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How to create a structure array?

2 ビュー (過去 30 日間)
I would like to associate to every name on the left an array with all the numbers on the right column associated to that name. Idk if this would be the best way of dividing the data according to the names, if you have other ideas I'm also open to new solutions
"NovanaModelTUDEAA" 19907
"NovanaModelB8" 0
"DK2013Hulbyrenden" 2456.9
"NovanaModelB53" 2000
"DKQNPAa62110222" 1349.5
"NovanaModelLINDESAADK1" 97
"DK20134780780" 0
"NovanaModelB18" 1469.5
"NovanaModelB28" 7581
"NovanaModelB18" 979.67
"DK2013Svenstruprenden" 2282
"NovanaModelB18" 1469.5
"NovanaModelB28" 2940
"NovanaModelTRYGGEVAELDEAA" 12481
"NovanaModelSKVL9" 617
"DK20132255838" 0
"DK20132255838" 489.83
"NovanaModelVAEREBROAA" 17752
"DK20132255838" 0
"NovanaModelBjoerupBaek" 1437.9
"NovanaModelFribroedreAA" 13249
"NovanaModelFribroedreAA" 13249
  2 件のコメント
Chiara Scarpellini
Chiara Scarpellini 2021 年 7 月 27 日
I would like to obtain something like this



Walter Roberson
Walter Roberson 2021 年 7 月 27 日
findgroups(), and use the group numbers to splitapply() where the work function is @(x){x} . The results should be a cell array.
Now take the second output from the findgroups, the IDs output, which should be a cell array
struct('name', IDs, 'Qvolume', results_of_splitapply)
In the case where the values are cell array, then struct will create a struct array.
  6 件のコメント
Walter Roberson
Walter Roberson 2021 年 7 月 29 日
C has entries of different lengths; it is not clear what it means to sort it? If you mean you want each one to be sorted within itself then change to @(x){sort(x)}


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