Find the coordinates of a cuboid using list comprehension in Python

Question

X, Y and Z are the three coordinates of a cuboid.

Now X=1,Y=1, Z=1 and N=2.

I have to generate a list of all possible coordinates on a 3D grid where the sum of Xi + Yi + Zi is not equal to N. If X=2, the possible values of Xi can be 0, 1 and 2. The same applies to Y and Z.

I have written this below code so far, and it is giving the output as :

[[0, 0, 0]]

however the expected output is

[[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]]

Below is my code, what is going wrong in my code?

[[x,y,z] for x in range(X) for y in range(Y) for z in range(Z) if x+y+z != N]

Solution 1

range is actually a half-closed function. So, the ending value will not be included in the resulting range.

If X=2, the possible values of Xi can be 0, 1 and 2

In your code, range(X) will give only 0 and 1, if X is 2. You should have used range(X + 1).

X, Y, Z, N = 1, 1, 1, 2
[[x,y,z] for x in range(X + 1) for y in range(Y + 1) for z in range(Z + 1) if x+y+z != N]

[[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]]

You can write the same, with itertools.product, like this

X, Y, Z, N = 1, 1, 1, 2
from itertools import product
[list(i) for i in product(range(X + 1), range(Y + 1), range(Z + 1)) if sum(i) != N]

[[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]]

Solution 2

Another approch with itertools.product and list comprehension:

In [91]: [list(l) for l in it.product([0,1], repeat=3) if sum(l) != 2]
Out[91]: [[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]]

Solution 3

try this:

    x = int(input())
    y = int(input())
    z = int(input())
    n = int(input())
    ans=[[i,j,k] for i in range(x+1) for j in range(y+1) for k in range(z+1) if i+j+k !=n]
    print(ans)

as range function doesn't include the ending value and stopped at (n-1)th position so if we use range(X) here it will result just 0 hence +1 should be used to print the list in lexicographic increasing order.

Solution 4

x = 1
y = 1
z = 2
n = 3

arr_3_D = []

for i in range(x+1):
    for j in range(y+1):
        for k in range(z+1):
            if i+j+k != n:
                arr = [i,j,k]
                arr_3_D.append(arr)


print(arr_3_D)

output: [[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 2]]

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Comments

[M4rtini]

Try to print range(1) and you may see the problem

[LordWilmore]

Please add explanation around your answer.

[Tasnuva Leeya]

as range function doesn't include the ending value and stopped at (n-1)th position so if we use range(X) here it will result just 0 hence +1 should be used to print the list in lexicographic increasing order. @LordWilmore

[MyNameIsCaleb]

The provided answer was flagged for review as a Low Quality Post. Here are some guidelines for How do I write a good answer?. This provided answer may be correct, but it could benefit from an explanation. Code only answers are not considered "good" answers. From review.

[Brian Tompsett - 汤莱恩]

While this code snippet may solve the problem, it doesn't explain why or how it answers the question. Please include an explanation for your code, as that really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. You can use the edit button to improve this answer to get more votes and reputation!

[Sillians]

Nice approach, well done. How can you approach this using a list comprehension rather than multiple loops.