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I fear that the hist or histogram function does not help that much, unless you find an easy way to create a histogram where every data value has its own bin and the midpoints of the intervals are exactly the data values.
Otherwise you would have rely on a small user-written function. I guess the creation of the correct vector of interval endpoints for "histogram" would be the same work as writung a routine which outputs exactly what you demanded.
Heres a quick hack which sure could be done in a more elegant way:
BTW, if you like to draw the histograms any way, I would suggest you use "histogram" instead of "hist" and use the first column of the output on the abscissa instead of the number of the bins.
I fear that the hist or histogram function does not help that much, unless you find an easy way to create a histogram where every data value has its own bin and the midpoints of the intervals are exactly the data values.
Otherwise you would have rely on a small user-written function. I guess the creation of the correct vector of interval endpoints for "histogram" would be the same work as writung a routine which outputs exactly what you demanded.
Heres a quick hack which sure could be done in a more elegant way:
BTW, if you like to draw the histograms any way, I would suggest you use "histogram" instead of "hist" and use the first column of the output on the abscissa instead of the number of the bins.
I am a bit puzzled by the argument of the if statement in the for loop: the column index of the matrix M starts by "O -1" which is negative, so the first element of the vector v is compared with something indefinite...
O-1 is not necessarily negative, but it sure would be an invalid matrix index.
As I wrote its a quick hack and could/should be done better. I am relying on the fact, that Mathcad uses short-circuit evaluation for boolean expressions. That means that the second argument is executed or evaluated only if the first argument does not suffice to determine the value of the expression.
So if i=O is true (looking at the first vector element) the second expression (which looks at the previous vector element) needs not to be evaluated and therefore Mathcad don't even look at it. If it would be evaluated, we would get an error because of the invalid index. Its convenient but admittedly not really good programming style and could easily done without by assigning the first value outside of the loop. But now we have to additionally consider the case that the vector consists of one element only (second statement in program fragment below).
... a variant of your Mode function using my setcount function. Handles scalars, single-element arrays, vectors, matrices and arbitrarily-deeply nested arrays.
There is an ORIGIN-independent version in the attached worksheet.
The function T (used in prefix mode) simply transposes any nested arrays in an array - useful for showing results in a more compact form (well, vertically at least!)
Stuart
