I need turnover for last month from my table

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Martin
Martin el 21 de Ag. de 2019
Comentada: Star Strider el 22 de Ag. de 2019
Hello, I have a table like the one below:
summary =
6×2 table
Time Monthly_Turnover
___________________ _______________
'start' 0
'01-Jul-2019 17:00:39' 76
'24-Jul-2019 14:00:18' 56
'01-Aug-2019 16:00:35' 76
'15-Aug-2019 15:40:24' 98
'21-Aug-2019 13:50:01' 10
I need to get the sum of the Monthly_Turnover-column for a full month based on the timestamp in the first column (from today*). If the timestamp-column does only represent, say, two weeks, I only need the sum of those two weeks.
*) the timeformat is the same at the following line of code:
datestr(datetime('now'))
ans =
'21-Aug-2019 21:22:58'
In this case the result should be: 56+76+98+10 = 240
Does anyone have an idea how to proceed with such a problem? Thanks in advance

Respuesta aceptada

Star Strider
Star Strider el 21 de Ag. de 2019
Editada: Star Strider el 21 de Ag. de 2019
Try this:
dv = datetime({'01-Jul-2019 17:00:39'; '24-Jul-2019 14:00:18'; '01-Aug-2019 16:00:35'; '15-Aug-2019 15:40:24'; '21-Aug-2019 13:50:01'});
tv = [76; 56; 76; 98; 10];
T = table(dv, tv, 'VariableNames',{'Time','Monthly_Turnover'});
TT = table2timetable(T);
TTR = retime(TT, 'monthly','sum')
producing:
TTR =
2×1 timetable
Time Monthly_Turnover
____________________ ________________
01-Jul-2019 00:00:00 132
01-Aug-2019 00:00:00 184
EDIT —
If you only want the month and year in ‘TTR’:
TT = table2timetable(T);
TT.Time.Format = 'MMM-yyyy';
TTR = retime(TT, 'monthly','sum')
produces:
TTR =
2×1 timetable
Time Monthly_Turnover
________ ________________
Jul-2019 132
Aug-2019 184
  11 comentarios
Martin
Martin el 22 de Ag. de 2019
Okay, I found that if I use your condition
TT.Something_Else(TT.Something_Else < 0) = 0;
and alter it to
TT.Something_Else(TT.Something_Else < 0) = NaN;
it actually fits my needs perfectly. Once again, thank you a lot for taking your time to help me! I appreciate it a lot
Star Strider
Star Strider el 22 de Ag. de 2019
As always, my pleasure.
I initially experimented with setting ‘Something_Else’ to NaN for values <0 , and using the nanmean function, however it did not give what I understood to be the desired result. I may not have completely understood what you wanted.

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