如何计算一组 x、y 坐标和位置变量之间的距离？

Posted 2023-03-14

【中文标题】如何计算一组 x、y 坐标和位置变量之间的距离？

【英文标题】：How do I calculate the distance between a set x, y coordinate and location variables?

【发布时间】：2021-07-07 02:44:19

【问题描述】：

对于上下文，我试图在 python 中转换此代码，该代码在地球的我的世界地图上获取一个位置，从地图周围获取多个经线，并向用户发送一个与最近的经线不一致的图像，其中一条线在输入位置和经线位置

import discord
import os
from discord.ext import commands
import math
from PIL import Image, ImageDraw
from io import BytesIO
import time


eco = []


NearestLand = ''
NearestCoords = []
sa = 0
side = ''

spawnPoints = 
    "Oceania Spawn": [16801, 2761],
    "Antartica Spawn": [8178, 8661],
    "Europe Spawn": [-386, -4782],
    "Asia Spawn": [12808, -3192],
    "Africa Spawn": [2420, 3738],
    "North America Spawn": [-10288, -4852],
    "South America Spawn": [-6487, 1360],
    "Soviet Nexus": [16507,-6595],
    "Ryvendor Warp-Pad":[9640,-2390],
    "Gulag Warp-Pad":[11741,-4596],
    "Soviet-Serbia Base Warp-Pad":[-10137,-5374],
    "KGB HQ Warp-Pad":[12958,-5627],


client = commands.Bot(command_prefix = '!')

@client.event
async def on_ready():
    print('Bot Started.')
    await client.change_presence(activity=discord.Activity(type=discord.ActivityType.watching, name="For Intruders"))
    
@client.command()
async def test(ctx):
    await ctx.send('All systems functioninal comrade!')

@client.command()
async def envoy(ctx, x, y):

    await ctx.send("Calculating Comrade!")
    
    time.sleep(2)

    lowest = 1000000000
    eco = [x, y]
    for i in range(0, 2):
        eco[i] = int(eco[i])
    print("Crate's Coordinates at X = x, Y = y".format(x=eco[0], y=eco[1]))
    # print(eco)
    # print(spawnPoints.get('Oceania'))
    for i in spawnPoints:
        z = math.dist(spawnPoints[i], eco)
        # print("From spawn, ".format(spawn = str(i)) + str(z) + " blocks away.")
        if lowest >= z:
            lowest = z
            # print("The lowest distance is at: " + str(lowest) + ", at "+ i)
            NearestLand = i
            NearestCoords = spawnPoints[i]
            print("newNearDist: " + str(math.dist(spawnPoints[i], eco)))

    await ctx.send("The nearest warp point is on nl, and the distance is z.".format
    (nl=NearestLand, z=round(lowest, 2)))
    print(NearestCoords)

    xmap = Image.open("map.jpg")
    draw = ImageDraw.Draw(xmap)
    draw.line(
        ((int(NearestCoords[0]) + 21472) / 10,
        (int(NearestCoords[1]) + 10735) / 10,
        (int(x) + 21472) / 10,
        (int(y) + 10735) / 10),
        fill = (255, 0, 0),
        width = 10)
    draw.ellipse(

    ((((int(x) + 21472) / 10) - 25),

    (((int(y) + 10735) / 10) - 25),

    (((int(x) + 21472) / 10) + 25),

    (((int(y) + 10735) / 10) + 25)),
    fill = (255, 0, 0),
    width = 25)



    #draw.line(((42975 - x)/10), ((21471 - y)/10))
    xmap.save("xmap.jpg")
    print("Saved! Uploading...")
    await ctx.send(file = discord.File("xmap.jpg"))
    print("Uploaded!")

@client.command()
async def envoys(ctx, x, y):

    await ctx.send("Calculating...")
    
    time.sleep(2)

    lowest = 1000000000
    eco = [x, y]
    for i in range(0, 2):
        eco[i] = int(eco[i])
    print("Crate's Coordinates at X = x, Y = y".format(x=eco[0], y=eco[1]))
    # print(eco)
    # print(spawnPoints.get('Oceania'))
    for i in spawnPoints:
        z = math.dist(spawnPoints[i], eco)
        # print("From spawn, ".format(spawn = str(i)) + str(z) + " blocks away.")
        if lowest >= z:
            lowest = z
            # print("The lowest distance is at: " + str(lowest) + ", at "+ i)
            NearestLand = i
            NearestCoords = spawnPoints[i]
            print("newNearDist: " + str(math.dist(spawnPoints[i], eco)))

    await ctx.send("The nearest warp point is on nl, and the distance is z.".format
    (nl=NearestLand, z=round(lowest, 2)))
    print(NearestCoords)

    xmap = Image.open("map.jpg")
    draw = ImageDraw.Draw(xmap)
    draw.line(
        ((int(NearestCoords[0]) + 21472) / 10,
        (int(NearestCoords[1]) + 10735) / 10,
        (int(x) + 21472) / 10,
        (int(y) + 10735) / 10),
        fill = (255, 0, 0),
        width = 10)
    draw.ellipse(

