I need some help related to plots in MATLAB.
I want to plot my data with respect to clock time on x-axis. I have an occupancy information data for every 15 minutes interval. I want to plot it against time. How can i do it? The problem is with the x-axis, how can i handle time and uniform intervals e.g data is of the form
data=[1 0 0 0 0 1 1 1 1 0 0 0 .............]
value of time is from 9 AM to 9 PM and the interval is 15 minutes
How can i set the intervals on the x-axis?
Thank you
The following code solves the problem. you enter Integer values for starting hour and minute, the ending time and the time steps between the measurements. Further enter/change the steps_x value. This shows how many time values are skipped on the x-axis. 7=skip6 values. Below the for-loop is my y-data. I just used random-function.
The resulting x-axis is a cell-array. This could be a problem for some applications. Further I used some 'unneccessary' variables. These i used for verifing my code. I didn't change it, because i thought that it would be easier to understand that way.
The biggest problem of my so
clc; clear all; close all;
%%//Variable declaration
start_hour = 9; %in hours
start_min = 5; %//in minutes
steps_min = 10; %//in minutes
end_hour = 21.0; %//in hours
end_min= 0; %//in minutes
steps_x = 7; %//how many times are displayed on the axis, doesn't change data
%%// code for computing internal values, steps and transforming to am/pm
%//changes the entries above to hour display(9.75=9h45min)
start_time= (start_hour+start_min/60);
%//changes the entries above to hour display(9.75=9h45min)
end_time=(end_hour+end_min/60);
%//changes steps to hour
steps_hour= steps_min/60;
%//array of hours
mytest_timeline = start_time:steps_hour:end_time;
%//array of minutes(copied and modified from #natan)
mytest_minutes = mod((0:steps_min:(steps_min*(length(mytest_timeline)-1)))+start_min,60);
%//cell array for am/pm display
mytest_timeline_ampm = num2cell(mytest_timeline);
%//if hour is smaller 12 write am otherwise pm
for k=1:length(mytest_timeline);
if mytest_timeline(k) < 12
mytest_ampm = {'am'};
%//converting the down rounded hour to str
helper_hour=num2str(mod(floor(mytest_timeline(k)),12));
%//converting the minute to str and giving it 2 digits eg. 05 for 5min
helper_minute = num2str(mytest_minutes(k),'%02d');
%//joining strings
mytest_timeline_ampm(k)= strcat(helper_hour,{'.'}, helper_minute, mytest_ampm);
%//same as for am just for pm
else
mytest_ampm = {'pm'};
helper_hour=num2str(mod(floor(mytest_timeline(k)),12));
helper_minute = num2str(mytest_minutes(k),'%02d');
mytest_timeline_ampm(k)= strcat(helper_hour,{'.'}, helper_minute, mytest_ampm);
end
end
%%// generated y data so that i could test the code
mytest_y = rand(size(mytest_timeline));
%%// changing display
%//x-coordinates(for displaying x,y)
mytest_x= 1:length(mytest_timeline);
%//x-axis label
mytest_x_axis = 1:steps_x:length(mytest_timeline);
%//plots the data mytest_y in uniform distances (mytest_x)
plot(mytest_x, mytest_y, 'b')
%//changes the x-label accordingly to mytest_x_axis to the am/pm timeline
set(gca, 'XTick',mytest_x_axis, 'XTickLabel',mytest_timeline_ampm(mytest_x_axis))
Related
I am trying to analyze timeseries of wheel turns that were sampled at 1 minute intervals for 10 days. t is a 1 x 14000 array that goes from .1666 hours to 240 hours. analysis.timeseries.(grp).(chs) is a 1 x 14000 array for each of my groups of interest and their specific channels that specifize activity at each minute sampled. I'm interested in collecting the maximum power and the frequency it occurs at. My problem is I'm not sure what units f is coming out in. I would like to have it return in cycles per hour and span to a maximum period of 30 hours. I tried to use the Galileo example in the documentation as a guide, but it didn't seem to work.
Below is my code:
groups = {'GFF' 'GMF' 'SFF' 'SMF'};
chgroups = {chnamesGF chnamesGM chnamesSF chnamesSM};
t1 = (t * 3600); %matlab treats this as seconds so convert it to an hour form
onehour = seconds(hours(1));
for i = 1:4
grp = groups{1,i};
chn = chgroups{1,i};
for channel = 1:length(chn)
chs = chn{channel,1};
[pxx,f]= plomb(analysis.timeseries.(grp).(chs),t, 30/onehour,'normalized');
analysis.pxx.(grp).(chs) = pxx;
analysis.f.(grp).(chs) = f;
analysis.lsp.power.(grp).(chs) = max(pxx);
[row,col,v] = find(analysis.pxx.(grp).(chs) == analysis.lsp.power.(grp).(chs));
analysis.lsp.tau.(grp).(chs) = analysis.f.(grp).(chs)(row);
end
end
Not really an answer but it is hard to put a image in a comment.
