I have montly macroeconomic data series and I am planning to use them for a weekly (every Monday) regression analysis. How can I match a data point which release once a month to my date template( 4 times during that month) since the new point release and so on.
for u=2:size(daily,1)
l=find(dailytemplate(u)==monthly)
%# when the monthly date is not equal to my daily template
if isempty(l)
%# I need a clearver code for this part to find the previous release
dailyclose(u)=dailyclose(u-1)
else
dailyclose(u)=monthlyclose(l)
end
end
UPDATE from comment
I have the following monthly macro data. I want to use them to feed the weekly dates. For example, at March 31/03/2012 the M-input was 2.7. So any weekly date after that date should be
W_output=2.7
until the April 30/04/2012. Then the weekly W_output will be 2.3 which is the new monthly point, M_input. The following table provides examples for the weekly W_ouput and monthly M_Input:
08/06/2012 1.7
30/06/2012 1.7
01/06/2012 1.7
31/05/2012 1.7
25/05/2012 2.3
30/04/2012 2.3
18/05/2012 2.3
31/03/2012 2.7
11/05/2012 2.3
29/02/2012 2.9
04/05/2012 2.3
31/01/2012 2.9
27/04/2012 2.7
31/12/2011 3
20/04/2012 2.7
format long g
%Create a vector of dates (what I am assuming your date template looks like, this is march 31 and the next 9 mondays that follow it)
datetemplate = [datenum('2012/03/31')];
for i = 1:10
datetemplate(i + 1) = datetemplate(i) + 7;
end
datetemplate';
%Your macro ecos input and dates
macrochangedate = [datenum('2012/03/31');datenum('2012/04/30')]
macrochangedate = [macrochangedate [2.7; 2.3]]
for i = 1:size(macrochangedate,1)
result(datetemplate >= macrochangedate(i,1)) = macrochangedate(i,2);
end
Results:
result =
2.7
2.7
2.7
2.7
2.7
2.3
2.3
2.3
2.3
2.3
2.3
datestr(datetemplate)
ans =
31-Mar-2012
07-Apr-2012
14-Apr-2012
21-Apr-2012
28-Apr-2012
05-May-2012
12-May-2012
19-May-2012
26-May-2012
02-Jun-2012
09-Jun-2012
Related
I found this tutorial online that helps convert ggplot2's geom_abline() to a Plotly graph: https://plotly.com/ggplot2/geom_abline/
It looks like we can simply make such conversion using ggplotly():
library(ggplot2)
library(plotly)
p <- ggplot(data, aes(x=x_val, y=y_val, colour=color_val)) +
geom_point() +
geom_vline(aes(xintercept=xintercept_val), colour=color_val)
ggplotly(p)
However, I cannot convert my ggplot2 graph into a Plotly graph with the following code:
# notice that both my x_val and xintercept_val are dates.
# here's my ggplot2 code:
gg <- ggplot(data) +
geom_point(aes(
x_val,
y_val,
color=color_val,
shape=shape_val
)) +
geom_vline(aes(
xintercept=xintercept_val,
color=color_val
))
ggplotly(gg)
Here's a screenshot of my ggplot2 graph (I cropped out the legends):
Here's a screenshot of my Plotly graph using ggplotly(gg):
Not sure why the vertical lines aren't showing up in Plotly.
Looks like you stumbled over a bug in ggplotly (perhaps you should raise an issue on github). The issue is that ggplotly internally converts the dates to numerics (same with categorical variables). However, inspecting the JSON representation via plotly_json shows that the xintercepts in geom_vline are not converted. That's why they don't show up. However, as a workaround you can make the conversion manually using as.numeric().
