QuickDraw 中的 python3 递归动画

我有一个文本文件，其中包含行星及其相应的卫星/卫星以及它们的轨道半径和周期，我想用它来创建动画quickdraw类似于下面的：

文本文件如下：

RootObject: Sun

Object: Sun
Satellites: Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus,Neptune,Ceres,Pluto,Haumea,Makemake,Eris
Radius: 20890260
Orbital Radius: 0

Object: Miranda
Orbital Radius: 5822550
Radius: 23500
Period: 1.413

Object: Ariel
Orbital Radius: 8595000
Radius: 60000
Period: 2.520379

Object: Umbriel
Orbital Radius: 11983500
Radius: 60000
Period: 4.144177

Object: Titania
Orbital Radius: 19575000
Radius: 75000
Period: 8.7058

Object: Oberon
Orbital Radius: 26235000
Radius: 75000
Period: 13.463

Object: Uranus
Orbital Radius: 453572956
Radius: 2555900
Period: 30799
Satellites: Puck,Miranda,Ariel,Umbriel,Titania,Oberon

Object: Neptune
Orbital Radius: 550000000
Radius: 2476400
Period: 60190
Satellites: Triton

Object: Triton
Orbital Radius: 40000000
Radius: 135300
Period: -5.8

Object: Mercury
Orbital Radius: 38001200
Period: 87.9691
Radius: 243900.7

Object: Venus
Orbital Radius: 57477000
Period: 224.698
Radius: 605100.8

Object: Earth
Orbital Radius: 77098290
Period: 365.256363004
Radius: 637100.0
Satellites: Moon

Object: Moon
Orbital Radius: 18128500
Radius: 173700.10
Period: 27.321582

Object: Mars
Orbital Radius: 106669000
Period: 686.971
Radius: 339600.2
Satellites: Phobos,Deimos

Object: Phobos
Orbital Radius: 3623500.6
Radius: 200000
Period: 0.31891023

Object: Deimos
Orbital Radius: 8346000
Period: 1.26244
Radius: 200000.2

Object: Jupiter
Orbital Radius: 210573600
Period: 4332.59
Radius: 7149200
Satellites: Io,Europa,Ganymede,Callisto

Object: Ceres
Orbital Radius: 130995855
Period: 1679.67
Radius: 48700

Object: Io
Orbital Radius: 22000000
Period: 1.7691377186
Radius: 182100.3

Object: Europa
Orbital Radius: 36486200
Period: 3.551181
Radius: 156000.8

Object: Ganymede
Orbital Radius: 47160000
Period: 7.15455296
Radius: 263400

Object: Callisto
Orbital Radius: 69700000
Period: 16.6890184
Radius: 241000

Object: Saturn
Orbital Radius: 353572956
Period: 10759.22
Radius: 6026800
Satellites: Mimas,Enceladus,Tethys,Dione,Rhea,Titan,Iapetus

Object: Mimas
Orbital Radius: 8433396
Radius: 20600
Period: 0.9

Object: Enceladus
Orbital Radius: 10706000
Radius: 25000
Period: 1.4

Object: Tethys
Orbital Radius: 13706000
Radius: 50000
Period: 1.9

Object: Dione
Orbital Radius: 17106000
Radius: 56000
Period: 2.7

Object: Rhea
Orbital Radius: 24000000
Radius: 75000
Period: 4.5

Object: Titan
Orbital Radius: 50706000
Radius: 257600
Period: 15.945

Object: Iapetus
Radius: 75000
Orbital Radius: 72285891
Period: 79

我已经将原来的代码（长得离奇）改为了这个较短的代码：（归功于 sudo_O 的出色表现help https://stackoverflow.com/questions/13519350/inputing-data-into-dictionary-python/13519883#13519883)

file = open("data1.txt","r")

def data(file):
    d = {}
    for line in file:
        if line.strip() != '':
            key,value = line.split(":")
            if key == 'RootObject':
                continue
            if key == 'Object':                
                obj = value.strip()
                d[obj]={}
            else:
                d[obj][key] = value.strip()
    return d

planets = data(file)

print(planets)

我最大的问题是我不知道如何编写从文件导入数据并通过以下方式使用它的代码recursion创建与所示类似的动画。我不断被告知解决方案非常简单，代码实际上非常短，但我不知道如何做到这一点，这实际上非常令人沮丧。

这是我绕圆旋转的代码，但看起来太复杂了......

import math

print("circle",400,300,50)

print("flush false")

centreX=400
centreY=300
radius=100
angle=math.pi/2

while True:
    print("color 0 0 0")
    print("clear")
    print("color 255 0 255")
    print("circle",centreX,centreY,radius)
    childX=math.sin(angle)*radius*1.5+centreX
    childY=math.cos(angle)*radius*1.5+centreY
    print("circle",childX,childY,radius/2)
    print("refresh")
    angle+=0.01

