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QuickDraw中的python3遞歸動畫

[英]python3 recursion animation in QuickDraw

我有一個有行星及其相應衛星的文本文件/他們的軌道半徑和時段,以及我沿着衛星想用這個來創建一個動畫quickdraw類似下面的一個:

太陽系動畫

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

我把原來的代碼(這個時間長得太長了)改成了這個較短的代碼:(感謝sudo_O提供了所有他非常棒的幫助

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)

我最大的問題是我不知道如何編寫從文件導入數據的代碼,並通過recursion利用它來創建類似於所示的動畫。 我不斷被告知解決方案非常簡單,而且代碼實際上非常簡短,但我不知道如何做到這一點,實際上非常令人沮喪。

這是我圍繞圓圈運行的圓圈的代碼,但它似乎太復雜了......

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

規模代碼:

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

軌道行星代碼:

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

這段代碼不能很好地工作,但作為一個初學者,我對這個問題的極限......

最后的努力!! 這會起作用嗎?

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

第一個問題遞歸 - 遞歸函數是一個自我調用的函數! 一個簡單的例子是倒計時功能,例如:

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

調用countdown(10)將打印10, 9, 8,.., 2, 1

您可以看到countdown傳遞了一個數字n ,它所做的就是打印該數字,然后自己調用,但這次傳遞n-1 只有在n=0才會傳遞它沒有任何事情要做,所以返回每個遞歸調用。 對於你的字典詞典案例(讓我們稱之為詞典庫以避免混淆)recursive打印方法將是:

  1. 將整個庫傳遞給打印例程。
  2. 打印第一本字典。
  3. 遞歸調用自身並傳遞庫減去第一個字典。
  4. 當庫沒有更多字典要打印時返回。

像這樣的東西:

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)

下一步是不是以當前格式打印字典,而是進行一些計算/轉換,因此輸出是有效的quickdraw輸入,就像你已經使用quickdraw打印圓圈的代碼quickdraw

您將需要擔心比例,以確保所有對象適合繪圖表面找到具有最大Orbital Radius的天體並使用它來計算比例。

使用類似於您的代碼的列表理解,我們可以找到最大的值:

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

>>> 8595000

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

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