How to generate all 2^n combinations of n choices from a 2-vector?

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Jerry Guern
Jerry Guern 2025-5-31
编辑： Matt J 2025-6-2
If I want all combinations of 3 choices of 0 or 1, I do this:
t = combinations([0 1],[0 1],[0 1]);
If I want all combinations of 4 choices, I have to do this:
t = combinations([0 1],[0 1],[0 1],[0,1])
If I want all combinations of 5 choices, I have to do this:
t = combinations([0 1],[0 1],[0 1],[0,1],[0 1])
...and so on. Is there any way to generate all combination of a variable number of choices?

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采纳的回答

John D'Errico
John D'Errico 2025-5-31
编辑：John D'Errico 2025-5-31
Ran in:
Simple enough. Don't use combinations.
n = 3;
t = dec2bin(0:2^n - 1) - '0'
t = 8×3
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
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Easy enough to make that into a table if you so desire.
Sigh. Do you INSIST on using combinations? This will suffice.
v = repmat({[0 1]},1,n);
t2 = combinations(v{:})
t2 = 8×3 table
Var1 Var2 Var3 ____ ____ ____ 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
I can think of at least a couple other ways if pushed, but they shoiuld work.
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Steven Lord
Steven Lord 2025-6-2
Ran in:
And of course, if you want a matrix from combinations (assuming all the table variables can be concatenated together) you can do that by indexing into the output from the function call.
n = 3;
t = dec2bin(0:2^n - 1) - '0';
v = repmat({[0 1]},1,n);
t2 = combinations(v{:}).Variables
t2 = 8×3
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
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doTheyMatch = isequal(t, t2)
doTheyMatch = logical
1
You need to be careful, though, if you have mixed types.
t3 = combinations([0 1], ["apple", "banana"])
t3 = 4×2 table
Var1 Var2 ____ ________ 0 "apple" 0 "banana" 1 "apple" 1 "banana"
t4 = t3.Variables % Note that this is a string array, using "0" and "1" not 0 and 1
t4 = 4×2 string array
"0" "apple" "0" "banana" "1" "apple" "1" "banana"
Matt J
Matt J 2025-6-2
编辑：Matt J 2025-6-2
Ran in:
The spped performance as a function of n is interesting. combinations() starts to overtake ndgrid() only for n>=20 or so. I would have expected combinations() to benefit from avoiding a cat() operation.
Timing(15)
ans = 0.0017
ans = 8.8375e-04
ans = 6.7575e-04
Timing(20)
ans = 0.1078
ans = 0.0808
ans = 0.1195
function Timing(n)
timeit(@() methodDEC2BIN(n))
timeit(@() methodCOMBINATIONS(n))
timeit(@() methodNDGRID(n))
end
function methodDEC2BIN(n)
t = dec2bin(0:2^n - 1) - '0';
end
function methodCOMBINATIONS(n)
v = repmat({[0 1]},1,n);
t = combinations(v{:});
end
function methodNDGRID(n)
[t{n:-1:1}]=ndgrid([0,1]);
t=reshape( cat(n+1, t{:}) , [],n);
end

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更多回答（1 个）

Matt J
Matt J 2025-5-31
编辑：Matt J 2025-5-31
Ran in:
n=3;
clear t
[t{n:-1:1}]=ndgrid([0,1]);
t=reshape( cat(n+1, t{:}) , [],n)
t = 8×3
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
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