如何使用另一个字典筛选字典？

huangapple
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英文:

How to filter a dictionary using another dictionary?

问题

我有两组字典。我们称它们为A和B。我需要能够创建第三个字典C，其中包含A的所有键和相关值，除了在B中的那些键值对。这是我到目前为止尝试过的：

  1. A = {"A": {"B": 6, "C": 5}, "B": {"A": 6, "D": 7, "E": 5}, "C": {"A": 5}, "D": {"B": 7, "E": 10, "H": 25},
  2.      "E": {"D": 10, "F": 14, "B": 5}, "F": {"E": 14}, "H": {"D": 25}}
  3. B = {"A": "C"}
  5. C = A
  7. for key, value in A.items():
  8.      C.update({key: value})
  9.      if key in B:
  10.           for i in value:
  11.                if i not in B.values():
  12.                     C[key].pop(i)
  14. print(C)

思路是，给定这两个字典，得到输出：

  1. A = {"A": {"B": 6}, "B": {"A": 6, "D": 7, "E": 5}, "C": {"A": 5}, "D": {"B": 7, "E": 10, "H": 25}, "E": {"D": 10, "F": 14, "B": 5}, "F": {"E": 14}, "H": {"D": 25}}

但是，终端返回错误，说在迭代过程中字典大小已经改变。我在这里到底做错了什么？

英文:

I have two sets of dictionaries. Let's call them A and B. I need to be able to create a third dictionary, C, which has all the keys and associated values of A, except for the pairs that are in B. This is what I have tried so far:

  1. A = {"A": {"B": 6, "C": 5}, "B": {"A": 6, "D": 7, "E": 5}, "C": {"A": 5}, "D": {"B": 7, "E": 10, "H": 25},
  2.      "E": {"D": 10, "F": 14, "B": 5}, "F": {"E": 14}, "H": {"D": 25}}
  3. B = {"A": "C"}
  5. C = A
  7. for key, value in A.items():
  8.      C.update({key: value})
  9.      if key in B:
  10.           for i in value:
  11.                if i not in B.values():
  12.                     C[key].pop(i)
  14. print(C)

The idea is, given the two dictionaries, to obtain the output:

  1. A = {"A": {"B": 6}, "B": {"A": 6, "D": 7, "E": 5}, "C": {"A": 5}, "D": {"B": 7, "E": 10, "H": 25}, "E": {"D": 10, "F": 14, "B": 5}, "F": {"E": 14}, "H": {"D": 25}}

Instead, the terminal returns an error, saying that the dictionary has changed size during the iteration. What exactly am I doing wrong here?

答案1

得分: 1

你可以使用字典推导式：

  1. A = {
  2.     "A": {"B": 6, "C": 5},
  3.     "B": {"A": 6, "D": 7, "E": 5},
  4.     "C": {"A": 5},
  5.     "D": {"B": 7, "E": 10, "H": 25},
  6.     "E": {"D": 10, "F": 14, "B": 5},
  7.     "F": {"E": 14},
  8.     "H": {"D": 25},
  9. }
  10. B = {"A": "C"}
  12. C = {
  13.     k: {kk: vv for kk, vv in v.items() if not (k in B and B[k] == kk)}
  14.     for k, v in A.items()
  15. }
  16. print(C)

输出结果：

  1. {
  2.     "A": {"B": 6},
  3.     "B": {"A": 6, "D": 7, "E": 5},
  4.     "C": {"A": 5},
  5.     "D": {"B": 7, "E": 10, "H": 25},
  6.     "E": {"D": 10, "F": 14, "B": 5},
  7.     "F": {"E": 14},
  8.     "H": {"D": 25},
  9. }
英文:

You can use dict comprehension:

  1. A = {
  2.     "A": {"B": 6, "C": 5},
  3.     "B": {"A": 6, "D": 7, "E": 5},
  4.     "C": {"A": 5},
  5.     "D": {"B": 7, "E": 10, "H": 25},
  6.     "E": {"D": 10, "F": 14, "B": 5},
  7.     "F": {"E": 14},
  8.     "H": {"D": 25},
  9. }
  10. B = {"A": "C"}
  12. C = {
  13.     k: {kk: vv for kk, vv in v.items() if not (k in B and B[k] == kk)}
  14.     for k, v in A.items()
  15. }
  16. print(C)

Prints:

  1. {
  2.     "A": {"B": 6},
  3.     "B": {"A": 6, "D": 7, "E": 5},
  4.     "C": {"A": 5},
  5.     "D": {"B": 7, "E": 10, "H": 25},
  6.     "E": {"D": 10, "F": 14, "B": 5},
  7.     "F": {"E": 14},
  8.     "H": {"D": 25},
  9. }

答案2

得分: 0

我认为你想要的是：

  1. import pprint
  3. A = {
  4.     "A": {"B": 6, "C": 5},
  5.     "B": {"A": 6, "D": 7, "E": 5},
  6.     "C": {"A": 5},
  7.     "D": {"B": 7, "E": 10, "H": 25},
  8.     "E": {"D": 10, "F": 14, "B": 5},
  9.     "F": {"E": 14},
  10.     "H": {"D": 25},
  11. }
  12. B = {"A": "C"}
  13. C = {}
  15. for k, v in A.items():
  16.     for kk, vv in v.items():
  17.         if k in B and B[k] == kk:
  18.             continue
  20.         C.setdefault(k, {})[kk] = vv
  22. pprint.pprint(C)

根据你的示例输入，会产生以下输出：

  1. {'A': {'B': 6},
  2.  'B': {'A': 6, 'D': 7, 'E': 5},
  3.  'C': {'A': 5},
  4.  'D': {'B': 7, 'E': 10, 'H': 25},
  5.  'E': {'B': 5, 'D': 10, 'F': 14},
  6.  'F': {'E': 14},
  7.  'H': {'D': 25}}
英文:

I think you want:

  1. import pprint
  3. A = {
  4.     "A": {"B": 6, "C": 5},
  5.     "B": {"A": 6, "D": 7, "E": 5},
  6.     "C": {"A": 5},
  7.     "D": {"B": 7, "E": 10, "H": 25},
  8.     "E": {"D": 10, "F": 14, "B": 5},
  9.     "F": {"E": 14},
  10.     "H": {"D": 25},
  11. }
  12. B = {"A": "C"}
  13. C = {}
  15. for k, v in A.items():
  16.     for kk, vv in v.items():
  17.         if k in B and B[k] == kk:
  18.             continue
  20.         C.setdefault(k, {})[kk] = vv
  22. pprint.pprint(C)

Which given your sample input produces:

  1. {'A': {'B': 6},
  2.  'B': {'A': 6, 'D': 7, 'E': 5},
  3.  'C': {'A': 5},
  4.  'D': {'B': 7, 'E': 10, 'H': 25},
  5.  'E': {'B': 5, 'D': 10, 'F': 14},
  6.  'F': {'E': 14},
  7.  'H': {'D': 25}}

huangapple
  • 本文由 发表于 2023年6月5日 00:34:14
  • 转载请务必保留本文链接：https://go.coder-hub.com/76401402.html
  • dictionary
  • iteration
  • loops
  • python
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