How to compare two lists

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Umer Khitab 2018-3-27

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评论： Jason 2018-5-8

i have two lists lets say a=[1 2 3 ] and b =[1 2 3 4 5 ] and i want to compare each element of ' a ' with all elements of "b" it means that 1 is compared with whole list "b" then 2 is compared then 3 and if the element of "a" is member of "b" then output "a"else 0.

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Walter Roberson 2018-3-27

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a .* ismember(a, b)

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Walter Roberson 2018-3-29

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Assuming you have a vector c, but since I do not know the range of values or data types involved:

if isa(c, 'uint8') || isa(c, 'int16') || isa(c, 'int32')
 scaled_c = c;
elseif isa(c, 'uint16') & all(c <= intmax('int16'))
 scaled_c = int16(c * 2 - intmax('int16'));
elseif isa(c, 'uint32') & all(c <= intmax('int32'))
 scaled_c = int32(c * 2 - intmax('int32'));
elseif all(c >= 0)
 scaled_c = double(c) ./ double(max(c)) * 2 - 1;
elseif all(c >= -1 & c <= 1)
 scaled_c = double(c);
else
 cmax = double(max(c));
 cmin = double(min(c));
 scaled_c = (double(c) - cmin) ./ (cmax - cmin) * 2 - 1;
end

Now audiowrite scaled_c with appropriate filename and sampling frequency.

Notes about the conversion I do above:

if the datatypes are uint8, int16, or int32, then the code assumes you knew what you were doing with the range of values and leaves them completely unchanged
if the range of values is -1 to +1 then the code assumes you knew what you were doing with the range of values and leaves them unchanged, but alters the datatype to double. The cases where the data type could end up altered are single() that are in the right range, and int64(). The way the code is set up cannot affect the other integer data types
if the values are all non-negative, then the code assumes you knew what you were doing about the lower end of the values, and rescales [0 max(c)] to [-1 +1] . So, for example, your range of values happened to be 30000 to 50000 then this will be mapped first to [30000/50000, 50000/50000] and then the [0 1] range would be mapped to [-1 +1] so the [30000/50000, 50000/50000] would end up mapping to [1/5 1]. It would also have been entirely reasonable to instead map [min(c) max(c)] to [-1 +1]: in the absence of information from you about what should be done, a decision had to be made
otherwise, in the case where there was a mix of positive and negative values but the range was outside [-1 +1], the code scales [min(c) max(c)] to [-1 +1]

For any particular case where you know the range and the data type in advance, then the code can be rewritten to at most three lines.

Walter Roberson 2018-4-5

https://www.mnn.com/health/fitness-well-being/stories/most-people-over-18-cant-hear-these-sounds

With that test (and the speakers I was using) I could only hear a couple of pops in the mid 20's of Hz, and could not clearly hear a sound until around 33 Hz. Without headphones I would probably not be able to hear your 128 point 5 second sound, and possibly not even with headphones either. I would recommend you at least double your sampling frequency. (But 256 Hz should be fine.)

In what respect are the two files different?

Jason 2018-5-8

Greetings, I’ve a similar question, but I'm trying to create a third matrix based on matching pairs of numbers in two other matrices. Current example:

a = [17,9;23,3;24,3;23,3;24,3;9,6;10,6;23,3;24,3;23,3]

b = [17,1;18,1;21,1;23,3;24,3;23,4;24,4;9,6;10,6;14,12;23,12;24,12;23,14]

My goal is to create 'c' where both columns in 'a' and 'b' are the same. So, row 1 of 'a' would be excluded, as would the first 2 rows of 'b', etc. etc.. Any help with accomplishing this task would be greatly appreciated.

Cheers, Jason

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