    ((((int(x) + 21472) / 10) - 25),

    (((int(y) + 10735) / 10) - 25),

    (((int(x) + 21472) / 10) + 25),

    (((int(y) + 10735) / 10) + 25)),
    fill = (255, 0, 0),
    width = 25)



    #draw.line(((42975 - x)/10), ((21471 - y)/10))
    xmap.save("xmap.jpg")
    print("Saved! Uploading...")
    await ctx.send(file = discord.File("xmap.jpg"))
    print("Uploaded!")



client.run(os.getenv('TOKEN'))

进入节点，这是我目前所拥有的：

const Discord = require('discord.js');
const fs = require('file-system');
require('dotenv').config();
const client = new Discord.Client();
const  prefix  = require('config.json');
client.commands = new Discord.Collection();

client.on('ready', () => console.log('The Bot is ready!'));

client.user.setActivity("For Intruders",  type: "WATCHING" );

const commandFiles = fs.readdirSync('./commands').filter(file => file.endsWith('.js'));

for (const file of commandFiles) 
  const command = require(`./commands/$file`);
  client.commands.set(command.name, command);




client.login(process.env.TOKEN)

我难过的是如何计算位置与输入位置之间的距离 这是我的 config.json。

  "prefix": "!",
  "spawnPoints": 
    "Oceania Spawn": "16801, 2761",
    "Antartica Spawn": "8178, 8661",
    "Europe Spawn": "-386, -4782",
    "Asia Spawn": "12808, -3192",
    "Africa Spawn": "2420, 3738",
    "North America Spawn": "-10288, -4852",
    "South America Spawn": "-6487, 1360",
    "Soviet Nexus": "16507,-6595",
    "Ryvendor Warp-Pad": "9640,-2390",
    "Gulag Warp-Pad": "11741,-4596",
    "Soviet-Serbia Base Warp-Pad": "-10137,-5374",
    "KGB HQ Warp-Pad": "12958,-5627",
  ,

提前致谢！

【问题讨论】：

我投票结束这个问题，因为“如何计算位置与输入位置之间的距离”是一道数学题，而不是编程题。尝试将distance between points formula 放入搜索引擎，或专门搜索math.stackexchange.com。 （我也不明白你为什么说“x、y、z 坐标”，但显示的数据每点只有两个坐标。）

为清晰起见进行了编辑

【参考方案1】：

使用距离公式，将其变成一个函数。

例如，这是我的解释：

import math

def distanceFormula(x1,y1,x2,y2):
  distance = math.sqrt((x1 - x2)**2 + (y1 - y2)**2)
  return distance

如果是3维空间，加上z，做同样的事情，只是再加一个z1和z2，像这样：

import math

def distanceFormula(x1,y1,x2,y2, z1,z2):
  distance = math.sqrt((x1 - x2)**2 + (y1 - y2)**2 + (z1 - z2)**2)
  return distance

或者对于广义的 N 维空间，您可以传递两个包含 N 个值的列表或元组，并通过以下方式获取它们之间的距离：

import math

def distanceFormula(p1, p2):
  distance = math.sqrt(sum((x - y)**2 for x, y in zip(p1, p2)))
  return distance

如果你想使用模块，那么：

import numpy as np

point_a = np.array((x,y,z))
point_b = np.array((x2,y2,z2))

print(np.linalg.norm(point_a - point_b))

【讨论】：

欧几里得距离可以用scipy.spatial.distance.euclidean(x, y) 完成，但它返回一个numpy.float64 实例而不是float 实例

@BatWannaBe 如果你想使用 numpy，你可以改为使用 linalg.norm，它会返回一个数量。您可以使用任何尺寸的变量分配一个点。

我不确定你的意思，scipy 已经使用了 numpy 数组，并且linalg.norm 也返回了一个numpy.float64 实例。这不是什么大问题，因为numpy.float64 和float 存储相同的数据。但是，用户应该注意，因为类及其方法，例如__repr__ 会有所不同。

哦，我以为你的意思是它实际上返回了语句'numpy.float64'而不是数字。我的错。