python - 在 Python 中浏览字典

Question

我有一本名为“位置”的字典。我为字典编写的函数以(d，描述，当前)格式设置，其中“d”引用字典，“描述”引用描述我们正在查找的位置的字符串，“当前”引用发现我们当前在字典中是一对坐标(x，y)。

基本上，每个位置在字典中都有多个位置，每个位置都有自己的坐标对，我的目标是找到距离我们当前在字典中的位置(当前)最近的位置。策略是使用距离公式来计算。

例如，如果我们正在寻找最近的加油站并且当前位于 (2,2)，则如果两个加油站分别位于 (3,1) 和 (2,2)，则该函数应返回 (3,1) 作为最近的加油站(1,4) 因为 (3,1) 更接近 (2,2) 任何有关我当前代码的建议将不胜感激。

代码:

def closest(d, description, current):
    current_location = (x, y)
    d = {(3,1):'gas', (1,4):'gas', (2,1):'food', (5,5):'food'}
    distancesFromCurrent = [distanceFormula(z, current_location) for z in places]
    for z in d:
    if z < minimum float:
        return z

我当前的代码没有错误，但肯定无法正常工作。它只是返回 0,0，我不确定如何修复它以返回距离我们当前位置最近的位置的坐标。

Answer 1

考虑了评论后，这是我的解决方案。

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Sun Nov  6 21:42:22 2016

@author: michaelcurrin
"""

import math

def findDistance(A, B):
    """
    In 2D space find the distance between two co-orinates is 
    known as Eucliciean distance.
    Args
        A: tuple or list of x and y co-ordinates 
            e.g. (1,2) e.g. [1,2]
        B: as A.
    Retuns
        distance: float. Decimal value for shortest between A and B
    """
    x = (A[0] - B[0])
    y = (A[1] - B[1])
    distance = math.sqrt(x**2 + y**2) # square root

    # remove comment if you want to see this outputted
    # print distance 

    return distance


def GetClosestPlace(places, loc, feature):
    """find shortest distance between current location and each locations 
    but only ones which have the desired feature"""

    # add distance from current location to each location
    for index in range(len(places)):

        # only continue if feature exists at place
        if feature in places[index]['features']:

            # calculate
            distance = findDistance(loc,
                                    places[index]['location'])
        else:
            # this is to represent n/a for now as every location needs a distance
            # for this version, so that it will not be chosen
            distance = 1000 

            # add calculated distance to existing dictionary for location
        places[index]['distance'] = distance    


    # find shortest distance and return details for that place

    allDistances = [x['distance'] for x in places]
    shortestDistance = min(allDistances)

    for place in places:
        if place['distance'] == shortestDistance:
            return place


placesList = [dict(name='foo',location=(0,3), features=['gas', 'food']),
              dict(name='bar',location=(4,6), features=['food', 'hospital']),
              dict(name='abc',location=(0,9), features=['gas','barber']),
              dict(name='xyz',location=(2,2), features=['food','barber'])
              ]

currentLocation = (5,9)
desiredFeature='food'

closestPlace = GetClosestPlace(placesList, currentLocation, desiredFeature)

print 'Current location: %s' % str(currentLocation)
print 'Desired feature: %s ' % desiredFeature
print
print 'The closest place is...'
print 'Name: %s' % closestPlace['name']
print 'Location %s' % str(closestPlace['location'])
print 'Distance %f' % closestPlace['distance']
# join multiple features in the list with commas
print 'Features: %s' % ', '.join(closestPlace['features'])

"""
OUTPUT

Current location: (5, 9)
Desired feature: food 

The closest place is...
Name: bar
Location (4, 6)
Distance 3.162278
Features: food, hospital
"""