计算 python 中基向量和的所有排列

Question

I'm trying to break down a vector,

[a,b,c]

into [a,0,0], [0,b,0] and [0,0,c] and then calculate all possible sums that can be created with these vectors.

For example, it should return

\[a,0,0\],
\[0,b,0\]
\[0,0,c\]
\[a,b,0\]
\[a,0,c\]
\[0,b,c\]
\[a,b,c\]

but for any length vector.

Ive tried itertool permutations, but it doesnt seem to work for this problem. Any ideas?

Answer 1

from itertools import combinations

# define the vector
vec = [a,b,c]

# create a list of vectors where each element of vec is set to zero
# except for one element, which is set to the corresponding value from vec
zero_vecs = [[int(i == j) * val for j, val in enumerate(vec)] for i in range(len(vec))]

# create all possible combinations of the zero_vecs list
combos = combinations(zero_vecs, r=len(vec))

# create a list of all possible sums by adding the vectors in each combination
sums = [sum(x) for x in combos]

# print the results
print("Zero vectors:")
for v in zero_vecs:
    print(v)
print("\nAll possible sums:")
for s in sums:
    print(s)

Answer 2

Here's a very simple way to do this using binary logic:

inp = [1, 2, 3]

result = []

for i in range(1, 2**len(inp)):
    elem = []
    for j in range(len(inp)):
        elem.append(inp[j] if ((1 << j) & i) > 0 else 0)
    result.append(elem)

print(result)

The code iterates through all of the integers between 1 and 2**<vector length>-1, treating each integer as a binary mask.

You can do the same thing in a single line of code with nested list comprehensions:

result = [[inp[j] if ((1 << j) & i) > 0 else 0 for j in range(len(inp))] for i in range(1, 2**len(inp))]

In either case, the result is:

[[1, 0, 0], [0, 2, 0], [1, 2, 0], [0, 0, 3], [1, 0, 3], [0, 2, 3], [1, 2, 3]]