have a new problem. Can someone help me with this?
A part of my datafile is
06-07-2013,13461,0,7464,0,
07-07-2013,14494,0,5584,0,
08-07-2013,149,13394,3412,2471,
09-07-2013,295,14058,3005,3201,
10-07-2013,194,8308,3264,4026,
11-07-2013,14,4597,3162,1945,
12-07-2013,113,4447,3009,2268,
And want to have the name of the
month and the year (in this case july 2013) in the legend or the title of the plot. How can I get first the month(value) out of the datafile and second make a text of it?
Woow, it is a beauty and almost perfect, look to the circle in the plot.
The Year is some late (lol).
http://ccvd.eu/downloads/Jul.png
I got him, the tricker was fmt = '%d-%m-%Y' in place of fmt = '%d-%m-%y'
http://ccvd.eu/downloads/Jul.png
Thank you so much!! I was for more then a day looking for the solution and now . . . YES.
In principle, to extract a certain value from a data file, you can use the stats command like shown in Gnuplot: How to load and display single numeric value from data file (requires at least version 4.6.0).
But stats does not work directly with time data using set xdata time, you need a little trick. In the using statement you must access the first column as string and format it to a time stamp with strptime.
After that you can format this time stamp with strftime and put it in the title or the legend:
stats 'file.txt' using (strptime('%d-%m-%Y', stringcolumn(1))) every ::0::0
set title strftime('%B %Y', STATS_max)
Only after calling stats you can switch to time data with set xdata time.
So a more complete script could look like
set datafile separator ','
fmt = '%d-%m-%Y'
file = 'file.txt'
stats file using (strptime(fmt, stringcolumn(1))) every ::0::0 nooutput
date = strftime('%B %Y', STATS_max)
set title 'Data of '.date
set timefmt fmt
set xdata time
plot file using 1:2 title 'again '.date
Related
* date is in %td format
gen date1 = real(string(mofd(daily(date, "DMY")), "%tmCYN"))
* type mismatch error
tostring date, gen(dt)
gen date1 = real(string(mofd(daily(dt, "DMY")), "%tmCYN"))
* the code runs but generates no results
tostring date, gen(dt)
gen date2=date(dt, "YMD")
* the code runs but generates no results
If a date variable has a display format %td it must be numeric and stored as some kind of integer. The display format is, and is only, an instruction to Stata on how to display such integers. Confusions about conversion often seem to hinge on a misunderstanding about what format means, as format is an overloaded word in computing, referring variously to file format (as in graphics file format, .png or jpg or whatever); data layout (as in wide or long layout, structure or format); variable or storage type; and (here) display format. There could well be yet other meanings.
A date displayed as 30jan2015 is stored as an integer, namely
. display mdy(1, 30, 2015)
20118
and a glance at help data types shows that your variable date could be stored as an int, float, long or double. All would work, although int is least demanding of memory. You would need (e.g.) to run describe date to find out which type is being used in your case, but nothing to come in this answer depends on knowing that type. Note that finding out what Stata is doing and thinking can be illuminated by running display with simple, single examples.
Your question is ambiguous.
Want to change display format? If you wish merely to see your dates in a display format exemplified by 20150130 then consulting help datetime display formats shows that the display format is as tested here with display, which can be abbreviated all the way down to di
. di %tdCCYYNNDD 20118
20150130
so
format date %tdCCYYNNDD
is what you need. That instructs Stata to change the display format, but the numbers stored remain precisely as they were.
Want such dates as variables held as integers? If you want the dates to be held as integers like 20150130 then you could convert it to string using the display format above, and then to a real value. A minimal sandbox dataset shows this:
. clear
. set obs 1
Number of observations (_N) was 0, now 1.
. gen date = 20118
. gen wanted = real(strofreal(date, "%tdCCYYNNDD"))
. format wanted %8.0f
. l
+------------------+
| date wanted |
|------------------|
1. | 20118 20150130 |
+------------------+
A display format such as %8.0f is needed to see such values directly.
Another method is to generate a large integer directly. You need to be explicit about a suitable storage type and (as just mentioned) need to set an appropriate format, but it can be got to work:
. gen long also = 10000 * year(date) + 100 * month(date) + day(date)
. format also %8.0f
Want such dates as variables held as strings? This is the previous solution, but leave off the real(). The default display format will work fine.
. gen WANTED = strofreal(date, "%tdCCYYNNDD")
. l
+-----------------------------+
| date wanted WANTED |
|-----------------------------|
1. | 20118 20150130 20150130 |
+-----------------------------+
I have not used tostring here but as its original author I have no bias against it. The principles needed here are better illustrated using the underlying function strofreal(). The older name string() will still work.
