Merging multiple dictionaries with dictionaries

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James
James 2025-7-12
评论： James 2025-7-12
Ran in:
Related to this question, how can one merge two dictionaries that have dictionaries as the keys and values?
A = dictionary(dictionary(["key1"], {1}), dictionary(["key2"], {2}));
A(dictionary(["key3"], {3})) = dictionary(["key4"], {4})
A = dictionary (dictionary --> dictionary) with 2 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
B = dictionary(dictionary(["key5"], {5}), dictionary(["key6"], {6}));
B(dictionary(["key7"], {7})) = dictionary(["key8"], {8})
B = dictionary (dictionary --> dictionary) with 2 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
C = dictionary(A.keys, A.values)
Error using dictionary/keys
Unable to combine keys of type 'dictionary'. Specify the format option as "cell".
C(B.keys) = B.values

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采纳的回答

Matt J
Matt J 2025-7-12
Ran in:
A = dictionary(dictionary(["key1"], {1}), dictionary(["key2"], {2}));
A(dictionary(["key3"], {3})) = dictionary(["key4"], {4});
B = dictionary(dictionary(["key5"], {5}), dictionary(["key6"], {6}));
B(dictionary(["key7"], {7})) = dictionary(["key8"], {8});
kv = [keys(A, "cell")', keys(B, "cell")';
values(A, "cell")', values(B, "cell")'];
C=dictionary(kv{:})
C = dictionary (dictionary --> dictionary) with 4 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
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Matt J
Matt J 2025-7-12
编辑：Matt J 2025-7-12
Ran in:
But understand that if you now want to loop over the entries of C, you will need an array of its keys. Because the keys are dictionaries, the array will have to be in cell form:
k=[keys(A, "cell"); keys(B, "cell")];
v=[values(A, "cell"); values(B, "cell")];
%% noncell key/values
kv=[k';v'];
C=dictionary(kv{:})
C = dictionary (dictionary --> dictionary) with 4 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
for i=1:numel(k)
d=C(k{i})
end
d = dictionary (string --> cell) with 1 entry: "key2" --> {[2]} d = dictionary (string --> cell) with 1 entry: "key4" --> {[4]} d = dictionary (string --> cell) with 1 entry: "key6" --> {[6]} d = dictionary (string --> cell) with 1 entry: "key8" --> {[8]}
But you could have implemented the same loop, with less code, by accepting cell-valued dictionary entries:
%% cell key/values
C=dictionary(k,v)
C = dictionary (cell --> cell) with 4 entries: {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]}
for i=1:numel(k)
d=C{k(i)}
end
d = dictionary (string --> cell) with 1 entry: "key2" --> {[2]} d = dictionary (string --> cell) with 1 entry: "key4" --> {[4]} d = dictionary (string --> cell) with 1 entry: "key6" --> {[6]} d = dictionary (string --> cell) with 1 entry: "key8" --> {[8]}

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更多回答（2 个）

Matt J
Matt J 2025-7-12
Ran in:
f=@(i) {dictionary(["key"+i], {i})}
f = function_handle with value:
@(i){dictionary(["key"+i],{i})}
A = dictionary(f(1), f(2));
A(f(3)) = f(4)
A = dictionary (cell --> cell) with 2 entries: {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]}
B = dictionary(f(5), f(6));
B(f(7)) = f(8)
B = dictionary (cell --> cell) with 2 entries: {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]}
C = dictionary(A.keys, A.values)
C = dictionary (cell --> cell) with 2 entries: {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]}
C(B.keys) = B.values
C = dictionary (cell --> cell) with 4 entries: {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]} {[dictionary (string --> cell) with 1 entry]} --> {[dictionary (string --> cell) with 1 entry]}
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Matt J
Matt J 2025-7-12
编辑：Matt J 2025-7-12
Maybe I don't have a necessary understanding of where the original dictionaries are coming from. Even before A, B are created, you apparently have a collection of multiple dictionaries from which A and B will be built. How are you holding this collection together?
It has to be with a cell array, becacuse again, dictionaries cannot be concatenated by themselves.There is no other way to maintain an array of dictionaries. And if the key/value dictionaries are already in cell form, why not continue to use them in that form?
James
James 2025-7-12
You gave me a lot to think about. I'm going to rework my dictionaries, however, to answer my question I posted a solution with a loop. Probably not optimal but it maintains the structure without the use of cell arrays.

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James
James 2025-7-12
Ran in:
A = dictionary(dictionary(["key1"], {1}), dictionary(["key2"], {2}));
A(dictionary(["key3"], {3})) = dictionary(["key4"], {4})
A = dictionary (dictionary --> dictionary) with 2 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
B = dictionary(dictionary(["key5"], {5}), dictionary(["key6"], {6}));
B(dictionary(["key7"], {7})) = dictionary(["key8"], {8})
B = dictionary (dictionary --> dictionary) with 2 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry
C = dictionary;
k = keys(A, "cell");
for i = 1:length(k)
C = insert(C, cell2mat(k(i)), A(cell2mat(k(i))));
end
k = keys(B, "cell");
for i = 1:length(k)
C = insert(C, cell2mat(k(i)), B(cell2mat(k(i))));
end
C
C = dictionary (dictionary --> dictionary) with 4 entries: dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry dictionary (string --> cell) with 1 entry --> dictionary (string --> cell) with 1 entry

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