Recreate "any()" function to find closest coordinates

René Martínez :

I have this problem where I´m given to list of numbers each list for coordinate X and Y and I have to find which of the coordinates is closest to the first coordinate of the list, for that I have the following code:

import math

list1 = [1, 2, 1, 1]
list2 = [4, 7, 6, 3]

if len(list2) == len(list1):
    print ("number of elements does match\n")
else:
    print ("number of elements doesnt match\n")

a=1
dist_list = []

for i in range(1, len(list1)):
    x1 = list1[0]
    y1 = list2[0]

    x2 = list1[a]
    y2 = list2[a]

    dist = abs(math.sqrt((x1-(x2)**2) - (y1-(y2)**2)))
    a=a+1
    dist_list.append(dist)

print ("distancias:", dist_list, "\n")

for element in iterable:
        if element:
            return
    return False

print(list1[b],",",list2[b], "es la coordenada mas cercana")
#                            "is the closest coordinate"

In this case, the distances between coordinates show that that coordinate 1,3 is the closest to 1,4.

distancias: [6.48074069840786, 5.656854249492381, 2.23606797749979]

I need to say that here: print(list1[b],",",list2[b], "es la coordenada mas cercana")

I have been searching and found the any() function and according to Python documentation the way I can recreate the formula is:

def any(iterable):
    for element in iterable:
        if element:
            return True
    return False

But I can´t see how to recreate it for me to get the desired result. Also, I must say that for this exercise I can only use for, in, while, sqer, abs and append.

furas :

any() is useless for this. You have to recreate min() (or maybe rather numpy.argmin())

You have to get first distance at start and compare with all other values and when other value is smaller then keep value and its index.

b = 0   # get first index
b_value = dist_list[0]   # get first value

# compare with other distances
for i in range(1, len(dist_list)):
    if dist_list[i] < b_value:  # if new distance is smaller
        b = i  # keep new index
        b_value = dist_list[i]   # keep new value

b = b+1  # because `dist_list[0]` has value for `list1[1]`, not `list1[0]`

(BTW: normally I would use enumerate() instead of range(len()) but I assume you don't know it)

EDIT: as @Tomerikoo pointed out formula for distance is wrong. It should be:

dist = math.sqrt((x1-x2)**2 + (y1-y2)**2)

Full code

import math

list1 = [1, 2, 1, 1]
list2 = [4, 7, 6, 3]

if len(list2) == len(list1):
    print ("number of elements does match\n")
else:
    print ("number of elements doesnt match\n")

dist_list = []

x1 = list1[0]
y1 = list2[0]

for i in range(1, len(list1)):

    x2 = list1[i]
    y2 = list2[i]

    #dist = abs(math.sqrt((x1-(x2)**2) - (y1-(y2)**2)))
    dist = math.sqrt((x1-x2)**2 + (y1-y2)**2)

    dist_list.append(dist)

print ("distancias:", dist_list, "\n")

b = 0
b_value = dist_list[0]

for i in range(1, len(dist_list)):
    if dist_list[i] < b_value:
        b = i
        b_value = dist_list[i]

b = b+1  # because `dist_list[0]` has value for `list1[1]`, not `list1[0]`

print(list1[b],",",list2[b], "es la coordenada mas cercana")
#                            "is the closest coordinate"

Recreate "any()" function to find closest coordinates

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