计算到平滑线的距离

Question

I'm trying to find the distance of a point (in 4 dimensions, only 2 are shown here) (any coloured crosses in the figure) to a supposed Pareto frontier (black line). This line represents the best Pareto frontier representation during an optimization process.

Pareto = [[0.3875575798354123, -2.4122340425531914], [0.37707675586149786, -2.398936170212766], [0.38176077842761763, -2.4069148936170213], [0.4080534133844003, -2.4914285714285715], [0.35963459448268725, -2.3631532329495126], [0.34395217638838566, -2.3579931972789114], [0.32203302106516224, -2.344858156028369], [0.36742404637441123, -2.3886054421768708], [0.40461156254852226, -2.4141156462585034], [0.36387868122767975, -2.375], [0.3393199109776927, -2.348404255319149]]

Right now, I calculate the distance from any point to the Pareto frontier like this:

def dominates(row, rowCandidate):
return all(r >= rc for r, rc in zip(row, rowCandidate))

def dist2Pareto(pareto,candidate):
    listDist = []

    dominateN = 0
    dominatePoss = 0
    if len(pareto) >= 2:
        for i in pareto:
            if i != candidate:
                dominatePoss += 1
                dominate = dominates(candidate,i)
                if dominate == True:
                    dominateN += 1
                listDist.append(np.linalg.norm(np.array(i)-np.array(candidate)))

        listDist.sort()

        if dominateN == len(pareto):
            print "beyond"            
            return listDist[0]  
        else:
            return listDist[0]

Where I calculate the distance to each point of the black line, and retrieve the shortest distance (distance to the closest point of the known Frontier).

However, I feel I should calculate the distance to the closest line segment instead. How would I go about achieving this?

Answer 1

The formula for the coordinates of the nearest point on the line is given here . Specifically, you are interested in the one called "line defined by two points". For posterity, the formula is:

Because the frontier is relatively simple, you can loop through each two-point line segment in the frontier, and calculate the closest distance for each, keeping the smallest. You could introduce other constraints / pre-computations to limit the number of calculations required.