to 2013

Answer 13

btw
`(gcdata2 < 2014) & (gcdata2 < 2015) & (gcdata2 < 2016)`
is very inefficient because saying

x < 2014 AND x < 2015 AND x < 2016
is the same as
x < 2014

even then, it's better to just say x == 2013 to get that specific year you want

Answer 14

I forgot about the year vector.

Answer 15

you have to access just the year vector or else you may pick up on data that has 2013 in it that isn't a year value. But that seems unlikely. Even then, it's a good idea to pull the data apart like this

Answer 16

% finds data for precipitation
precip_2013 = find(year == 2013);
year(precip_2013);

Answer 17

so that will just spit out a bunch of 2013's

Answer 18

Yes, I would agree. But is there a way to extract just 2013 from the gcdata2 = [year', month', day', gcdata1]

Answer 19

can i see your m file?

Answer 20

i forget how gcdata1 is set up

Answer 21

clc; clear % clears command and workspace window
load('gcdata'); % loads gcdata
format shortg; % formats gcdata

% imports excel file into Matlab
gcdata = xlsread('govt_camp.xlsx');

% new matrix B, eliminates last two columns
Ngcdata = gcdata(:, [1:end - 2]);

% converting first column into a string
[m, n] = size(Ngcdata);
for k = 1:1:m
DateString = sprintf('%d',Ngcdata(k,1)); % converts the date value from a number to string
year(k) = str2double(DateString(1:4)); % reads off the first 4 digits of DateString and stores in year matrix
month(k) = str2double(DateString(5:6)); % reads off next two digits
day(k) = str2double(DateString(7:8)); % reads off next one digit
end

gcdata1 = Ngcdata(:, [2:end]); % removes first colum in Ngcdata
gcdata2 = [year', month', day', gcdata1]; % new matrix gcdata2

% finds data for precipitation
precip_2013 = find(year == 2013);
year(precip_2013);

Answer 22

ok right, first column cut off

Answer 23

which column holds the precip data?

Answer 24

it's column 4

column1 = year
column2 = month
column3 = day
column4 = "Precipitation (inches)"
column5 = "Snow Depth (inches)"
column6 = "Snowfall (inches)"

Answer 25

so you'll look through just `gcdata2(:,4)`

basically every row and column4

Answer 26

here's how I'd do it

% finds indices (aka address) for precip data in 2013
precip_indices_2013 = find(year == 2013);

% store just the precipitation values in a temporary matrix
temp = gcdata2(:,4)

% pull out the precipitation values that correspond to year 2013
% uses the indices found previously
precip_2013 = temp(precip_indices_2013)

that should give you a vector `precip_2013` of precipitation values for the year 2013

Answer 27

I guess you could rename `temp` to `precipitation_values` to be more specific

Answer 28

okay well my precip_2013 vector doesn't match excels

Answer 29

what do you mean?

Answer 30

% finds data for precipitation
precip_indices_2013 = find(year == 2013); % locates the indices for years
year(precip_indices_2013); % extracts 2013 years from year vector
temp_v = gcdata(:,4) % temporar vector for precip measured in inches
precip_2013 = temp_v(precip_indices_2013); % actual precip_2013 values


precip_2013 =

0
0
0
0
0
0
1
0.1
0
6
2
0
0
0
0
0

Answer 31

Those are the first couple values of the output, but look at the excel file.

Answer 32

you're accessing the wrong gcdata matrix

Answer 33

my bad lol

Answer 34

somehow you ended up in the wrong column. You accessed column 6 instead of column 4

Answer 35

Okay it's fixed

Answer 36

okay I should be able to do the next following steps. It wants me to find the precip for 2014, 2015, and 2016 as well.

Answer 37

btw you don't need `year(precip_indices_2013);` in your code

Answer 38

yeah I guess you are right. I was using it to check something though.

Answer 39

well you suppressed the output so it doesn't do anything. I guess beforehand you didn't have the semicolon at the end

Answer 40

I see what you mean though

Answer 41

I will leave it and suppress the output.

Answer 42

we are using a bar graph right? My bar graph looks similar to the image displayed on the lab. Also I am working on getting a title, x label, and y label on the graph too.

Answer 43

in your pdf, it looks like a line graph

Answer 44

nvm I got the subplot to work correctly with the correct title, x label, and y label

Answer 45

% finds 2013 precip data
precip_indices_2013 = find(year == 2013); % locates the indices for years
year(precip_indices_2013); % extracts 2013 years from year vector
temp_13 = gcdata2(:, 4); % temporar vector for precip measured in inches
precip_2013 = temp_13(precip_indices_2013); % actual precip_2013 values from col. 4
bar2013 = bar(precip_2013);
subplot(2, 2, 1);
title('2013 precipitation (in)');
xlabel('years (t)'); ylabel('snowfall (in)');


% find 2014 precip data
precip_indices_2014 = find(year == 2014);
year(precip_indices_2014);
temp_14 = gcdata2(:, 4);
precip_2014 = temp_14(precip_indices_2014);
bar2014 = bar(precip_2014);
subplot(2, 2, 2);
title('2014 precipitation (in)');
xlabel('years (t)'); ylabel('snowfall (in)');

Answer 46

my subplots are reversed and the output is weird looking. Does the code look correct or do I have something backwards? This is for 2013 and 2014.

