This is probably because behaves more like C macros, so there is no additional function call overhead. And at the end of the documentation, they say that the built-in function is 60% faster than the anonymous function x+y. ", since they also only internalize the for loop. But then follows up with "If the function does accept array arguments it is better to call the function directly. MATLAB will recognize the loop even if every. 2 Answers Sorted by: 104 Looping through all the files in the folder is relatively easy: files dir ('.csv') for file files' csv load (file. I am more interested in the reasons why someone would use arrayfun over a for loop given that arrayfun is 2x slower.Įdit 1: Octave gives the following justification for arrayfun's existence: "This is useful for functions that do not accept array arguments.". It is not necessary to indent the body of a for loop as we have shown above. So let's say that I don't care about the output, it could as well be dumped after computation. For example, S as the output of arrayfun is a cell array in contrast to the n-by-2 matrix S we get from the for loop approach. Under this observation, what is the correct use case for the function arrayfun? If for loops are faster, why would someone use arrayfun? Especially when (arguably) for loops are easier to read? Apparently, this behaviour is expected since arrayfun just “hides” (internalizes) the for loop, and we have additional overhead introduced by the extra function handle. The time difference even more significant if S is initialized ahead of time and transposed (so it aligns with column-major ordering of MATLAB). S(i,:) = excerpt runs two times faster than the previous. I can replace these two lines with the following for loop: for i=1:length(u) Options is a structure to set the parameters of fsolve, u and v are column vectors of medium size (~8k or larger) and fsolve returns a vector (1 row by 2 columns) which necessitates the option (for arrayfun) UniformOutput to be set false. Without going into the details of the PDE problem (since it is irrelevant to my main question), the following lines are executed each time step: S =. Root-finding problems are commonly encountered in scientific computing in various shapes and forms but in this case, the goal is to solve a non-linear time-dependent PDE using operator splitting techniques. This may not be so much of a scientific computing question but more of a MATLAB question, if that is the case, please feel free to close or migrate the question.
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