Judging by this (plomb manual matlab),
I think that pxx is without dimension as for f is is the frequency so 1/(dimension of t) dimension. If your t is in hours I would say h^-1.
So I'd rather say try
[pxx,f]= plomb(analysis.timeseries.(grp).(chs),t*30.0/onehour,'normalized');
I'm trying to convert an array of milliseconds and its respective data. However I want to do so in hours and minutes.
Millis = [60000 120000 180000 240000....]
Power = [ 12 14 12 13 14 ...]
I've set it up so the data records every minute, hence the 60000 millis (= 1 minimte). I am trying to plot time on the x axis and power on the y. I would like to have the x axis displayed in hours and minutes with each respective power data corresponding to its respective time.
I've tried this
for i=2:length(Millis)
Conv2Min(i) = Millis(i) / 60000;
Time(i) = startTime + Conv2Min(i);
if (Time(i-1) > Time(i) + 60)
Time(i) + 100;
end
end
s = num2str(Time);
This in attempt to turn the milliseconds into hours starting at 08:00 and once 60 minutes have past going to 09:00, the problem is plotting this. I get a gap between 08:59 and 09:00. I also cannot maintain the 0=initial 0.
In this scenario it is preferable to work with datenum values and then use datetick to set the format of the tick labels of your plot to 'HH:MM'.
Let's suppose that you started taking measurements at t_1 = [HH_1, MM_1] and stopped taking measurements at t_2 = [HH_2, MM_2].
A cool trick to generate the array of datenum values is to use the following expression:
time_datenums = HH_1/24 + MM_1/1440 : 1/1440 : HH_2/24 + MM_2/1440;
Explanation:
We are creating a regularly-spaced vector time_datenums = A:B:C using the colon (:) operator, where A is the starting datenum value, B is the increment between datenum values and C is the ending datenum value.
Since your measurements have been taken every minute (60000 milliseconds), then the increment between datenum values should be of 1 minute too. As a day has 24 hours, that makes 1440 minutes a day, so use B = 1/1440 as the increment between vector elements, to get 1 minute increments.
For A and C we simply need to divide the hour digits by 24 and the minute digits by 1440 and sum them up like this:
A = HH_1/24 + MM_1/1440
C = HH_2/24 + MM_2/1440
So for example, if t_1 = [08, 00], then A = 08/24 + 00/1440. As simple as that.
Notice that this procedure doesn't use the datenum function at all, and still, it manages to generate a valid array of datenum values only taking into consideration the time of the datenum, without needing to bother about the date of the datenum. You can learn more about this here and here.
Going back to your original problem, let's have a look at the code:
time_millisec = 0:60000:9e6; % Time array in milliseconds.
power = 10*rand(size(time_millisec)); % Random power data.
% Elapsed time in milliseconds.
elapsed_millisec = time_millisec(end) - time_millisec(1);
% Integer part of elapsed hours.
elapsed_hours_int = fix(elapsed_millisec/(1000*60*60));
% Fractional part of elapsed hours.
elapsed_hours_frac = (elapsed_millisec/(1000*60*60)) - elapsed_hours_int;
t_1 = [08, 00]; % Start time 08:00
t_2 = [t_1(1) + elapsed_hours_int, t_1(2) + elapsed_hours_frac*60]; % Compute End time.
HH_1 = t_1(1); % Hour digits of t_1
MM_1 = t_1(2); % Minute digits of t_1
HH_2 = t_2(1); % Hour digits of t_2
MM_2 = t_2(2); % Minute digits of t_2
time_datenums = HH_1/24+MM_1/1440:1/1440:HH_2/24+MM_2/1440; % Array of datenums.
plot(time_datenums, power); % Plot data.
datetick('x', 'HH:MM'); % Set 'HH:MM' datetick format for the x axis.
