[已解决]python 字典

lou

{'red': [(50, 50), (50, 90), (90, 90)], 'green': [(160, 170), (160, 130), (120, 130)], 'blue': [(200, 30), (180, 30), (180, 50), (220, 50)], 'yellow': [(40, 100), (40, 140), (60, 140), (60, 120)], 'purple': [(210, 70), (230, 90), (270, 90), (270, 50), (230, 50)], 'olive': [(180, 130), (180, 170), (220, 210), (240, 190)], 'magenta': [(100, 200), (120, 180), (80, 140), (80, 180)]}
请问
如何把该字典中的值输出成这样呢？
{'red': [(0,40),(40,0),(- 40,- 40)], 'green': [(0, -40), (-40, 0), (-40, -40)], 'blue': [(-20, 0), (0, 20), (40, 0), (-20,-20)], 'yellow': [(0, 40), (20, 0), (0, -20), (-20, -20)], 'purple': [(20, 20), (40, 0), (0, -40), (40, 0), (-20, 20)], 'olive': [(0, 40), (40, 40), (20, 20), (-60, -60)], 'magenta': [(20, -20), (-40, -40), (0, 40), (20, 20)]}
就是由点左边————>向量， 后面的坐标减去前面的坐标值。。得到上面这个字典

最佳答案

月排行榜 / 总排行榜

冬雪雪冬

2019-4-23 15:06:23

1. d1 = {'red': [(50, 50), (50, 90), (90, 90)], 'green': [(160, 170), (160, 130), (120, 130)], 'blue': [(200, 30), (180, 30), (180, 50), (220, 50)], 'yellow': [(40, 100), (40, 140), (60, 140), (60, 120)], 'purple': [(210, 70), (230, 90), (270, 90), (270, 50), (230, 50)], 'olive': [(180, 130), (180, 170), (220, 210), (240, 190)], 'magenta': [(100, 200), (120, 180), (80, 140), (80, 180)]}
   
2. d2 = {}
   
3. for i, j in d1.items():
   
4.     data = []
   
5.     len1 = len(j)
   
6.     for n in range(len1):
   
7.         data.append((j[(n + 1) % len1][0] - j[n][0], j[(n + 1) % len1][1] - j[n][1]))
   
8.         d2[i] = data

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跳转到最佳答案楼层

冬雪雪冬

1. d1 = {'red': [(50, 50), (50, 90), (90, 90)], 'green': [(160, 170), (160, 130), (120, 130)], 'blue': [(200, 30), (180, 30), (180, 50), (220, 50)], 'yellow': [(40, 100), (40, 140), (60, 140), (60, 120)], 'purple': [(210, 70), (230, 90), (270, 90), (270, 50), (230, 50)], 'olive': [(180, 130), (180, 170), (220, 210), (240, 190)], 'magenta': [(100, 200), (120, 180), (80, 140), (80, 180)]}
   
2. d2 = {}
   
3. for i, j in d1.items():
   
4.     data = []
   
5.     len1 = len(j)
   
6.     for n in range(len1):
   
7.         data.append((j[(n + 1) % len1][0] - j[n][0], j[(n + 1) % len1][1] - j[n][1]))
   
8.         d2[i] = data

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凌九霄

1. import collections
   
2. 
   
3. d = {'red': [(50, 50), (50, 90), (90, 90)], 'green': [(160, 170), (160, 130), (120, 130)],
   
4.      'blue': [(200, 30), (180, 30), (180, 50), (220, 50)], 'yellow': [(40, 100), (40, 140), (60, 140), (60, 120)],
   
5.      'purple': [(210, 70), (230, 90), (270, 90), (270, 50), (230, 50)], 'olive': [(180, 130), (180, 170), (220, 210), (240, 190)],
   
6.      'magenta': [(100, 200), (120, 180), (80, 140), (80, 180)]}
   
7. 
   
8. new = {}
   
9. 
   
10. for i in d.items():
    
11.     tp = []
    
12.     dl = collections.deque()
    
13.     dl.extend(i[1])
    
14.     for k in range(len(dl)):
    
15.         x, y = dl[1][0] - dl[0][0], dl[1][1] - dl[0][1]
    
16.         tp.append((x, y))
    
17.         dl.rotate(-1)
    
18.     new[i[0]] = tp[:]
    
19. 
    
20. print(new)

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lou

凌九霄 发表于 2019-4-23 15:12

谢谢您，不过不好意思，楼上的老哥先解决了我的问题，最佳答案能设置多次就好了 感谢！学习了

wp231957

2楼 3楼  实在好厉害 佩服的很

lou

wp231957 发表于 2019-4-23 15:27
2楼 3楼  实在好厉害 佩服的很

学习到了， 我还在想着用numpy转了再减。。晕了