As you provided no data I use a simple example dataset from the plotly website to which I added some dates. Try this:
dat <- read.table(header=TRUE, text='
cond xval yval
control 11.5 10.8
control 9.3 12.9
control 8.0 9.9
control 11.5 10.1
control 8.6 8.3
control 9.9 9.5
control 8.8 8.7
control 11.7 10.1
control 9.7 9.3
control 9.8 12.0
treatment 10.4 10.6
treatment 12.1 8.6
treatment 11.2 11.0
treatment 10.0 8.8
treatment 12.9 9.5
treatment 9.1 10.0
treatment 13.4 9.6
treatment 11.6 9.8
treatment 11.5 9.8
treatment 12.0 10.6
')
dat$xval <- rep(as.Date(paste0("2020-", 1:10, "-01")), 2)
max_date1 <- dat[dat$cond == "control", "xval"][which.max(dat[dat$cond == "control", "yval"])]
max_date2 <- dat[dat$cond == "treatment", "xval"][which.max(dat[dat$cond == "treatment", "yval"])]
# The basic scatterplot
p <- ggplot(dat, aes(x=xval, y=yval, colour=cond)) +
geom_point()
# Add colored lines for the date of the max yval of each group
p <- p +
geom_vline(aes(xintercept=as.numeric(max_date1)), colour="green") +
geom_vline(aes(xintercept=as.numeric(max_date2)), colour="lightblue")
p
fig <- ggplotly(p)
fig
Gives me this plot:
I have some weather data stored in a csv file in the form of: „id, date, temperature, rainfall“, with id being the weather station and, obviously, date being the date of measurement. The file contains the data of 3 different stations over a period of 10 years.
What I'd like to do is analyze the data of each station and each year. For example: I'd like to calculate day-to-day differences in temperature [abs((n+1)-n)] for each station and each year.
I thought while-loops could be a possibility, with the loop calculating something as long as the id value is equal to the one in the next row.
But I’ve no idea how to do it.
Best regards
If you still need assistance, I would consider importing the .csv file data using "readtable". So long as only the first row are text, MATLAB will create a 'table' variable (this shouldn't be an issue for a .csv file). The individual columns can be accessed via "tablename.header" and can be reestablished as double data type (ex variable_1=tablename.header). You can then concatenate your dataset as you like. As for sorting by date and station id, I would advocate using "sortrows". For example, if the station id is the first column, sortrow(data,1) will sort "data" by the station id. sortrow(data, [1 2]) will sort "data" by the first column, then by the second column. From there, you can write an if statement to compare the station id's and perform the required calculations. I hope my brief answer is somewhat helpful.
A basic code structure would be:
path=['copy and paste file path here']; % show matlab where to look
data=readtable([path '\filename.csv'], 'ReadVariableNames',1); % read the file from csv format to table
variable1=data.header1 % general example of making double type variable from table
variable2=data.header2
variable3=data.header3
double_data=[variable1 variable2 variable3]; % concatenates the three columns together
sorted_data=sortrows(double_data, [1 2]); % sorts double_data by column 1 then column 2
It always helps to have actual data to work on and specifics as to what kind of output format is expected. Basically, ins and outs :) With the little info provided, I figured I would generate random data for you in the first section, and then calculate some stats in the second. I include the loop as an example since that's what you asked, but I highly recommend using vectorized calculations whenever available, such as the one done in summary stats.
