taichi1.4.1里有双线性插值和三线性插值函数吗？还是说要自己定义这种函数。

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#1

taichi1.4.1里有双线性插值和三线性插值函数吗？还是说要自己定义这种函数。

#2

我好像在文档里没找到 :face_with_head_bandage:

#3 
@ti.func
def bilinear_interpolate(val:ti.template(), x: ti.f32, y: ti.f32)->ti.f32:
    '''
    Bilinear interpolation of a 2D scalar field

    Args:
        val (ti.template()): 2D scalar field
        x (ti.f32): x coordinate of the point
        y (ti.f32): y coordinate of the point

    Returns:
        ti.f32: interpolated value
    '''
    shape = val.shape
    i = int(x)
    j = int(y)
    if i < 0 or i >= shape[0]-1 or j < 0 or j >= shape[1]-1:
        return 0.0
    s = x - i
    t = y - j
    return (1-s)*(1-t)*val[i, j] + s*(1-t)*val[i+1, j] + (1-s)*t*val[i, j+1] + s*t*val[i+1, j+1]

@ti.func
def trilinear_interpolate(val:ti.template(), x: ti.f32, y: ti.f32, z: ti.f32)->ti.f32:
    '''
    Trilinear interpolation of a 3D scalar field

    Args:
        val (ti.template()): 3D scalar field
        x (ti.f32): x coordinate of the point
        y (ti.f32): y coordinate of the point
        z (ti.f32): z coordinate of the point

    Returns:
        ti.f32: interpolated value
    '''
    shape = val.shape
    i = int(x)
    j = int(y)
    k = int(z)
    if i < 0 or i >= shape[0]-1 or j < 0 or j >= shape[1]-1 or k < 0 or k >= shape[2]-1:
        return 0.0
    s = x - i
    t = y - j
    u = z - k
    return (1-s)*(1-t)*(1-u)*val[i, j, k] + s*(1-t)*(1-u)*val[i+1, j, k] + (1-s)*t*(1-u)*val[i, j+1, k] + s*t*(1-u)*val[i+1, j+1, k] + (1-s)*(1-t)*u*val[i, j, k+1] + s*(1-t)*u*val[i+1, j, k+1] + (1-s)*t*u*val[i, j+1, k+1] + s*t*u*val[i+1, j+1, k+1]

试试这个？
有BUG跟我说哈。

3 Likes
#4

谢谢啦，代码能跑 :handshake:

#5

我们要不要把这个加到ti.math库里面？