规模代码：

scale=250/max([dict[planets]["Orbital Radius"] for x in dict if "Orbital Radius" in dict[planets]])

绕轨道运行的行星代码：

print("flush false")
scale=250/max([dict[planets]["Orbital Radius"] for x in dict if "Orbital Radius" in dict[planets]])
t=0
x=400
y=300
while True:
    print("refresh")
    print("colour 0 0 0")
    print("clear")
    print("colour 255 255 255")
    print("fillcircle",x,y,dict['Sun']['Radius']*scale)
    print("text ", "\"Sun\"",x+dict['Sun']['Radius']*scale,y)
    r_earth=dict['Earth']['Orbital Radius']*scale;
    print("circle",x,y,r_earth)
    r_X=x+math.sin(t*2*math.pi/dict['Earth']['Period'])*r_earth
    r_Y=y+math.cos(t*2*math.pi/dict['Earth']['Period'])*r_earth
    print("fillcircle",r_X,r_Y,dict['Earth']['Radius']*scale)
    print("text ", "\"Earth\"",r_X+dict['Earth']['Radius']*scale,r_Y)
    t+=0.02

这段代码不能很好地工作，但是作为初学者，我对这个问题已经处于极限......

最后的努力！！这行得通吗？

print("flush false")
scale = 250/max([dict[planets]["Orbital Radius"] for x in dict if "Orbital Radius" in dict[planets]])
t = 0
x = 400
y = 300
print("fillcircle",400,300,dict['Sun']['Radius']*scale)
print("text ", "\"Sun\"",x+dict['Sun']['Radius']*scale,y)


while True:
    r_new = dict['Object']['Orbital Radius']*scale
    print("circle",x,y,r_new)
    r_X = x + math.sin(t*2*math.pi/dict['Object']['Period'])*r_new
    r_Y = y + math.cos(t*2*math.pi/dict['Object']['Period'])*r_new
    print("fillcircle",r_X,r_Y,dict['Object']['Radius']*scale)
    print("text ",Object,r_X+dict['Object']['Radius']*scale,r_Y)
    t += 0.02
    if planets['Object']['Satellites'] = 0
        return
    else:
        r_sat = dict['Object']['Satellites']['Orbital Radius']*scale
        print("circle",x,y,r_sat)
        r_satX = x + math.sin(t*2*math.pi/dict['Object']['Satellites']['Period'])*r_sat
        r_satY = y + math.cos(t*2*math.pi/dict['Object']['Satellites']['Period'])*r_sat
        print("fillcircle",r_satX,r_satY,dict['Object']['Satellites']['Radius']*scale)
        print("text ",['Object']['Satellites'],r_satX+dict['Object']['Satellites']['Radius']*scale,r_satY)
        t += 0.02

第一个问题递归- 递归函数是调用自身的函数！一个简单的例子是倒计时函数，例如：

def countdown(n):
    # counting down the recursive way! 
    if n > 0:
        print n
        countdown(n-1)
    else:
        return

Calling countdown(10)将打印10, 9, 8,.., 2, 1.

你可以看到countdown传递了一个数字n它所做的就是打印该数字，然后调用自身但传递n-1这次。只有一次n=0如果传递了，那么它就没有什么可做的了，因此每个递归调用都会返回。适合您的词典词典案例（我们称其为词典库以避免混淆） a recursive打印方法是：

1. 将整个库传递给打印例程。
2. 打印第一本词典。
3. 递归调用自身并传递减去第一个字典的库。
4. 当图书馆没有更多词典可供打印时返回。

像这样的事情：

def recursive_print(dic):

    if len(dic) > 0:                  # If dictionaries in library > 0 
        print dic.keys()[0]           # Print the key i.e Earth
        print dic[dic.keys()[0]]      # Print the dictionary value for i.e Earth
        dic.popitem()                 # Remove the Earth dictionary from library
        recursive_print(dic)          # Recursive call 
    else:
        return                        # Printed all, return up the stack.


planets = data(file)
recursive_print(planets)

下一步不是以当前格式打印字典，而是进行一些计算/转换，以便输出有效quickdraw输入，就像您已经使用的用于打印圆圈的代码一样quickdraw.

您需要担心比例，以确保所有物体都适合绘图表面，找到最大的天体Orbital Radius并用它来计算比例。

使用类似于您的代码的列表理解，我们可以找到最大值：

max([planet[key]['Orbital Radius'] for key in planet])

>>> 8595000

scale = gridsize/max([planet[key]['Orbital Radius'] for key in planet])