Turning to your code,
tostring date, gen(dt)
will just put integers like 20118 in string form, so "20118", but there is no way that Stata can understand that alone to be a daily date. You could have run tostring with a format argument, which would have been equivalent to the code above. The advantage of tostring would only be if you had several such variables you wished to convert at once, as tostring would loop over such variables for you.
I can't follow why you thought that conversion to a monthly date or use of a monthly date display format was needed or helpful, as at best you'd lose the information on day of the month. Thus at best Stata can only map a monthly date back to the first day of that month, and at worst a monthly date (here 660) could not be understood as anything you want.
. di mofd(20118)
660
. di %td mofd(20118)
22oct1961
. di %td dofm(mofd(20118))
01jan2015
There is no shortcut to understanding how Stata thinks about dates that doesn't involve reading the needed parts of help datetime and help datetime display formats.
Yet more explanation and examples can be found at https://www.stata-journal.com/article.html?article=dm0067
I have a .csv file with the first column containing dates, a snippet of which looks like the following:
date,values
03/11/2020,1
03/12/2020,2
3/14/20,3
3/15/20,4
3/16/20,5
04/01/2020,6
I would like to import this data into Matlab (I think the best way would probably be using the readtable() function, see here). My goal is to bring the dates into Matlab as a datetime array. As you can see above, the problem is that the dates in the original .csv file are not consistently formatted. Some of them are in the format mm/dd/yyyy and some of them are mm/dd/yy.
Simply calling data = readtable('myfile.csv') on the .csv file results in the following, which is not correct:
'03/11/2020' 1
'03/12/2020' 2
'03/14/0020' 3
'03/15/0020' 4
'03/16/0020' 5
'04/01/2020' 6
Does anyone know a way to automatically account for this type of data in the import?
Thank you!
My version: Matlab R2017a
EDIT ---------------------------------------
Following the suggestion of Max, I have tried specifiying some of the input options for the read command using the following:
T = readtable('example.csv',...
'Format','%{dd/MM/yyyy}D %d',...
'Delimiter', ',',...
'HeaderLines', 0,...
'ReadVariableNames', true)
which results in:
date values
__________ ______
03/11/2020 1
03/12/2020 2
NaT 3
NaT 4
NaT 5
04/01/2020 6
and you can see that this is not working either.
If you are sure all the dates involved do not go back more than 100 years, you can easily apply the pivot method which was in use in the last century (before th 2K bug warned the world of the danger of the method).
They used to code dates in 2 digits only, knowing that 87 actually meant 1987. A user (or a computer) would add the missing years automatically.
In your case, you can read the full table, parse the dates, then it is easy to detect which dates are inconsistent. Identify them, correct them, and you are good to go.
With your example:
a = readtable(tfile) ; % read the file
dates = datetime(a.date) ; % extract first column and convert to [datetime]
idx2change = dates.Year < 2000 ; % Find which dates where on short format
dates.Year(idx2change) = dates.Year(idx2change) + 2000 ; % Correct truncated years
a.date = dates % reinject corrected [datetime] array into the table
yields:
a =
date values
___________ ______
11-Mar-2020 1
12-Mar-2020 2
14-Mar-2020 3
15-Mar-2020 4
16-Mar-2020 5
01-Apr-2020 6
Instead of specifying the format explicitly (as I also suggested before), one should use the delimiterImportoptions and in the case of a csv-file, use the delimitedTextImportOptions
opts = delimitedTextImportOptions('NumVariables',2,...% how many variables per row?
'VariableNamesLine',1,... % is there a header? If yes, in which line are the variable names?
'DataLines',2,... % in which line does the actual data starts?
'VariableTypes',{'datetime','double'})% as what data types should the variables be read
readtable('myfile.csv',opts)
because the neat little feature recognizes the format of the datetime automatically, as it knows that it must be a datetime-object =)
I want to assign the current year in a YY format to either a macro or data set variable.
I am able to use the automatic macro variables &sysdate or &sysdate9 to get the current date. However, extracting the year in a YY format is proving to be a nightmare. Below are some examples of what I've been trying.
There exists the YEARw. format. But when I try to use it I get errors or weird results. For instance, running
data _null_;
yy = year(input("&sysdate9.", year2.));
put yy=;
run;
produces the error
ERROR 48-59: The informat YEAR was not found or could not be loaded.
If I try to format the variable in the output, I get 1965 instead of the current year. The following
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
outputs
yy=2016 65
Please help.
