Calculating discharge over multiple years

Need a function that finds the monthly maximum discharge. It should output two values, the discharge itself and the row in which each value occurs. The code should extract results for all months in all years.
I have a data set that has discharge in first column, day, month then year in 2,3 and 4 columns.

2 comentarios

Rik
Rik el 20 de En. de 2021
Editada: Rik el 20 de En. de 2021
Edit restored from Google cache. This thread is now also added to the Wayback Machine.
@George Sands Please don't edit away parts of your question after receiving an answer. That is very rude. It also looks like you deleted a comment, but unfortunately I can't restore that.
@dpb Could you flag this thread if there is more deleted content I could not restore?
dpb
dpb el 20 de En. de 2021
@Rik I don't think there was any further comment...George never got back with sample data or additional questions so presume must have gotten his results from the hints given earlier.

Iniciar sesión para comentar.

 Respuesta aceptada

dpb
dpb el 16 de En. de 2021
Editada: dpb el 16 de En. de 2021

0 votos

Use a table or timetable and rowfun with grouping variables...a couple lines of code and an anonymous function will do it.
For the second desired output you may find augmenting the table with the row number may be the expeditious solution.

1 comentario

dpb
dpb el 17 de En. de 2021
Look at doc/examples for timetable, rowfun for starters. Give it a shot and show us what you have done and where get stuck. Look at section in importing data, too...
Attach a smallish sample datafile would help folks write specific code; you have a very generic description.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Creating and Concatenating Matrices en Centro de ayuda y File Exchange.

Etiquetas

Preguntada:

el 16 de En. de 2021

Comentada:

dpb
el 20 de En. de 2021

Community Treasure Hunt

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

Start Hunting!

Translated by