Azzera filtri
Azzera filtri

HELP: Find All Rows of a table which meet a specific condition

641 visualizzazioni (ultimi 30 giorni)
Hello everyone and thanks a lot in advance for your assistance.
Please allow me to introduce you to the problem:
I have an excel file where in the first column there are different students' names. Also, in the second and third column, there are the names of their hometowns and their class level, respectively.
For example:
George London Class B1
George Liverpoool Class B2
Andrew London Class C1
... ... ...
Mairy London Class C2
All these data were imported into Matlab via readable command (because with the xlsread I was getting NAN values).
Now what I need to do is a formula which will find all those students that live in 1 specific town. Let that town be London.
Next, from the previously created table, I need to find those students which belong to Class B1, B2, C1 and C2 respectively. Let B2 to be selected class for this example:
My code is as below:
Alldata = readtable('alldata.xlsx'); Hometown = alldata(:, 14); ==> The hometowns are in column 14.
Class = alldata(:, 13); ==> The classes are in column 13.
a = 'London';
b = 'Class B2';
Result1 = find(alldata.HOMETOWNS == 'a');
Relust 2 = find(alldata.CLASSES == 'b');
FinalTable = table(Relult1, Relust2);
I though that this should worked fine but I get the error: Undefined operator '==' for input arguments of type 'table'.
I also changed parenthesis to braces, the table(s) to cell arrays but nothing worked for me. Unfortunately, I have not too much experience with Matlab sofware, and hence, you are kindly requested to help me with this one cause I really don't have a clue about what is wrong.
I look forward to your reply.
Thanks a lot in advance guys,
Nick A.
  1 Commento
Emil Stobbe
Emil Stobbe il 3 Apr 2020
Hi,
this was already really useful thanks! However I want to do the same thing but my condition is not a string but a subject number like 101. I cant get it to work. I figured that strcmp is not the right function but I could not find any other alternative.
Thanks in advance,
Best Wishes

Accedi per commentare.

Risposta accettata

Guillaume
Guillaume il 8 Mag 2018
find is overused by newcomers to matlab. find is very rarely needed and it is certainly not needed in your case. That's unrelated to your problem though, it makes your code less efficient.
Your problem is that you cannot use == to compare char arrays. The proper way to compare char arrays is to use strcmp
FinalTable = alldata(strcmp(alldata.HOMETOWNS, 'London') & strcmp(alldata.CLASSES', 'Class B2'), :)
Note that if your intent is to split your table into several tables according to the class and hometown, then a) you shouldn't, and b) there are lot more efficient ways to do it than just copy/pasting the above and changing the condition each time.
  3 Commenti
Emil Stobbe
Emil Stobbe il 3 Apr 2020
Hi,
this was already really useful thanks! However I want to do the same thing but my condition is not a string but a subject number like 101. I cant get it to work. I figured that strcmp is not the right function but I could not find any other alternative.
Thanks in advance,
Best Wishes
Guillaume
Guillaume il 3 Apr 2020
If you're trying to match numerical, logical or categorical values, then simply use the standard == comparison
FinalTable = yourtable(yourtable.SubjectNumber == 101, :)

Accedi per commentare.

Più risposte (3)

Peter Perkins
Peter Perkins il 14 Mag 2018
Convert thos variables to categorical, something like
alldata.HOMETOWNS = categorical(alldata.HOMETOWNS);
Then
i = (alldata.HOMETOWNS == 'a') && (alldata.CLASSES == 'b');
alldata(i,:)

ASJ
ASJ il 8 Mag 2018
Hi, try something like this:
Result01=alldata(alldata(:,14) =='London',:)
Result02=alldata(alldata(:,13) =='Class B2',:)

Alon Zaharony
Alon Zaharony il 18 Lug 2022
Hi,
See the following correspondance. It works for me:

Categorie

Scopri di più su Characters and Strings in Help Center e File Exchange

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by