python - Parse dictionary with Tuple as key

Question

My dictionary with tuple as a key is as follows:

Key represents the x and y coordinates. (x, y)

D1 = {(10,12): 23, (8,14): 45, (12, 9): 29}
D2 = {(2, 8) : 67, (12, 10): 23, (14, 8): 56}

Now, from the above Dictionary, I would like to perform the following

1. Sort dictionary D1 based on Keys
2. Sort dictionary D2 based on Keys
3. Simply, print the keys and values in D1
4. Simply, print the keys and values in D2
5. Then do the following:

(pseudocode)

total = 0
For each key (x, y) in D1,  
   if D2.has_key((y, x)):
      total = total + D1[(x, y)] * D2[(y, x)]
   Print total

Answer 1

• （1/2）要对您必须使用的collections.OrderedDict字典进行排序（因为未对普通字典进行排序）代码：

  from collections import OrderedDict
  D1 = {(10,12): 23, (8,14): 45, (12, 9): 29}
  D1_sorted = OrderedDict(sorted(D1.items()))
  

• (3/4) 代码：print(D1)

• (5) 将你的代码转换成python

  total = 0
  for x, y in D1.keys():
      try:
          total = total + D1[(x, y)] * D2[(y, x)]
      except KeyError:
          pass
  print total
  

Answer 2

我想你在（对于你的伪代码）：

D1 = {(10,12): 23, (8,14): 45, (12, 9): 29}
D2 = {(2, 8) : 67, (12, 10): 23, (14, 8): 56}

total = sum(D1[k] * D2.get(k[::-1], 0) for k in D1.iterkeys())
# 3049

Answer 3

>>> from collections import OrderedDict
>>> D1 = {(10,12): 23, (8,14): 45, (12, 9): 29}
>>> D2 = {(2, 8) : 67, (12, 10): 23, (14, 8): 56}
>>> D1 = OrderedDict(sorted(D1.items()))
>>> D2 = OrderedDict(sorted(D2.items()))
>>> print D1
OrderedDict([((8, 14), 45), ((10, 12), 23), ((12, 9), 29)])
>>> print D2
OrderedDict([((2, 8), 67), ((12, 10), 23), ((14, 8), 56)])
>>> sum(D1[(x, y)] * D2.get((y, x), 0) for x, y in D1)
3049