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Jan 27, 2018 · Max[list] (* 32/39 *) From the documentation for Position: "Position returns a list of positions in a form suitable for use in Extract, ReplacePart, and MapAt. The form is different from the one used in Part." pos = Position[list, Max[list]] (* {{1, 1}} *) Extract[list, pos] (* {32/39} *) If you want the Max of each of the sublists of list

Given a table of values, I know how to find find maximum values, but how do I find the position of the maximum value in each column. For example, given { …

Hi, I want to find the position of the 1st,2nd and 3rd maximum value of a matrix.I know that I can find the position of the max value using find() function ...

findLargest1[list_List] := Module[ {max, poss}, (* Find largest element in list *) {max} = TakeLargest[list, 1]; (* Find all positions of largest element in list *) poss = Flatten[Position[list, max]]; (* Return *) {max, poss} ] absTime = AbsoluteTime[]; {max, poss} = findLargest1[list]; absTime = AbsoluteTime[] - absTime; show[absTime, max, poss, findLargest1, "TakeLargest & Position", list]; findLargest1 (TakeLargest & Position): 0.923 sek.

In Mathematica, Max[] is the most efficient function to get the maximum number in a list of numbers, but how do I find the list with the maximum ...

Position returns a list of positions in a form suitable for use in Extract, ReplacePart, and MapAt. The form is different from the one used in Part. Position looks only for matches to values in Association objects that appear in expr. It returns any part specifications within Association objects in the form Key [ k].

Position[expr, pattern] gives a list of the positions at which objects matching pattern appear in expr. Position[expr, pattern, levelspec] finds only objects that appear on levels specified by levelspec. Position[expr, pattern, levelspec, n] gives the positions of the first n objects found. Position[pattern] represents an operator form of Position that can be applied to an expression.

Sep 14, 2014 · For example, here is a random list of lists of of lists of reals: data = Table [RandomReal [], {i, 1, RandomInteger [ {4, 8}]}, {j, 1, RandomInteger [ {4, 8}]}, {k, 1, RandomInteger [ {4, 8}]} ]; You can just do: m = Max [data] Position [data, m] This will tell you the position of the maximum value.

Ordering [ list, - n] gives the positions of the n largest elements. If there are several smallest elements in list, Ordering [ list, 1] will give only the position of the one that appears first. list [ [ Ordering [ list]]] is the same as Sort [ list]. Ordering [ list, seq] is equivalent to Take [ …

You can try with lists functions Max or Min or Position. ... most direct approach (and usually the one function approach is most optimized in Mathematica).

FindMaximum returns a list of the form { f max, { x -> x max } }, where f max is the maximum value of f found, and x max is the value of x for which it is found. If the starting point for a variable is given as a list, the values of the variable are taken to be lists with the same dimensions.

14.09.2014 · Given some matrix, tensor, or basically any list-of-lists of real numbers, you can simply use the Max function to determine the maximum value and then Position to say where it is. Assuming your data isn't enormous (requiring some conservative/careful approach to save time/memory), this should be fine.

We can instead use FirstPosition to avoid search the entire list. Now the time is 6 seconds. It's still not great. {# ...

Will give positions of minimum and maximum, respectively. Computation times order[n_] := Block[{}, list = RandomReal[1, n]; t1 = (Position[list, Min@list]; // RepeatedTiming // First); t2 = (Ordering[list, 1]; // RepeatedTiming // First); {{n, t1}, {n, t2}}] tab = ParallelTable[order[Floor[1.1^n]], {n, 1, 100, 1}]; ListLogLogPlot[{tab[[All, 1]], tab[[All, 2]]}]

Position (Position[list,Max[list]]) seems the most direct approach (and usually the one function approach is most optimized in Mathematica). Can you explain why Pick is faster? Does this mean that the code for Position needs work? Thank you. POSTED BY: Neil Singer. Answer.

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Position[list,_?(#==Max[list]&)]. or shorter (per your comment...) Position[list,Max[list]]. will do the trick, obviously change Max to Min for minimum.

Roughly speaking, goes through the list element by element, comparing the best answer (i.e. the one with the largest y) found so far with every element in the list in turn. If on comparison, the element being compared to the best answer so far has a …