Python 在dict中查找不可散列键和值的公共键：值对

我有一个字典列表如下：

    [{'EchoTime': 0,
      'FlipAngle': 90,
      'MRAcquisitionType': '2D',
      'MagneticFieldStrength': 3,
      'Manufacturer': 'SIEMENS',
      'ManufacturerModelName': 'TrioTim',
      'RepetitionTime': 2,
      'ScanOptions': 'FS',
      'ScanningSequence': 'EP',
      'SequenceName': 'epfid2d1_64',
      'SequenceVariant': 'SK',
      'SliceTiming': [[0.0025000000000000022],
       [0.5],
       [-0.030000000000000027],
       [0.46625],
       [0.168749999995],
       [-0.36124999999999996],
       [-0.46124999999500005],
       [0.036250000005000005]],
      'TaskName': 'tom'},
     {'EchoTime': 0,
      'FlipAngle': 90,
      'MRAcquisitionType': '2D',
      'MagneticFieldStrength': 3,
      'Manufacturer': 'SIEMENS',
      'ScanningSequence': 'EP',
      'SequenceName': 'epfid2d1_64',
      'SequenceVariant': 'SK',
      'SliceTiming': [[0.0037500000050000315],
       [0.5],
       [-0.030000000000000027],
       [0.46750000001500003],
       [-0.0625],
       [0.36750000001],
       [-0.16249999998999998],
       [0.33500000000499996],
       [-0.19499999998999995],
       [0.301250000005]],
      'TaskName': 'tom'},
       {'EchoTime': 0,
  'FlipAngle': 90,
  'MRAcquisitionType': '2D',
  'MagneticFieldStrength': 3,
  'RepetitionTime': 2,
  'ScanOptions': 'FS',
  'ScanningSequence': 'EP',
  'SequenceName': 'epfid2d1_64',
  'SequenceVariant': 'SK',
  'SliceTiming': [[0.0025000000000000022],
   [0.5],
   [-0.030000000000000027],
   [0.46625],
   [-0.19625000000000004],
   [0.301250000005],
   [-0.22875],
   [0.26749999999999996],
   [-0.26249999999500007]],
  'TaskName': 't1'},
  {'EchoTime': 0,
  'FlipAngle': 90,
  'MRAcquisitionType': '2D',
  'Manufacturer': 'SIEMENS',
  'ManufacturerModelName': 'TrioTim',
  'SequenceVariant': 'SK',
  'SliceTiming': [[0.0025000000000000022],
   [0.5],
   [-0.030000000000000027],
   [0.46625],
   [-0.06374999999999997],
   [0.43375000000499997],
   [-0.09624999999999995],
   [0.4],
   [-0.22875],
   [0.26749999999999996],
   [-0.26249999999500007],
   [0.235]],
  'TaskName': 'fb'}]

名单上大约有2000个dict。 现在我想在列表中的所有dict中找到公共key:value对。 同样，我使用了以下函数：

def twodict_intersec(json1, json2):
    d = {x:json1[x] for x in json1 if x in json2}
    return d

for i in range(0,10,1):
    print(i)
    if i == 0:
        res = twodict_intersec(json_list[i], json_list[i+1])
    else:
        res = dict_intersect(res, json_list[i+1])
print(res)

但输出类似于下面的内容，其中每个键包含所有值的列表，而不是公共值

res = {'EchoTime': [[[[[[[[[0, 0], 0], 0], 0], 0], 0], 0], 0], 0], 'FlipAngle': [[[[[[[[[90, 90], 90], 90], 90], 90], 90], 90], 90], 90], 'ManufacturerModelName': [[[[[[[[['TrioTim', 'TrioTim'], 'TrioTim'], 'TrioTim'], 'TrioTim'], 'TrioTim'], 'TrioTim'], 'TrioTim'], 'TrioTim']}

---

我无法获得每个键只有一个值的输出，这在所有dict中都是通用的，而不是上述情况。有人能指出我的错误吗？

你可以这样比较字典列表中的所有元素：

import copy
li = [{
        "keep this value":5,
        "remove this":59
    },{
        "keep this value":5,
        "remove this":12
    }]

def compareList(li):
    result = copy.deepcopy(li[0])
    for element in li:
        for key in element:
            if key in result and result[key] != element[key]:
                del result[key]
    return result

print (compareList(li))

---

您可以通过以下方式比较字典列表中的所有元素：

import copy
li = [{
        "keep this value":5,
        "remove this":59
    },{
        "keep this value":5,
        "remove this":12
    }]

def compareList(li):
    result = copy.deepcopy(li[0])
    for element in li:
        for key in element:
            if key in result and result[key] != element[key]:
                del result[key]
    return result

print (compareList(li))

---

你的意思是我需要用它来计算从上述函数得到的

结果

？更新了我的答案。记住向上投票并标记为解决方案，这非常有帮助。你的意思是我需要将此用于

结果

我从上述函数获得的结果？更新了我的答案。记住把投票和标记作为解决方案，这会很有帮助。