This is the output:
I would use datenums:
Millis = [60000 120000 180000 240000 360000];
Power = [ 12 14 12 13 14 ];
d = [2017 05 01 08 00 00]; %starting point (y,m,d,h,m,s)
d = repmat(d,[length(Millis),1]);
d(:,6)=Millis/1000; %add them as seconds
D=datenum(d); %convert to datenums
plot(D,Power) %plot
datetick('x','HH:MM') %set the x-axis to datenums with HH:MM as format
an even shorter approach would be: (thanks to codeaviator for the idea)
Millis = [60000 120000 180000 240000 360000];
Power = [ 12 14 12 13 14 ];
D = 8/24+Millis/86400000; %24h / day, 86400000ms / day
plot(D,Power) %plot
datetick('x','HH:MM') %set the x-axis to datenums with HH:MM as format
I guess, there is an easier way using datetick and datenum, but I couldn't figure it out. This should solve your problem for now:
Millis=6e4:6e4:6e6;
power=randi([5 15],1,numel(Millis));
hours=floor(Millis/(6e4*60))+8; minutes=mod(Millis,(6e4*60))/6e4; % Calculate the hours and minutes of your Millisecond vector.
plot(Millis,power)
xlabels=arrayfun(#(x,y) sprintf('%d:%d',x,y),hours,minutes,'UniformOutput',0); % Create time-strings of the format HH:MM for your XTickLabels
tickDist=10; % define how often you want your XTicks (e.g. 1 if you want the ticks every minute)
set(gca,'XTick',Millis(tickDist:tickDist:end),'XTickLabel',xlabels(tickDist:tickDist:end))
I have a dataset of trajectories of users: every current location of the traiectories has these fields:_ [userId year month day hour minute second latitude longitude regionId]. Based on the field day, I want to divide trajectories based on daily-scale in interval of different hours: 3 hours, 4 hours, 2 hours. I have realized this code that run for interval of 4 hours
% decomposedTraj is a struct that contains the trajectories based on daily scale
for i=1:size(decomposedTraj,2)
if ~isempty(decomposedTraj(i).dailyScaled)
% find the intervals
% interval [0-4]hours
Interval(i).interval_1=(decomposedTraj(i).dailyScaled(:,5)>=0&decomposedTraj(i).dailyScaled(:,5)<4);
% interval [4-8]hours
Interval(i).interval_2=(decomposedTraj(i).dailyScaled(:,5)>=4&decomposedTraj(i).dailyScaled(:,5)<8);
% interval [8-12]hours
Interval(i).interval_3=(decomposedTraj(i).dailyScaled(:,5)>=8&decomposedTraj(i).dailyScaled(:,5)<12);
% interval [12-16]hours
Interval(i).interval_4=(decomposedTraj(i).dailyScaled(:,5)>=12&decomposedTraj(i).dailyScaled(:,5)<16);
% interval [16-20]hours
Interval(i).interval_5=(decomposedTraj(i).dailyScaled(:,5)>=16&decomposedTraj(i).dailyScaled(:,5)<20);
% interval [20-0]hours
Interval(i).interval_6=(decomposedTraj(i).dailyScaled(:,5)>=20);
end
end
or more easily to understand the logic of the code:
A=[22;19;15;15;0;20;22;19;15;15;0;20;20;0;22;21;17;23;22]';
A(A>=0&A<4)
A(A>=4&A<8)
A(A>=8&A<12)
A(A>=12&A<16)
A(A>=16&A<20)
A(A>=20)
It runs and gives the right answer but it's not smart: if I want to change the interval, I have to change all the code... can you help me to find a smart solution more dinamical of this? thanks
0 Comments
Interval k is defined as [(k-1)*N k*N] where N=4 in your example. Therefore you can do the same using a for loop:
for k=1:floor(24/N)
Interval(k) = A(A>=(k-1)*N & A<k*N);
end
Note that in this example A(A>=(k-1)*N & A<k*N) is not necessarily the same size for each k so Interval should be a cell array.
Hello everyone I have a new small problem:
The data I am using have a weird trade time that goes from 17.00 of one day to 16.15 of the day after.
That means that, e.g., for the day 09-27-2013 The source I am using registers the transactions occurred as follows:
DATE , TIME , PRICE
09/27/2013,17:19:42,3225.00,1 #%first obs of the vector
09/27/2013,18:37:59,3225.00,1 #%second obs of the vector
09/27/2013,08:31:32,3200.00,1
09/27/2013,08:36:17,3203.00,1
09/27/2013,09:21:34,3210.50,1 #%fifth obs of the vector
Now first and second obs are incorrect for me: they belong to 9/27 trading day but they have been executed on 9/26. Since I am working on some functions in matlab that relies on non-decremental times I need to solve this issue. The date format I am using is actually the datenum Matlab format so I am trying to solve the problem just subtracting one from the incorrect observations:
%#Call time the time vector, I can identify the 'incorrect' observations
idx=find(diff(time)<0);
time(idx)=time(idx)-1;
It is easy to tell that this will only fix the 'last' incorrect observations of a series. In the previous example this would only correct the second element. And I should run the code several times (I thought about a while loop) until idx will be empty. This is not a big issue when working with small series but I have up to 20millions observations and probably hundred of thousands consecutively incorrect ones.