%% example for weather stations
% generation of random data to correspond to what your csv file looks like
rng(1); % keeps the random seed for testing purposes
nbDates = 1000; % number of days of data
nbStations = 3; % number of weather stations
measureDates = repmat((now()-(nbDates-1):now())',nbStations,1); % nbDates days of data ending today
stationIds = kron((1:nbStations)',ones(nbDates,1)); % assuming 3 weather stations with IDs [1,2,3]
temp = rand(nbStations*nbDates,1)*70+30; % temperatures are in F and vary between 30 and 100 degrees
rain = max(rand(nbStations*nbDates,1)*40-20,0); % rain fall is 0 approximately half the time, and between 0mm and 20mm the rest of the time
csv = table(measureDates, stationIds, temp, rain);
clear measureDates stationIds temps rain;
% augment the original dataset as needed
years = year(csv.measureDates);
data = [csv,array2table(years)];
sorted = sortrows( data, {'stationIds', 'measureDates'}, {'ascend', 'ascend'} );
% example looping through your data
for i = 1 : size( sorted, 1 )
fprintf( 'Id=%d, year=%d, temp=%g, rain=%g', sorted.stationIds( i ), sorted.years( i ), sorted.temp( i ), sorted.rain( i ) );
if( i > 1 && sorted.stationIds( i )==sorted.stationIds( i-1 ) && sorted.years( i )==sorted.years( i-1 ) )
fprintf( ' => absolute difference with day before: %g', abs( sorted.temp( i ) - sorted.temp( i-1 ) ) );
end
fprintf( '\n' ); % new line
end
% depending on the statistics you wish to do, other more efficient ways of
% accessing summary stats might be accessible, for example:
grpstats( data ...
, {'stationIds','years'} ... % group by categories
, {'mean','min','max','meanci'} ... % statistics we want
, 'dataVars', {'temp','rain'} ... % variables on which to calculate stats
) % doesn't require data to be sorted or any looping
This produces one line printed for each row of data (and only calculates difference in temperature when there is no year or station change). It also produces some summary stats at the end, here's what I get:
stationIds years GroupCount mean_temp min_temp max_temp meanci_temp mean_rain min_rain max_rain meanci_rain
__________ _____ __________ _________ ________ ________ ________________ _________ ________ ________ ________________
1_2016 1 2016 82 63.13 30.008 99.22 58.543 67.717 6.1181 0 19.729 4.6284 7.6078
1_2017 1 2017 365 65.914 30.028 99.813 63.783 68.045 5.0075 0 19.933 4.3441 5.6708
1_2018 1 2018 365 65.322 30.218 99.773 63.275 67.369 4.7039 0 19.884 4.0615 5.3462
1_2019 1 2019 188 63.642 31.16 99.654 60.835 66.449 5.9186 0 19.864 4.9834 6.8538
2_2016 2 2016 82 65.821 31.078 98.144 61.179 70.463 4.7633 0 19.688 3.4369 6.0898
2_2017 2 2017 365 66.002 30.054 99.896 63.902 68.102 4.5902 0 19.902 3.9267 5.2537
2_2018 2 2018 365 66.524 30.072 99.852 64.359 68.69 4.9649 0 19.812 4.2967 5.6331
2_2019 2 2019 188 66.481 30.249 99.889 63.647 69.315 5.2711 0 19.811 4.3234 6.2189
3_2016 3 2016 82 61.996 32.067 98.802 57.831 66.161 4.5445 0 19.898 3.1523 5.9366
3_2017 3 2017 365 63.914 30.176 99.902 61.932 65.896 4.8879 0 19.934 4.246 5.5298
3_2018 3 2018 365 63.653 30.137 99.991 61.595 65.712 5.3728 0 19.909 4.6943 6.0514
3_2019 3 2019 188 64.201 30.078 99.8 61.319 67.082 5.3926 0 19.874 4.4541 6.3312
Here is my code
---
date: "7 December 2018"
output: html_document
---
## 7 December 2018
```{r, echo=FALSE}
library(leaflet)
library(jsonlite)
citibike <- fromJSON("http://citibikenyc.com/stations/json")
stations <- citibike$stationBeanList
m = leaflet(stations) %>% addTiles(urlTemplate = 'http://{s}.tile.openstreetmap.org/{z}/{x}/{y}.png') %>% addCircles(lat = ~latitude, lng = ~longitude, weight = 5, radius = ~availableBikes, popup = paste("Station:", stations$stationName, "<br>", stations$availableBikes, "available bikes", "<br>", stations$availableDocks, "available docks")) %>% addControl("Available Bikes in NYC on 12/07/2018", position = "topright")
m
```
Using Knit, the html document created only shows the date but not the map. The map is created without any problem when using that code in the console of RStudio.