This works to get you the 2-digit year number of the current year:
DATA _NULL_;
YEAR = PUT(TODAY(),YEAR2.);
PUT YEAR;
RUN;
/* Returns: 16 */
To breakdown what I am doing here:
I use TODAY() to get the current date as a DATE type. &SASDATE needs to be converted to a DATE, but also it is the date that the SAS session started. TODAY() is the current date.
PUT allows us to pass in a non-character (numeric/date) value, which is why it is used with TODAY() as opposed to INPUT.
I think it is worth exploring the issues here in more detail.
First, Formats are patterns for converting numeric values to a human readable format. That's what you want to do here: convert a date value to a human readable format, in this case to a year.
Informats, on the other hand, convert human readable information to numeric values. That's not what you're doing here; you have a value already.
Second, put matches with Formats, and input matches with informats, exclusively.
Third, you get close in your last try: but you misuse the year format. Formats are basically value mappings, so they map every possible numeric value in their range (sometimes "all values" is the range, sometimes not) to a display value (string). You need to know what kind of value is expected on the input. YEARw. expects a date value as input, not a year value: meaning input is "number of days from 1/1/1960", mapped to "year". So you cannot take a value you've already mapped to a year value and map it again with that method; it will not make any sense.
Let's look at it:
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
yy contains the result of the year function - 2016. Good so far. Now, you need the 2 digit year (16); you can get that through mod function, if you like, or put/substr/input:
data _null_;
yy = input(substr(put(year(input("&sysdate9.", date9.)),4.),3,2),2.);
put yy=;
run;
mod is probably easier though since it's a number. But of course you could've used year:
data _null_;
yy = put(input("&sysdate9.", date9.),year2.);
put yy=;
run;
Now, yy is character, so you could wrap that with input(...,2.) or leave it character depending on your purposes.
Finally - a use note on &sysdate9.. You can easily make this a date without input:
"&sysdate9."d
So:
yy = put("&sysdate9."d,year2.);
That's called a date literal (and "..."dt and "..."t also work for datetime,time). They require things in the standard SAS formats to work properly.
And as pointed out in Nicarus' answer, today() is a bit better than &sysdate9 since it is guaranteed to be today. If you're running this in batch or restart your session daily, this won't matter, but it will if you have a long-running session.
Apply the year function to the date variable
Convert to string
Take last 2 digits
EDIT: change input to PUT
Year = substr(put(year(today()), 4.), 3);
If I have a group of .wav files and Im trying to pick only month wise or do daily/only night psd(power spectral density) averages etc or choose 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.
Many thanks
You need to be more specific.
Do all files always start with "AMAR168.1." for instance?
Anyway, here's a general approach to get you started:
AllFilenames = fileread ('filenames.dat');
FileNames = strsplit (AllFilenames, '\n');
for i = FileNames
if ~isempty (strfind (i{:}, '20150823')); disp(i{:}); end
end
Your filename examples aren't very useful because they all have the same date, but, anyway, you get the point.
Alternatively, if the filenames always have the same format and size, you could do, e.g.:
AllFilenames = fileread ('filenames.dat');
AllFilenames = strvcat (strsplit (AllFilenames, '\n'));
LogicalIndices = categorical (cellstr (AllFilenames(:,15:16))) == '08';
to obtain all rows where the month is '08' for instance. This assumes that the month is always at position 15 to 16 in the string
At the moment I am using xlsread to open a set of data that I have in excel with given timestamps. But when these values are placed in matlab it changes the formatting of the timestamp.
In excel it is:
dd/mm/yyyy HH:MM
but when it puts it into matlab it changes it to
mm/dd/yyyy HH:MM
which ruins my other code. I have tried using formatIn and specifying it, but then it returns an error if no value for midnight is given.
Any help would be appreciated.
You can use datenum and datestr to convert the format to what you want. In the following example I'm assuming your timestamps are contained in a cell array of strings, but it also works if it's a char matrix:
>> timestamps = {'08/25/2014 13:14'; '08/26/2014 14:15'} %// mm/dd/yyyy HH:MM
>> result = datestr(datenum(timestamps, 'mm/dd/yyyy HH:MM'), 'dd/mm/yyyy HH:MM')
result =
25/08/2014 13:14
26/08/2014 14:15
What Luis recommended should help you to get any format that you like. However there is something important to realize here:
Excel does not 'have' the date in your format. It has the date stored as a number like 123546.123 and presents it to you in a certain way.
If you want to get the date in exactly the way that excel presents it, the trick is to avoid importing the relevant column as a date, but just import it as text instead.
How to do this depends on your import method, but it should not be very hard.