Is there a way to fix this in a vectorized way?
idx=find(diff(time)<0);
while idx
However, given that the computation would not be so complex I thought that a for loop could efficiently solve the issue and my idea was the following:
[N]=size(time,1);
for i=N:-1:1
if diff(time(i,:)<0)
time(i,:)=time(i,:)-1;
end
end
sadly it does not seems to work.
Here is an example of data I am actually using.
735504.591157407
735507.708030093 %# I made this up to give you an example of two consecutively wrong observations
735507.708564815 %# This is an incorrect observation
735507.160138889
735507.185358796
735507.356562500
Thanks everyone in advance
Sensible version -
for count = 1:numel(time)
dtime = diff([0 ;time]);
ind1 = find(dtime<0,1,'last')-1;
time(ind1) = time(ind1)-1;
end
Faster-but-crazier version -
dtime = diff([0 ;time]);
for count = 1:numel(time)
ind1 = find(dtime<0,1,'last')-1;
time(ind1) = time(ind1)-1;
dtime(ind1+1) = 0;
dtime(ind1) = dtime(ind1)-1;
end
More Crazier version -
dtime = diff([0 ;time]);
ind1 = numel(dtime);
for count = 1:numel(time)
ind1 = find(dtime(1:ind1)<0,1,'last')-1;
time(ind1) = time(ind1)-1;
dtime(ind1) = dtime(ind1)-1;
end
Some average computation runtimes for these versions with various datasizes -
Datasize 1: 3432 elements
Version 1 - 0.069 sec
Version 2 - 0.042 sec
Version 3 - 0.034 sec
Datasize 2: 20 Million elements
Version 1 - 37029 sec
Version 2 - 23303 sec
Version 3 - 20040 sec
So apparently I had 3 other different problems in the data source that I think could have stucked the routine Divakar proposed. Anyway I thought it was being too slow so I started thinking to another solution and came up with a super quick vectorized one.
Given that the observations I wanted to modify fall in a determined known interval of time the function just look for every observation falling in that interval and modifies it as I want (-1 in my case).
function [ datetime ] = correct_date( datetime,starttime, endtime)
%#datetime is my vector of dates and times in matlab numerical format
%#starttime is the starting hour of the interval expressed in datestr format. e.g. '17:00:00'
%#endtime is the ending hour of the interval expressed in datestr format. e.g. '23:59:59'
if (nargin < 1) || (nargin > 3),
error('Requires 1 to 3 input arguments.')
end
% default values
if nargin == 1,
starttime='17:00';
endtime='23:59:59';
elseif nargin == 2,
endtime='23:59:59';
end
tvec=[datenum(starttime) datenum(endtime)];
tvec=tvec-floor(tvec); %#As I am working on multiples days I need to isolate only HH:MM:SS for my interval limits
temp=datetime-floor(datetime); %#same motivation as in the previous line
idx=find(temp>=tvec(1)&temp<=tvec(2)); %#logical find the indices
datetime(idx)=datetime(idx)-1; %#modify them as I want
clear tvec temp idx
end
I have this code as part of my program to calculate the solar time. But my calculation using decimal time. And now, I need to display the result in clock time.
%To calculate solar time
stime = time + (((4*(Lloc-Lsm))+ Et)/60);
disp(['The true solar time is ' num2str(stime),'hr']); %To display the solar time
The answer goes like this:
The true solar time is 13.0501hr
How can I convert the time to clock time (12hr or 24 hr format). i.e the minute should multiply by 60 (0.0501*60min = 3.006) and the time should display as 13:03 or 1:03PM.
Appreciate your help. Thank you.
Regards, -researcher-
you can use the following
my_hour = floor(stime);
my_minute = round(mod(stime, 1) * 60);
disp(['The true solar time is ', num2str(my_hour), ':', num2str(my_minute)]);
Use datenum to convert to a serial date number and then datestr to build the string with appropriate format:
h = 13.0501; %// your computed decimal time
string = datestr(datenum([0 0 0 h 0 0]),14); %// or change "14" for other formats
The two formats you specify correspond to formats 13 through 16 in datestr (see link to the documentation above). For example, with format 14 the string is
>> disp(string)
1:03:00 PM