I'm trying to create a test for my students. I put the teacher's name, date, start time and duration time on the same line using the minipage environment. I don't know why the date field is not aligned with the other fields. If I write anything other than the date, the text lines up perfectly.
The code I am using in overleaf is:
\documentclass[paper=a4, fontsize=10pt]{scrartcl}
\usepackage[T1]{fontenc}
\usepackage{times}
\usepackage[brazil]{babel}
\usepackage{amsmath,amsfonts,amsthm}
\usepackage[utf8]{inputenc}
\usepackage{ae}
\usepackage{times}
\usepackage[american voltages, american currents]{circuitikz}
\ctikzset{bipoles/length=2.5em}
\usepackage{anysize}
\marginsize{2cm}{2cm}{1cm}{1cm}
\usepackage{caption}
%\usepackage{columns}
\begin{document}
\begin{minipage}{0.3\textwidth}
Teacher: Guilherme
\end{minipage}
\begin{minipage}{0.2\textwidth}
Date: 22/06/2022
\end{minipage}
\begin{minipage}{0.15\textwidth}
Start: 19:00
\end{minipage}
\begin{minipage}{0.2\textwidth}
Duration: 120 min
\end{minipage}
\vspace{1.5cm}
\begin{minipage}{0.5\textwidth}
Student:\underline{\hspace{7cm}}
\end{minipage}
\end{document}
The result is:
How could I solve the problem?
You can bottom-align the minipages:
\begin{minipage}[b]{0.3\textwidth}
Teacher: Guilherme
\end{minipage}
\begin{minipage}[b]{0.2\textwidth}
Date: 22/06/2022
\end{minipage}
\begin{minipage}[b]{0.15\textwidth}
Start: 19:00
\end{minipage}
\begin{minipage}[b]{0.2\textwidth}
Duration: 120 min
\end{minipage}
Result:
The reason for the different alignment of the date are the / signs which go a bit below the baseline. You can make sure that all your fields have the same height by adding \strut to them:
\documentclass[paper=a4, fontsize=10pt]{scrartcl}
\usepackage[T1]{fontenc}
\usepackage{times}
\usepackage[brazil]{babel}
\usepackage{amsmath,amsfonts,amsthm}
\usepackage[utf8]{inputenc}
\usepackage{ae}
\usepackage{times}
\usepackage[american voltages, american currents]{circuitikz}
\ctikzset{bipoles/length=2.5em}
\usepackage{anysize}
\marginsize{2cm}{2cm}{1cm}{1cm}
\usepackage{caption}
%\usepackage{columns}
\begin{document}
\begin{minipage}{0.3\textwidth}
Teacher: Guilherme\strut
\end{minipage}
\begin{minipage}{0.2\textwidth}
Date: 22/06/2022\strut
\end{minipage}
\begin{minipage}{0.15\textwidth}
Start: 19:00\strut
\end{minipage}
\begin{minipage}{0.2\textwidth}
Duration: 120 min\strut
\end{minipage}
\vspace{1.5cm}
\begin{minipage}{0.5\textwidth}
Student:\underline{\hspace{7cm}}
\end{minipage}
\end{document}
Related
Is there any formula that I can use to show up each month according to start & end date in spreadsheet.
Example:
Start Date:2022-07-22
End Date:2022-10-22
I expected formula to extract value something like this
Jul - Aug - Sep - Oct
I've tried formula
=IF(A2="","",IF(TEXT(B2,"MM")-TEXT(A2,"MM")>1,CONCATENATE(TEXT(A2,"MMM")&" - "&text(EDATE(A2,1),"MMM")&" - "&TEXT(B2,"MMM")),IF(TEXT(A2,"MMM")=TEXT(B2,"MMM"),TEXT(A2,"MMM"),CONCATENATE(TEXT(A2,"MMM")&" - "&TEXT(B2,"MMM"))))) but it only give me correct value if there is up to 3 month period between start & end date.
Here's a link to the sample spreadsheet
For single cell can try-
=JOIN("-",UNIQUE(INDEX(TEXT(SEQUENCE(B2-A2+1,1,A2),"mmm"))))
For spill array-
=BYROW(A2:INDEX(B2:B,MATCH(9^9,B2:B)),LAMBDA(x,JOIN("-",UNIQUE(INDEX(TEXT(SEQUENCE(INDEX(x,2)-INDEX(x,1)+1,1,INDEX(x,1)),"mmm"))))))
See your sheet.
Get the difference in dates in months using DATEDIF and get dates in each intervening month using EOMONTH+SEQUENCE and convert the end of month dates to TEXT:
Start Date
End Date
Months
2022-07-01
2022-10-30
Jul - Aug - Sep - Oct
2022-08-02
2022-08-31
Aug
2022-07-03
2022-11-01
Jul - Aug - Sep - Oct - Nov
Drag fill formula:
=ARRAYFORMULA(JOIN(" - ",TEXT(EOMONTH(A2,SEQUENCE(DATEDIF(A2,EOMONTH(B2,),"M")+1)-1),"mmm")))
Or as a self adjusting array formula:
=MAP(A2:INDEX(A:A,COUNTA(A:A)),LAMBDA(a, ARRAYFORMULA(JOIN(" - ",TEXT(EOMONTH(a,SEQUENCE(DATEDIF(a,EOMONTH(OFFSET(a,0,1),),"M")+1)-1),"mmm")))))
This should be faster and efficient than getting all the dates and filtering them out one by one, thereby reducing space and time complexity.