I have downloaded the latest version of leaflet from github. I use Windows 10.
platform x86_64-w64-mingw32 arch x86_64
os mingw32 system x86_64,
mingw32 status major
3 minor 5.1
year 2018 month 07
day 02 svn rev 74947
language R version.string R version
3.5.1 (2018-07-02) nickname Feather Spray
I have a matrix that has 3months of data or so..Its a 952x1 matrix with the elements in the following format(3 hourly )
Aug-05-2015 03:00:00
Aug-05-2015 06:00:00
Aug-05-2015 09:00:00
Aug-05-2015 12:00:00
Aug-05-2015 15:00:00
Aug-05-2015 18:00:00
Aug-05-2015 21:00:00
Aug-06-2015 00:00:00
Aug-06-2015 03:00:00
Aug-06-2015 06:00:00
I would want to choose say only day timings/ only night or say for august month alone. How do i do that.
Further to my question, if I have a group of .wav files and Im trying to pick only month wise or do daily psd averages etc or chose files belonging to a month how to go about? The following are first 10 .wav files in a .txt file that are read into matlab code-
AMAR168.1.20150823T200235Z.wav
AMAR168.1.20150823T201040Z.wav
AMAR168.1.20150823T201845Z.wav
AMAR168.1.20150823T202650Z.wav
AMAR168.1.20150823T203455Z.wav
AMAR168.1.20150823T204300Z.wav
AMAR168.1.20150823T205105Z.wav
AMAR168.1.20150823T205910Z.wav
AMAR168.1.20150823T210715Z.wav
yyyymmddTHHMMSSZ.wav is part of the format to get sense of some parameters.
Thanks.
Are these datetimes? If so, you can use logical indexing here if you make use of some of the datetime functions. To get the times in August:
t = datetime(2015, 8, 1, 3, 0, 0) + hours(3:3:3000)';
t(month(t) == 8) % Times in August
To get the times that are during the day or night:
t(hour(t) < 12) % Day times
t(hour(t) >= 12) % Night times
I have to open some logs that have a date in the filename. So I am trying to open all the files through a certain date.
I'm using DateTime. I do:
do
{
$datechoice = $today->strftime('%G%m%d'); #YearMonthDay
$date_for_graph = $today->strftime('%d/%m/%G');
# unshift #Log_Period_Time, "$date_for_graph";
print $datechoice." - ".$date_for_graph."<br>";
$today->subtract(days => 1);
} while($datechoice > 20141107);
But the output shows the wrong year for dates near the end of a year:
20160109 - 09/01/2016
20160108 - 08/01/2016
20160107 - 07/01/2016
20160106 - 06/01/2016
20160105 - 05/01/2016
20160104 - 04/01/2016
20150103 - 03/01/2015 <-- Should be 2016
20150102 - 02/01/2015
20150101 - 01/01/2015
20151231 - 31/12/2015
20151230 - 30/12/2015
20151229 - 29/12/2015
...
20150103 - 03/01/2015
20150102 - 02/01/2015
20150101 - 01/01/2015
20151231 - 31/12/2015
20151230 - 30/12/2015
20151229 - 29/12/2015
20141228 - 28/12/2014 <-- Should be 2015
20141227 - 27/12/2014
20141226 - 26/12/2014
Why is this happening?
Use %Y, not %G, unless you specifically mean to display the date according to the ISO 8601 week number calendar.
(In the ISO calendar, every year is a whole number of weeks running Monday-Sunday. So whenever January 1st isn't a Monday there will be up to three days on one side of it that fall in the "wrong" year by ISO reckoning. For instance, ISO year 2021 started on Monday, January 4th; January 1st through 3rd fell in the last week of 2020. Going the other way, ISO year 2026 will start on Monday, December 29th, 2025, so the last 3 days of that December are already the first week of the next ISO year.)