英文:
Combine two dicts and replace missing values
问题
我正在寻求将两个字典组合起来，通过将具有相同键的元素分组，但我也希望考虑那些不共享在两个字典之间的键。例如，给定以下两个字典。
d1 = {'a': 1, 'b': 2, 'c': 3, 'e': 5}
d2 = {'a': 11, 'b': 22, 'c': 33, 'd': 44}
预期的代码将输出
df = {'a': [1, 11], 'b': [2, 22], 'c': [3, 33], 'd': [0, 44], 'e': [5, 0]}
或类似的数组：
df = [['a', 1, 11], ['b', 2, 22], ['c', 3, 33], ['d', 0, 44], ['e', 5, 0]]
具体使用 0 来表示不存在的条目并不重要，可以使用任何字符来表示缺失的值。
我尝试使用以下代码：
df = defaultdict(list)
for d in (d1, d2):
for key, value in d.items():
df[key].append(value)
但得到以下结果：
df = {'a': [1, 11], 'b': [2, 22], 'c': [3, 33], 'd': [44], 'e': [5]}
这并没有告诉我哪个字典缺少了条目。我可以回头查看它们两个，但我正在寻找一个更优雅的解决方案。
英文:
I am looking to combine two dictionaries by grouping elements that share common keys, but I would also like to account for keys that are not shared between the two dictionaries. For instance given the following two dictionaries.
d1 = {'a':1, 'b':2, 'c': 3, 'e':5}
d2 = {'a':11, 'b':22, 'c': 33, 'd':44}
The intended code would output
df = {'a':[1,11] ,'b':[2,22] ,'c':[3,33] ,'d':[0,44] ,'e':[5,0]}
Or some array like:
df = [[a,1,11] , [b,2,22] , [c,3,33] , [d,0,44] , [e,5,0]]
The fact that I used 0 specifically to denote an entry not existing is not important per se. Just any character to denote the missing value.
I have tried using the following code
df = defaultdict(list)
for d in (d1, d2):
for key, value in d.items():
df[key].append(value)
But get the following result:
df = {'a':[1,11] ,'b':[2,22] ,'c':[3,33] ,'d':[44] ,'e':[5]}
Which does not tell me which dict was missing the entry.
I could go back and look through both of them, but was looking for a more elegant solution
答案1
得分: 3
你可以使用字典推导式来实现，代码如下：
d1 = {'a': 1, 'b': 2, 'c': 3, 'e': 5}
d2 = {'a': 11, 'b': 22, 'c': 33, 'd': 44}
res = {k: [d1.get(k, 0), d2.get(k, 0)] for k in set(d1).union(d2)}
print(res)
英文:
You can use a dict comprehension like so:
d1 = {'a':1, 'b':2, 'c': 3, 'e':5}
d2 = {'a':11, 'b':22, 'c': 33, 'd':44}
res = {k: [d1.get(k, 0), d2.get(k, 0)] for k in set(d1).union(d2)}
print(res)
答案2
得分: 2
另一种解决方案：
d1 = {"a": 1, "b": 2, "c": 3, "e": 5}
d2 = {"a": 11, "b": 22, "c": 33, "d": 44}
df = [[k, d1.get(k, 0), d2.get(k, 0)] for k in sorted(d1.keys() | d2.keys())]
print(df)
输出：
[['a', 1, 11], ['b', 2, 22], ['c', 3, 33], ['d', 0, 44], ['e', 5, 0]]
如果不需要排序的结果，可以去掉 sorted()。
英文:
Another solution:
d1 = {"a": 1, "b": 2, "c": 3, "e": 5}
d2 = {"a": 11, "b": 22, "c": 33, "d": 44}
df = [[k, d1.get(k, 0), d2.get(k, 0)] for k in sorted(d1.keys() | d2.keys())]
print(df)
Prints:
[['a', 1, 11], ['b', 2, 22], ['c', 3, 33], ['d', 0, 44], ['e', 5, 0]]
If you do not want sorted results, leave the sorted() out.
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