Use sequence(), edate() and join(), like this:
=arrayformula( map(
A2:A, B2:B,
lambda(
start, end,
if(
isdate(start) * isdate(end),
join(
" - ",
text(
edate(
start,
sequence(
12 * (year(end) - year(start)) + month(end) - month(start) + 1,
1, 0
)
),
"MMM"
)
),
iferror(1/0)
)
)
) )
I need to extract the date from this text:
Mellisoni 2014 Malbec (Columbia Valley (WA))
Okapi 2013 Estate Cabernet Sauvignon (Napa Valley)
Podere dal Nespoli 2015 Prugneto Sangiovese (Romagna)
Simonnet-Febvre 2015 Chablis
Lagler 2012 1000 Eimerberg Smaragd Neuburger (Wachau)
I use this code:
vino<-mutate(vino, year1=sub("^.*([0-9]{4}).*", "\\1", vino$title))
It works, but I have the last value extract on 1000 instead of 2012, how can I fix it if have another numbers?
I have some saved dates in JavaScript using new Date() that looks like:
"Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)"
I'm trying to parse these to Elixir DateTime; I didn't find anything in "timex" that can help and I already know that I can use DateTime.from_iso8601 but for dates saved using new Date().toISOString() but what i need is to parse the above string.
Thanks in advance
You can use elixir binary pattern matching to extract the date parts and parse using Timex's RFC1123 format. The RFC1123 is the format e.g Tue, 05 Mar 2013 23:25:19 +0200. Run h Timex.Format.DateTime.Formatters.Default in iex to see other formats.
iex> date_string = "Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)"
iex> <<day_name::binary-3,_,month_name::binary-3,_,day::binary-2,_,year::binary-4,_,time::binary-8,_::binary-4,offset::binary-5,_,rest::binary>> = date_string
iex> Timex.parse("#{day_name}, #{day} #{month_name} #{year} #{time} #{offset}", "{RFC1123}")
iex> {:ok, #DateTime<2019-02-24 14:44:20+02:00 +02 Etc/GMT-2>}
Pattern matching:
The binary-size are in byte sizes. 1 byte == 1 character. For instance to get
3-character day_name the size is 3. Underscores (_) is used to pattern match the spaces in the date format
Updated answer to use binary-size rather than bitstring-size for simplicity
I didn't find anything in "timex" that can help
The Timex Parsing docs say that you can use strftime sequences, e.g %H:%M:%S, for parsing. Here's a list of strftime characters and what they match.
Here's a format string that I think should work on javascript Dates:
def parse_js_date() do
Timex.parse!("Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)",
"%a %b %d %Y %H:%M:%S GMT%z (%Z)",
:strftime)
end
Unfortunately, %Z doesn't want to match the time zone name, which causes Timex.parse!() to spit out an error. It looks like %Z in Elixir only matches one word, e.g. a timezone abbreviation EET. Therefore, my simple, clean solution is spoiled.
What you can do is chop off the time zone name before parsing the date string:
def parse_js_date_string() do
[date_str|_tz_name] = String.split(
"Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)",
" (",
parts: 2
)
Timex.parse!(date_str,
"%a %b %d %Y %H:%M:%S GMT%z",
:strftime)
end
In iex:
~/elixir_programs/my$ iex -S mix
Erlang/OTP 20 [erts-9.3] [source] [64-bit] [smp:4:4] [ds:4:4:10] [async-threads:10] [hipe] [kernel-poll:false]
Compiling 1 file (.ex)
Interactive Elixir (1.6.6) - press Ctrl+C to exit (type h() ENTER for help)
iex(1)> My.parse_js_date_string()
#DateTime<2019-02-24 14:44:20+02:00 +02 Etc/GMT-2>
iex(2)>
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'd like to construct a graph like the following which is done with excel:
I've entered my data into matlab using the following lines:
year = [1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010
]
fix_No = [-9.449167466 -11.19432509 -8.500517848 -5.644813211 -2.608063866 2.614370892 6.461752833 7.035549084 8.542521755 12.11070577 6.476900841 8.029225388 4.315820526 4.165349512 5.34593031 7.510812752 -2.629044124 -5.713139529 -8.626773532 -11.83226415 . -8.821345246 -6.396197293 -5.187823611 -1.79008821 3.34099288 5.545228048 7.013763711 6.580524638 4.256524275
]
fix_No_and_mean = [11.1610424 5.437315474 5.833032482 4.591658232 1.578021362 -1.572756298 -1.03351595 -2.250991302 -3.222969261 -5.734621837 . 12.96685642 10.95095066 10.2207684 5.654017602 1.753259697 -2.596143576 -7.155087995 -9.687001589 -8.700979283 -4.290434459 . 2.299711172 1.640802028 1.714407543 0.8360893 -0.425484303 -1.160053823 -0.858530711 0.123787867 0.782208621
]
In 1985 and 1996 the lines should be interrupted as in the excel graph. But how can I do this using Matlab? I've put a "." (point) where the interruptions should be but matlab doesn't accept points.
The lines for the construction of the graph look like this:
plot(year,fix_No, 'color', 'k', 'LineWidth',2, 'LineSmoothing','on')
line(year,fix_No_and_mean, 'color', 'r', 'LineWidth',2, 'LineSmoothing','on')
xlabel('year')
legend('fixed number', 'fixed number')
You Should put "NaN" instead of '.' ,i.e. in the places on the vectors fix_No_and_mean and fix_No that correspond to 1985 and 1996, put NaN.
This will plot different lines as in the figure you attached from excel.