How to create a structure array?
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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
"NOVANAMODELLERKENFELDAA" 29066
"NOVANAMODELVOLDBAEK" 4225
"NOVANAMODELLYNGBYGAARDSAA" 16316
"NOVANAMODELMODEBROBAEK" 4100
"DK20132255838" 0
"DK20132255838" 489.83
"NovanaModelVAEREBROAA" 17752
"DK20132255838" 0
"NovanaModelBjoerupBaek" 1437.9
"NovanaModelFribroedreAA" 13249
"NovanaModelFribroedreAA" 13249
2 个评论
Matt J
2021-7-27
You have duplicate instances of names on the left with different values on the right. How would you wish to resolve that?
Chiara Scarpellini
2021-7-27
I would like to obtain something like this
采纳的回答
Walter Roberson
2021-7-27
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 个评论
Chiara Scarpellini
2021-7-28
编辑：Chiara Scarpellini
2021-7-28
Walter Roberson
2021-7-28
T=table(string,ID_number);
[G, IDs] = findgroups(T.string);
C = splitapply(@(x){x},T.ID_number,G);
R = struct('name', IDs, 'Code', C);
Chiara Scarpellini
2021-7-28
I obtain the same, for every cello of the name vector I have all the names of the measurements points instead of having just the one relative to the measures
Walter Roberson
2021-7-29
T=table(string,ID_number);
[G, IDs] = findgroups(T.string);
C = splitapply(@(x){x},T.ID_number,G);
R = struct('name', cellstr(IDs), 'Code', C);
Chiara Scarpellini
2021-7-29
Walter Roberson
2021-7-29
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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