筛选所有组合以匹配前两个选择。从剩余的组合中获取所有已使用的选择。筛选z下拉选项，使其不包括该组合的任何已使用的选择。

// Dropdown options
let x = ['Juice', 'Milkshake'];
let y = ['Medium Sugar', 'Regular'];
let z = ['Banana', 'Almond', 'Papaya'];

// Previously selected options
let rows = [
    {x: 'Juice', y: 'Medium Sugar', z: 'Banana'},
    {x: 'Juice', y: 'Medium Sugar', z: 'Papaya'},
    // More selections...
]

// Selected from dropdown
let selectedX = 'Juice';
let selectedY = 'Medium Sugar'

// Get already used z selections for the previous two selections
let usedZs = rows.filter(row => row.x === selectedX && row.y === selectedY).map(row => row.z);
// ['Banana', 'Papaya']

// Remove used options from z
let newZ = z.filter(option => ! usedZs.includes(option));
// ['Almond']
console.log(newZ)

我猜您可能希望对“y”选择做同样的事情，或者处理选择顺序不对的情况，这会稍微复杂一些。