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calculating the number of days within each month for a range of dates

조회 수: 7 (최근 30일)
Nat
Nat 2019년 10월 7일
댓글: Nat 2019년 10월 7일
CB = [datenum([2011,07,13]), datenum([2012,02,19]);
datenum([2012,05,03]), datenum([2013,03,21]);
datenum([2013,06,06]),datenum([2013,09,05]);
datenum([2013,09,05]),datenum([2014,04,28]);
datenum([2014,04,29]),datenum([2014,09,09]);
datenum([2014,09,09]),datenum([2015,05,02]);
datenum([2015,05,01]),datenum([2015,09,06])
datenum([2017,04,30]),datenum([2017,09,12])];
PT = [datenum([2012,09,09]), datenum([2013,06,10]);
datenum([2013,06,11]), datenum([2013,08,20]);
datenum([2013,09,03]),datenum([2014,03,21]);
datenum([2014,04,30]),datenum([2014,09,10])];
QN = [datenum([2013,06,10]), datenum([2013,09,11]);
datenum([2013,09,11]), datenum([2014,04,16]);
datenum([2014,09,10]),datenum([2015,05,02]);
datenum([2015,05,02]),datenum([2015,08,18]);
datenum([2017,04,30]),datenum([2017,09,14])];
AB = [datenum([2017,04,28]), datenum([2017,09,14])];
KS = [datenum([2010,06,03]), datenum([2010,07,20])];
BD = [ datenum([2010,08,27]), datenum([2011,05,26]);
datenum([2011,05,31]),datenum([2012,08,26])];
OCNMS = [ datenum([2007,07,03]),datenum([2008,06,15]);
datenum([2011,01,27]),datenum([2011,10,07]);
datenum([2011,12,07]),datenum([2012,07,11]);
datenum([2012,09,14]),datenum([2013,06,30]);
datenum([2013,07,17]),datenum([2014,05,02])];
DCPP01C = [datenum([2012,11,07]), datenum([2013,03,19])];
CCE = [datenum([2016,10,10]), datenum([2017,11,08])];
HOKE = [datenum([2008,09,15]), datenum([2009,06,06])];
CORC = [datenum([2014,05,01]),datenum([2015,12,04])];
CBtext = repmat({'CB'},size(CB,1),1);
CB2 = [CBtext num2cell(CB)];
PTtext = repmat({'PT'},size(PT,1),1);
PT2 = [PTtext num2cell(PT)];
QNtext = repmat({'QN'},size(QN,1),1);
QN2 = [QNtext num2cell(QN)];
ABtext = repmat({'AB'},size(AB,1),1);
AB2 = [ABtext num2cell(AB)];
KStext = repmat({'KS'},size(KS,1),1);
KS2 = [KStext num2cell(KS)];
BDtext = repmat({'BD'},size(BD,1),1);
BD2 = [BDtext num2cell(BD)];
OCNMStext = repmat({'OCNMS'},size(OCNMS,1),1);
OCNMS2 = [OCNMStext num2cell(OCNMS)];
DCPP01Ctext = repmat({'DCPP01C'},size(DCPP01C,1),1);
DCPP01C2 = [DCPP01Ctext num2cell(DCPP01C)];
CCEtext = repmat({'CCE'},size(CCE,1),1);
CCE2 = [CCEtext num2cell(CCE)];
HOKEtext = repmat({'HOKE'},size(HOKE,1),1);
HOKE2 = [HOKEtext num2cell(HOKE)];
CORCtext = repmat({'CORC'},size(CORC,1),1);
CORC2 = [CORCtext num2cell(CORC)];
D = [CB2;PT2;QN2;AB2;KS2;BD2;OCNMS2;DCPP01C2;CCE2;HOKE2;CORC2];
Dtab = cell2table(D);
datetime.setDefaultFormats('default','MM-dd-yyy');
Dtab.D2 = datetime(Dtab.D2,'ConvertFrom','datenum');
Dtab.D3 = datetime(Dtab.D3,'ConvertFrom','datenum');
Dtab.Total_Days = days(Dtab.D3-Dtab.D2);
I have a table with a range of dates (Dtab) and I'm trying to find out how many days fall within each month. For example, the first row has 221 days between 7/13/11 and 2/19/2012. I want to know how many days in January, February, etc. Eventually, I want to sort it by season (Winter = Dec - Feb) to find out how much effort I had in each season. Thanks in advance for any help!
  댓글 수: 2
the cyclist
the cyclist 2019년 10월 7일
Can you upload the data, instead of an image of the data? That saves us the work of recreating something to test an algorithm on.
Nat
Nat 2019년 10월 7일
Sorry about that, I uploaded the data!

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채택된 답변

Guillaume
Guillaume 2019년 10월 7일
You've got to learn to work with datetime arrays. There's no need to use datenum to construct datetimes, and adjusting the default formatting is not really the proper way of setting the format of individual arrays.
Anyway, assuming R2018a for splitvars:
Dtab = [Dtab, ...
splitvars(rowfun(@(s, t) accumarray(month(s:t)', 1, [12 1])', Dtab, 'InputVariables', {'D2', 'D3'}), ...
1, 'NewVariableNames', datetime.MonthsOfYear.Short)]
I'm also using the undocumented hidden constant property MonthsOfYear of datetime to get the months name. If you don't like that you can hardcode them or get the names another way.
  댓글 수: 1
Nat
Nat 2019년 10월 7일
This worked, thanks! I'm going to work on making the code more streamlined and will take your notes into consideration.

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