Question

我有一个嵌套的字典，如何获得具有常见键值（例如37、74等）的值：

myDict = {0: {37.0:  {'C23H27O9': 'C22H23O10'}},
          1: {74.0:  {'C23H27O9': 'C21H19O11'}},
          2: {111.0: {'C23H27O9': 'C20H15O12'}},
          3: {148.0: {'C23H27O9': 'C19H11O13'}},
          4: {37.0:  {'C22H23O10': 'C21H19O11'}},
          5: {74.0:  {'C22H23O10': 'C20H15O12'}},
          6: {111.0: {'C22H23O10': 'C19H11O13'}},
          7: {37.0:  {'C21H19O11': 'C20H15O12'}},
          8: {74.0:  {'C21H19O11': 'C19H11O13'}},
          9: {37.0:  {'C20H15O12': 'C19H11O13'}}
         }

所需的输出：

37.0 --> C23H27O9: C22H23O10: C21H19O11 : C20H15O12 : C19H11O13
74.0 --> C23H27O9 : C21H19O11 : C19H11O13
...

Answer 1

您似乎希望从最里面的字典中获得所有键和值的集合，以保留它们遇到的顺序。

这似乎可以完成工作：

from collections import defaultdict

result = defaultdict(list)
for _, outerDict in myDict.items():
    for innerKey, innerDict in outerDict.items():
        for k, v in innerDict.items():
            if k not in result[innerKey]:
                result[innerKey].append(k)
            if v not in result:
                result[innerKey].append(v)

此后，result是dict中的list：

>>> for key, values in result.items():
>>>    print(key, values)

37.0 ['C23H27O9', 'C22H23O10', 'C21H19O11', 'C20H15O12', 'C19H11O13']
74.0 ['C23H27O9', 'C21H19O11', 'C22H23O10', 'C20H15O12', 'C19H11O13']
111.0 ['C23H27O9', 'C20H15O12', 'C22H23O10', 'C19H11O13']
148.0 ['C23H27O9', 'C19H11O13']

您可以根据需要设置输出的格式，我的假设是您只需要数据。

Answer 2

您可以使用itertools.groupby：

from itertools import groupby as gb
myDict = {0: {37.0: {'C23H27O9': 'C22H23O10'}}, 1: {74.0: {'C23H27O9': 'C21H19O11'}}, 2: {111.0: {'C23H27O9': 'C20H15O12'}}, 3: {148.0: {'C23H27O9': 'C19H11O13'}}, 4: {37.0: {'C22H23O10': 'C21H19O11'}}, 5: {74.0: {'C22H23O10': 'C20H15O12'}}, 6: {111.0: {'C22H23O10': 'C19H11O13'}}, 7: {37.0: {'C21H19O11': 'C20H15O12'}}, 8: {74.0: {'C21H19O11': 'C19H11O13'}}, 9: {37.0: {'C20H15O12': 'C19H11O13'}}}
d = sorted([(a, [j for k in b.items() for j in k]) for i in myDict.values() for a, b in i.items()], key=lambda x:x[0])
r = [(a, [*{i for _, j in b for i in j}]) for a, b in gb(d, key=lambda x:x[0])]

输出：

[(37.0, ['C19H11O13', 'C20H15O12', 'C21H19O11', 'C22H23O10', 'C23H27O9']), (74.0, ['C19H11O13', 'C20H15O12', 'C21H19O11', 'C22H23O10', 'C23H27O9']), (111.0, ['C20H15O12', 'C19H11O13', 'C23H27O9', 'C22H23O10']), (148.0, ['C23H27O9', 'C19H11O13'])]

在嵌套字典中取通用值

2 个答案: