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Raymond MacNeil


Last seen: 2 meses hace |  Con actividad desde 2018

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Sequential, row-wise reshape of an M–by–N matrix to form a 1–by–N vector
Dumb question. Let's say I have the following matrix: A = [2 1 4 3 6 5 8 7 10 9]; I...

más de 4 años hace | 1 respuesta | 0

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Compare Data from two tables and show how close the comparison is as a percentage
How are you loading in the data? If you're currently using readtable(), I strongly encourage you to instead use textscan(). In d...

más de 4 años hace | 1

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Removing zeros from a binary array if length of zeros is lesser than a value
k = [0;0;0;0;0;0;0;1;0;1;1;1;0;0;0;0;0;0;1;1;0;0;0;0]; y = regexp(num2str(k)', '(?<=[0]{5,})0*', 'split'); z = strjoin(cellfun...

casi 5 años hace | 0

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My school needs to upgrade its license. I do not know what toolboxes to request. Who should I ask for advice?
Hi Valerie: I was at one point a lab technichian, and though I did not have the higher-level adminstrative role that you seem ...

casi 5 años hace | 0

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Pregunta


regexp: what am I missing from the documentation?
I have tried to carefully read the regexp documentation, and I am able to sucessfully implement regexp in the simplest cases. F...

casi 5 años hace | 1 respuesta | 0

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Convert String array into Cell array
Yes, this also helped me. Thanks a lot.

alrededor de 5 años hace | 0

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Append a char vector to a subset of cells (whose indices are known) in cell array of character vectors
Alright, this was easily achieved with the following (continuing from the example given abvove): mynewcell = strcat('cal_'), m...

más de 5 años hace | 0

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Append a char vector to a subset of cells (whose indices are known) in cell array of character vectors
Hello MATLAB community: I am trying to determine the most effective way of going about the following task. I have a cell array...

más de 5 años hace | 2 respuestas | 0

2

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Get range of integer values from adjacent column elements (integers) for multiple rows.
Note: (inadvertent early submission with question incomplete) Dear MATLAB Community: There is perhaps an even more elegant so...

alrededor de 6 años hace | 1 respuesta | 0

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