suppose i have a matrix x and the second row has values ranging from 0.2 to 102 and weightage has to be assigned to these depending on the mean. the values close to the mean should have highest weightage and progressively decrease as it goes farther.

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madhan ravi
madhan ravi 2018년 11월 12일
give an example and your desired output
johnson saldanha
johnson saldanha 2018년 11월 12일
the example is the same. the values ranging from 0.2 to 102 in randomn order. totally there are more than 1000 values. the highest is 102 and lowest is 102. the mean is 67. the weightage should be given to these values. the weightage should be highest for the values close to the mean. it is okay if the values are put in bins and weightage is given depending the bins. the output should be such that 60-70 should have highest weightage and decrease as it goes farther.
johnson saldanha
johnson saldanha 2018년 11월 12일
after this i need to find the time for which values are constant.
johnson saldanha
johnson saldanha 2018년 11월 12일
as there are many set of values where the values are constant, i need to give different weightage to each set
Stephen23
Stephen23 2018년 11월 12일
Not difficult at all, but what function should the weighting have? Linear down to some minimum value? Logarithmic? Square root? Some other function?
Walter Roberson
Walter Roberson 2018년 11월 12일
What is to be done with the first row of the matrix?
johnson saldanha
johnson saldanha 2018년 11월 12일
the weighing factor could be in terms of 0.1 to 0.9 where later i can multiplty the time for which the values are constant. i also need the time for which it is constant
johnson saldanha
johnson saldanha 2018년 11월 12일
the values start from the first row itself
Walter Roberson
Walter Roberson 2018년 11월 12일
The values start from the first row but you have only defined processing the second row. Should we just discard the first row?
johnson saldanha
johnson saldanha 2018년 11월 12일
my bad. im sorry. theres a correction in the question. it should be second column instead of second row
johnson saldanha
johnson saldanha 2018년 11월 12일
the operation is to be done on the second column only
Walter Roberson
Walter Roberson 2018년 11월 12일
Okay what should be done with the first column?
johnson saldanha
johnson saldanha 2018년 11월 12일
discard it. i mean we have to access only the second column. and the description given is for the second column. thank you
johnson saldanha
johnson saldanha 2018년 11월 12일
any updates?
Walter Roberson
Walter Roberson 2018년 11월 12일
sort() on abs(x-mean(x)) and take the second output of sort to tell you the order.
johnson saldanha
johnson saldanha 2018년 11월 12일
second output of sort as in? its a column matrix ryt
[sorted_distance, sort_order] = sort(abs(x-mean(x))) ;
johnson saldanha
johnson saldanha 2018년 11월 12일
im sorry. i havent understood how its related to my problem statement
Walter Roberson
Walter Roberson 2018년 11월 12일
You said that the values closest to the mean should have the highest weight . sorted_distance tells you exactly how far values are from the mean, arranged in increasing distance . sort_order tells you what the original order was. For example if the first entry of sorted_distance is 0.063 and the first entry of sort_order is 19 then that tells you that the entry that was closest to the mean was aa distance of 0.063 from the mean and that it was the 19th row in the original data that had that entry .
Now it is up to you to decide how a particular distance from the mean shold be transformed into a weight .
johnson saldanha
johnson saldanha 2018년 11월 13일
yeah got it. thank you

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