Using cellfun to extract specific rows from a cell array?

7 visualizaciones (últimos 30 días)
omidm
omidm el 22 de Dic. de 2021
Comentada: omidm el 1 de Abr. de 2022
Hello,
I have a 1x6 cell array, 'Peaks':
Peaks = {199x4500} {192x4500} {167x4500} {171x1950} {157x1500} {191x1500}
I would like to create a new cell array, 'New_Peaks' using only 16 rows from each cell. The row numbers I want to choose are are in a 16X6 double array, 'Indices'.
Indices =
2 1 1 1 4 1
16 23 7 72 133 24
44 30 19 15 13 63
41 15 28 9 18 58
35 26 31 3 15 4
8 49 39 19 53 32
11 72 40 96 29 64
14 39 55 73 111 84
153 33 57 129 12 71
70 116 63 92 75 60
144 104 95 94 68 83
137 137 97 108 77 74
91 155 107 140 120 97
36 40 119 49 144 56
80 64 127 105 43 59
160 94 145 66 149 49
I tried the following line:
New_peaks = cellfun(@(x) x(Indices), all_peaks, 'UniformOutput', false)
But the output is a 1x6 cell array consisting of 16x6 doubles.
The output I want is:
New_peaks = {16x4500} {16x4500} {16x4500} {16x1950} {16x1500} {16x1500}
Is cellfun able to extract the contents of a cell array in this way?
  3 comentarios
omidm
omidm el 22 de Dic. de 2021
Hi, sure, uploaded the 'peaks' and 'indices' variables.
the cyclist
the cyclist el 23 de Dic. de 2021
Separate piece of advice: Don't name a variable "peaks", which shadows the name of a function. That can cause issues and confusion.

Iniciar sesión para comentar.

Respuesta aceptada

Matt J
Matt J el 22 de Dic. de 2021
Editada: Matt J el 23 de Dic. de 2021
New_peaks =cellfun(@(a,b) a(b,:), Peaks,num2cell(Indices,1) ,'uni',0);
  6 comentarios
Matt J
Matt J el 1 de Abr. de 2022
Editada: Matt J el 1 de Abr. de 2022
In that case, you can just use Indices in its current form.
New_peaks =cellfun(@(a,b) a(b,:), Peaks,Indices ,'uni',0);
I would also encourage you to read the cellfun documentation to understand what this is doing.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Structures en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by