python索引器：超出范围？

所以我使用了这段代码，显然索引超出了范围？我不明白为什么，但它给了我错误

Grid = [[0,0,0,0],[0,0,0,0]]

def SpawnNumber(Grid):
    # Check is true when the random grid space is a 0
    check = False
    # Loop until an empty space is chosen
    while check == False:
        rn1 = randint(0,3)
        rn2 = randint(0,3)
        print(rn1)
        print(rn2)
        # If the space is empty, fill it
        if Grid[rn1][rn2] == 0:
            Grid[rn1][rn2] = 2
            check = True
    return(Grid)

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网格应该有两个维度，每个维度从0到3，那么为什么randint（0,3）超出范围？

网格只有两个元素，索引为0和1。您可以通过计算len（网格）进行快速检查。也就是说，rn1只能有值0和1。

在这种情况下，您发送的“网格”在您期望的索引中没有项目

要使函数更通用，并消化传入的各种网格，可以执行以下操作：

def SpawnNumber(Grid):
    # Check is true when the random grid space is a 0
    check = False
    # Loop until an empty space is chosen
    while check == False:
        rn1 = randint(0,len(Grid)-1)
        rn2 = randint(0,len(Grid[rn1])-1)
        print(rn1)
        print(rn2)
        # If the space is empty, fill it
        if Grid[rn1][rn2] == 0:
            Grid[rn1][rn2] = 2
            check = True
    return(Grid)

def SpawnNumber(Grid):
    # Check is true when the random grid space is a 0
    check = False
    # Check if there are any open spots
    for row in Grid:
        if all(row):
            check = True
    # Loop until an empty space is chosen
    while check == False:
        rn1 = randint(0,len(Grid)-1)
        rn2 = randint(0,len(Grid[rn1])-1)
        print(rn1)
        print(rn2)
        # If the space is empty, fill it
        if Grid[rn1][rn2] == 0:
            Grid[rn1][rn2] = 2
            check = True
    return(Grid)

由于RandInt的包容性逻辑，-1在其中。根据文件：

random.randint(a, b)
    Return a random integer N such that a <= N <= b

---

rn1

只能从0到1。您只有两个子列表。简言之，只需更改代码，使rn1被约束为0或1，您就可以开始了。