Answer 47

why are you doing a bar graph?

Answer 48

I see a line graph on page 2

Answer 49

i am reading the lab. I don't see where it says "use a line graph".

Answer 50

well if I am using a plot(x, y) command I know why my subplots won't work correctly.

Answer 51

I'm looking at the examples given

Answer 52

top of page 2
`Matlab Figure 1: You will create a Matlab Figure 1 that will end up looking like this document’s Figure 1.`
then it shows a bunch of line graphs

Answer 53

okay, thanks. I am trying to get the subplots to work.

Answer 54

subplot need to be before or after plot(x,y)?

Answer 55

before plot

Answer 56

subplot sets up the empty slots for the plots to go in

Answer 57

% finds 2013 precip data
precip_indices_2013 = find(year == 2013); % locates the indices for years
year(precip_indices_2013); % extracts 2013 years from year vector
temp_13 = gcdata2(:, 4); % temporar vector for precip measured in inches
precip_2013 = temp_13(precip_indices_2013); % actual precip_2013 values from col. 4
subplot(2, 2, 1);
plot(year, temp_13);
title('2013 precipitation (in)');
xlabel('years (t)'); ylabel('snowfall (in)');


% find 2014 precip data
precip_indices_2014 = find(year == 2014);
year(precip_indices_2014);
temp_14 = gcdata2(:, 4);
precip_2014 = temp_14(precip_indices_2014);
subplot(2, 2, 2);
plot(year, temp_14);
title('2014 precipitation (in)');
xlabel('years (t)'); ylabel('snowfall (in)');

Answer 58

How does this look? It seems to run fine.

Answer 59

you didn't set up the 2013 plot correctly

Answer 60

you created precip_2013 just fine, but you forgot to use it in the plot

Answer 61

same for 2014

Answer 62

I used temp_13 and temp_14 as my y in plot(x,y) because we are plotting snowfall with respect to years I thought.

Answer 63

temp_13 and temp_14 is my data for precip measured in inches.

Answer 64

here's how I'd do the 2013 plot


% finds indices (aka address) for precip data in 2013
precip_indices_2013 = find(year == 2013);

% store just the precipitation values in a temporary matrix
temp = gcdata2(:,4);

% pull out the precipitation values that correspond to year 2013
% uses the indices found previously
precip_2013 = temp(precip_indices_2013);
% pulls out the corresponding day values for the year 2013
days_2013 = day(precip_indices_2013);

% set up plot
subplot(2, 2, 1);
plot(days_2013, precip_2013);
title('2013 Daily Precipitation (in)');
xlabel('day of year'); ylabel('inches');

Answer 65

why can't I use my code? lol

Answer 66

you can. I'm just showing you how I did it. The way you set it up, you put the year along the x axis when it should be the day value.

Also when you wrote `plot(year, temp_13);` you are using ALL the precip values when you want the 2013 precip values only

Answer 67

so you're mixing up variables

Answer 68

I am thinking.

Answer 69

take all the time you need

Answer 70

day isn't a vector? Don't you mean years

Answer 71

I see the lab says days

Answer 72

remember how we broke up the first column into 3 columns?

years, months, days

each column is a vector

Answer 73

hmm it's either `day` or `days`. I forget what we named it

Answer 74

ugh i see i see lol

Answer 75

hmm now that I think about it, the days vector won't work. Because after 31, the days shouldn't wrap back to 1. It should keep counting up

Answer 76

okay good because my graph looks really weird.

Answer 77

hmm let me think

Answer 78

I am just formatting my code

Answer 79

we have to create a vector starting from 1 and going through to 365 to represent the 365 days of the year. So this vector will be a 365x1 matrix

Answer 80

ok

Answer 81

this might do the trick

for k=1:1:365
day_of_year(k) = k;
end

though for 2016 it won't because 2016 isn't over yet. The graph is like 100 days into 2016

Answer 82

I see what you are doing there. Why couldn't I use k = 1:1:365;

Answer 83

you mean for the `for` loop?

Answer 84

day_of_year is just a vector right?

Answer 85

yes spanning from 1 to 365 (incrementing by 1)

Answer 86

why do we need to use a for loop though?

Answer 87

oh you mean

day_of_year = 1:1:365;

??

Answer 88

yeahhh

Answer 89

hmm that might actually work, let me test

Answer 90

yes it works, it makes a 1x365 matrix

Answer 91

so you may have to transpose that

Answer 92

okay, sorry I was thinking.

Answer 93

day_of_year = [1:1:365]'

will transpose it if you need it to be a column vector

Answer 94

what is our y in plot(x,y) ?

Answer 95

the precip data still