CODEWITHSUNDEEP
pythonpython-3.xfunctionnumpy
querty
querty

Reputation: 205

Programming a distance

First of all: I am by no means a Python expert, so this question is probably rather easy. Secondly, I worked over more than 2 hours on this and now I've decided I needed help. I want to implement a distance in Python. The distance is given as follows:

enter image description here

Where ui and uj are vectors that are given. d is the dimension of that vector. For example: if ui = (1,2,3), then ui_0= 1.

Now, this is what I've come up with so far: (here, xi = ui and xj = uj)

def dist(xi, xj, k):
    distances = np.zeros(len(xi))
    min1 = (0, 0)
    min2 = (0, 0)
    for dim in [0, len(xi)]:
        for s in [-k, k]:
            min1 = abs(xi[dim] - xj[dim + s])
            min1[-k, k].min()
            min2 = min(abs(xj[dim] - xi[dim + s]))
            min2[-k, k].min()
            distances = max(min1,min2)

but it doesn't work. Does anybody know where I've gone wrong?

Traceback:

Traceback (most recent call last):   File "<input>", line 1, in
<module>   File "<input>", line 8, in dist IndexError: invalid index
to scalar variable.

EDIT:

Ok, I tried to look at the case where k is maximal and I've taken the comments of Riley and Wouda into account. I came up with this piece of code:

def dist1(xi, xj):
    for dim in range(len(xi)):
        for s in range(-dim, len(xi) - dim):  
            return max(min(abs(xi[dim] - xj[dim + s])), min(abs(xj[dim] - xi[dim + s])))

and I still get the error:

Traceback (most recent call last):
  File "<input>", line 1, in <module>
  File "<input>", line 4, in dist1
TypeError: 'numpy.int64' object is not iterable

What is going on?

Upvotes: 2

Views: 147

Answers (1)

Nelewout
Nelewout

Reputation: 6564

You appear to forget d is a parameter also (since you compute Dist_d^{ij}, where d is clearly a passed parameter). Together with k, I think the following represents the snippet of math you posted.

def dist(xi, xj, d, k):
    min1 = min(abs(xi[d] - xj[d + s]) for s in range(-k, k + 1))
    min2 = min(abs(xj[d] - xi[d + s]) for s in range(-k, k + 1))

    return max(min1, min2)

Notice that Python will index negative values from the end of the list, which may or may not be what you want (in math, generally, not). So getting the arguments to this function right is a bit tricky, and you should build-in some checks to make sure all cases are handled correctly.

Upvotes: 2

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