i have a set of data for couple of days and the names of the data files like this
name='Newyork20200915'
which is for the 15th of September and i want to export only the date to excel like shown below
So how can i get the date from the name string ?
Thanks in advance
Assuming that the other part of name will not contain any digits besides the date, you can use regexp to get all the digits from the character array:
name = 'Newyork20200915'
date_only = regexp(name, '\d*', 'match')
Next, you can convert this date string to a serial date number using datenum, by providing the format in which the date is currently. And then use datestr to format it to your desired format.
date_formatted = datestr(datenum(date_only, 'yyyymmdd'), 'dd. mmm')
date_formatted =
'15. Sep'
Related
how to separate date and time from datetime column if you have the format as below :
click here to view image
I am trying int(datetime column) for fetching date ; Datetime column - int(datetime column) for fetching time column
Your formula cannot work because your data is a text string (note that it has a letter included) and not a number.
So first convert the string into a "real" time with:
=substitute(a2,"T"," ")
You can then use:
Date: =INT(SUBSTITUTE(A2,"T"," "))
Time: =MOD(SUBSTITUTE(A2,"T"," "),1)
and be sure to format the results as desired:
If your column is formatted true date then use to separate date
=TEXT(A1,"yyyy-mm-dd")
For time
=TEXT(A1,"hh:mm:ss")
If data is in text string or output by TEXT() function then try below functions.
for date =TEXT(FILTERXML("<t><s>"&SUBSTITUTE(A1,"T","</s><s>")&"</s></t>","//s[1]"),"yyyy-mm-dd")
for time =TEXT(FILTERXML("<t><s>"&SUBSTITUTE(A1,"T","</s><s>")&"</s></t>","//s[last()]"),"hh:mm:ss")
For date
=LEFT(A2,FIND("T",A2)-1)
For time
=RIGHT(A2,LEN(A2)-FIND("T",A2))
Trying to find dates in a string, but code doesnt work when I use "/" as a date separator. Also if i enter multiple dates. it returns only one.
I'd like to use all valid date separators viz "/" "-" "." and get all the dates in the string.
Also i'd like to use all the date formats like ddmmyy mmddyy yymmdd yyyymmdd ddmmyyyy mmddyyyy.
str = " here is some text in 31-01-2019 my string 01/02/2019 for fun 02.02.2019"
match = re.search('\d{2}-\d{2}-\d{4}', str)
date = dt.strptime(match.group(), '%d-%m-%Y').date()
print(date)
re.findall(r'\b\d{2}[-\./]\d{2}[-\./]\d{2}\b|\b\d{4}[-\./]\d{2}[-\./]\d{2}\b|\b\d{2}[-\./]\d{2}[-\./]\d{4}\b',lsstr)
output : ['31-01-2019', '01/02/2019', '02.02.2019']
Use this regexp to find all the dates present in the string.
I have a piece of code that converts a unix date to standard U.S. date format. It displays properly in excel, however excel doesn't recognize it as a date.
if str(startdate).isdigit():
startdate = int(startdate)
date = datetime.datetime.fromtimestamp(int(startdate)).strftime('%m/%d/%Y')
ws4.cell(row=cell.row,column=2).value = date
ws4.cell(row=cell.row,column=2).number_format = 'MM DD YYYY'
Any idea how to get excel to see this as a date rather than text?
My mistake was assuming the line below created a date.
date = datetime.datetime.fromtimestamp(int(startdate)).strftime('%m/%d/%Y')
After venturing into the docs (Scary as hell for a noob. Does it get easier?) I realized .strftime('%m/%d/%Y') created a string not a date.
I converted that string to a date using:
date = datetime.datetime.strptime(date, '%m/%d/%Y').date()
Now excel recognizes it as a date.
Hopefully this helps someone in the future.
I have a series of dates and some corresponding values. The format of the data in Excel is "Custom" dd/mm/yyyy hh:mm.
When I try to convert this column into an array in Matlab, in order to use it as the x axis of a plot, I use:
a = datestr(xlsread('filename.xlsx',1,'A:A'), 'dd/mm/yyyy HH:MM');
But I get a Empty string: 0-by-16.
Therefore I am not able to convert it into a date array using the function datenum.
Where do I make a mistake? Edit: passing from hh:mm to HH:MM doesn't work neither. when I try only
a = xlsread('filename.xlsx',1,'A2')
I get: a = []
According to the documentation of datestr the syntax for minutes, months and hours is as follows:
HH -> Hour in two digits
MM -> Minute in two digits
mm -> Month in two digits
Therefore you have to change the syntax in the call for datestr. Because the serial date number format between Excel and Matlab differ, you have to add an offset of 693960 to the retrieved numbers from xlsread.
dateval = xlsread('test.xls',1,'A:A') + 693960;
datestring = datestr(dateval, 'dd/mm/yyyy HH:MM');
This will read the first column (A) of the first sheet (1) in the Excel-file. For better performance you can specify the range explicitly (for example 'A1:A20').
The code converts...
... to:
datestring =
22/06/2015 16:00
Edit: The following code should work for your provided Excel-file:
% read from file
tbl = readtable('data.xls','ReadVariableNames',false);
dateval = tbl.(1);
dateval = dateval + 693960;
datestring = datestr(dateval)
% plot with dateticks as x-axis
plot(dateval,tbl.(2))
datetick('x','mmm/yy')
%datetick('x','dd/mmm/yy') % this is maybe better than only the months
Minutes need to be called with a capital M to distinguish them from months.
Use a=datestr(xlsread('filename.xlsx',1,'A:A'),'dd/mm/yyyy HH:MM')
Edit: Corrected my original answer, where I had mixed up the cases needed.
I tried with this. It works but it is slow and I am not able to plot the dates at the end. Anyway:
table= readtable ('filename.xlsx');
dates = table(:,1);
dates = table2array (dates);
dates = datenum(dates);
dates = datestr (dates);
I'm new to Stata, and I'm wondering how can I change a string variable which contains a date to a date format.
The data in the variable looks like this:
yyyy-mm-dd
Should I first remove the dashes so that Stata can recognize the format in order to later use gen var = date() ?
Thank you for your help.
The Stata date function is smart about removing separator characters. See help datetime_translation under the section "the date function"
If your dates are in v1 and in the form yyyy-mm-dd you can specify the commands:
generate v2 = date(v1, "YMD")
format %td v2
The YMD is called a mask, and it tells Stata the order in which the parts of the date are specified. The second line will assign the variable the Stata daily date format, which means that when you look at that variable in the data, it will be shown in human readable form. The date is stored, however, as the number of days since January 1, 1960.
The best way to experiment with the date function is to use the display command. The first line will display an integer representing the number of days since January 1, 1960. The second line will display the date in a human readable format.
display date("2013-08-14", "YMD")
display %td date("2013-08-14", "YMD")
you can look here to see how to convert to data in Stata or do like this
tostring datedx, replace
generate str4 dxyr1= substr(datedx,1,4)
generate str2 dxmo1 = substr(datedx,6,7)
generate str2 dxda1 = substr(datedx,9,10)
destring dx*, replace
gen datedx1 = mdy(dxmo1, dxda1, dxyr1)