Question

我想我自己陷入了俄罗斯娃娃。

我试图制作一个规模较小的口袋妖怪＆＃34;引擎＆＃34;，并决定开展实施动作。一件事是，当我试图调动此举时，我卡住了。代码在这里（它不是很干净，大多数属性甚至还没有使用）

import math
import random
import time

class type (object):
        def __init__(self, name, relat):
                self.name = name
                self.relat = relat

#define types
fire = type("fire", None)
normal = type("normal", None)
water = type("water", None)
grass = type("grass", None)

#add relationships
fire.relat = {fire: 0.5, normal: 1, water: 0.5, grass: 2}
normal.relat = {fire: 1, normal: 1, water: 1, grass: 1}
water.relat = {fire: 2, normal: 1, water: 0.5, grass: 0.5}
grass.relat = {fire: 0.5, normal: 1, water: 2, grass: 0.5}

#call the variable of fire relationship with water
print fire.relat[water]

class move (object):
        def __init__(self, name, damage, type, ps):
                self.name = name
                self.damage = damage
                self.type = type
                self.ps = ps

tackle = move("Tackle", 5, normal, "physical")
searingburst = move("Searing Burst", 5, fire, "special")
pinesmack = move("Pine Smack", 5, grass, "special")
refresher = move("Refresher", 5, water, "special")

class Pokemon(object):
        # Creates a Pokemon!
        def __init__(self, speciesname, nickname, gender, type1, type2, chp, thp, exp, acc, moves):
                self.speciesname = speciesname
                self.nickname = nickname
                self.gender = gender
                self.type1 = type1
                self.type2 = type2
                self.chp = chp
                self.thp = thp
                self.exp = exp
                self.damage = 2
                self.bacc = 0
                self.acc = acc
                self.moves = moves
        def displayhp(self):
                print str(self.chp) + "/" + str(self.thp) + " HP"
        def damageenemy(self, opponent):
                if ((opponent.type1).upper() == 'GHOST'):
                        opponent.chp = opponent.chp
                        print("%s's attack passed through %s!" % (self.nickname, opponent.nickname))
                        opponent.displayhp()
                else:
                        opponent.chp = opponent.chp - self.damage
                        opponent.displayhp()
        #make a Pokemon by defining a variable = Pokemon(speciesname, nickname etc...)

treecko = Pokemon("Treecko", "Treecko", "male", grass, None, 20, 20, 0, 0, [tackle, pinesmack])
krabby = Pokemon("Krabby", "Krabby", "male", water, None, 20, 20, 0, 0, [tackle, refresher])

def choose(o1, o2):
        o1r = input("What will %s do? " % o1.speciesname)
        if o1r in o1.moves:
                print o1r.moves
        else:
                print "no"

choose(treecko, krabby)

当前代码将访问move类;我需要的是移动的名称。因此，我不想检查移动列表中的对象处理，而是检查字符串＆＃34; Tackle＆＃34;在移动类中，在对象内的Moves列表内。

Answer 1

我认为当你想要操作的选择对象时，一些技巧就足够了。

请不要使用type作为类名。它是内置函数的名称。

#!/usr/bin/env python
#-*- coding:utf-8 -*-

import math
import random
import time

class myType (object):
        def __init__(self, name, relat):
                self.name = name
                self.relat = relat

#define myTypes
fire = myType("fire", None)
normal = myType("normal", None)
water = myType("water", None)
grass = myType("grass", None)

#add relationships
fire.relat = {fire: 0.5, normal: 1, water: 0.5, grass: 2}
normal.relat = {fire: 1, normal: 1, water: 1, grass: 1}
water.relat = {fire: 2, normal: 1, water: 0.5, grass: 0.5}
grass.relat = {fire: 0.5, normal: 1, water: 2, grass: 0.5}

#call the variable of fire relationship with water
print fire.relat[water]

class move (object):
        def __init__(self, name, damage, myType, ps):
                self.name = name
                self.damage = damage
                self.myType = myType
                self.ps = ps


tackle = move("Tackle", 5, normal, "physical")
searingburst = move("Searing Burst", 5, fire, "special")
pinesmack = move("Pine Smack", 5, grass, "special")
refresher = move("Refresher", 5, water, "special")

class Pokemon(object):
        # Creates a Pokemon!
        def __init__(self, speciesname, nickname, gender, myType1, myType2, chp, thp, exp, acc, moves):
                self.speciesname = speciesname
                self.nickname = nickname
                self.gender = gender
                self.myType1 = myType1
                self.myType2 = myType2
                self.chp = chp
                self.thp = thp
                self.exp = exp
                self.damage = 2
                self.bacc = 0
                self.acc = acc
                self.moves = moves
        def displayhp(self):
                print str(self.chp) + "/" + str(self.thp) + " HP"
        def damageenemy(self, opponent):
                if ((opponent.myType1).upper() == 'GHOST'):
                        opponent.chp = opponent.chp
                        print("%s's attack passed through %s!" % (self.nickname, opponent.nickname))
                        opponent.displayhp()
                else:
                        opponent.chp = opponent.chp - self.damage
                        opponent.displayhp()
        #make a Pokemon by defining a variable = Pokemon(speciesname, nickname etc...)

treecko = Pokemon("Treecko", "Treecko", "male", grass, None, 20, 20, 0, 0, [tackle, pinesmack])
krabby = Pokemon("Krabby", "Krabby", "male", water, None, 20, 20, 0, 0, [tackle, refresher])

def choose(o1, o2):
        o1r = input("What will %s do? enter 1 for tackle, 2 for pinesmack: " % o1.speciesname)
        print type(o1r)
        if 1 == o1r:
            print o1.moves[0]
        elif 2 == o1r:
            print o1.moves[1]
        else:
            assert 0, 'wrong input!'

#         if o1r in o1.moves:
#                 print o1r.moves
#         else:
#                 print "no"

choose(treecko, krabby)

告诉我它是否适合你。

Answer 2

if o1r in (lambda move : move.name)(o1.moves):
    print o1r.moves

是你想要的。这将取出移动的名称

Answer 3

好的，这是我在评论中的意思：在口袋妖怪类init函数中：

self.moves = {move.name:move for move in moves}

然后在你的选择功能中：

if o1r in o1.moves:
    print 'Move {o} caused damage {1}'.format(o1r, o1.moves[o1r].damage)

如何在列表中访问类的属性...在另一个类中？

3 个答案: