Python代码效率

Question

该程序测试矩阵是否是单位矩阵.

我已将代码粘贴在下面,并想知道如何优化此代码的效率.另外我是python编程的新手,是否有一些内置函数可以解决这个问题呢？

    def is_identity_matrix(test):
    if (test == []):
        return False
    i = 0
    while (i < len(test)):
        if (len(test[i]) == len(test)):
            j = 0
            while(j < len(test[i])):
                if (j != i):
                    if(test[i][j] != 0):
                        return False
                else:
                    if(test[i][j] != 1):
                        return False
                if(j == (len(test[i]) - 1)):
                    break
                j += 1
            if(i == (len(test) - 1)):
                break
            i += 1
        else:
            return False
    if(i == j and i == (len(test) - 1)):
        return True

# Test Cases:

matrix1 = [[1,0,0,0],
           [0,1,0,0],
           [0,0,1,0],
           [0,0,0,1]]
print is_identity_matrix(matrix1)
#>>>True

matrix2 = [[1,0,0],
           [0,1,0],
           [0,0,0]]

print is_identity_matrix(matrix2)
#>>>False

matrix3 = [[2,0,0],
           [0,2,0],
           [0,0,2]]

print is_identity_matrix(matrix3)
#>>>False

matrix4 = [[1,0,0,0],
           [0,1,1,0],
           [0,0,0,1]]

print is_identity_matrix(matrix4)
#>>>False

matrix5 = [[1,0,0,0,0,0,0,0,0]]

print is_identity_matrix(matrix5)
#>>>False

matrix6 = [[1,0,0,0],  
           [0,1,0,2],  
           [0,0,1,0],  
           [0,0,0,1]]

print is_identity_matrix(matrix6)
#>>>False

Answer 1

def is_identity_matrix(listoflist):
    return all(val == (x == y) 
        for y, row in enumerate(listoflist)  
            for x, val in enumerate(row))

(但是,这不检查矩阵是否为正方形,并且对于空列表返回True)

说明:在内部,all我们有一个带有嵌套循环的生成器表达式,其中val循环遍历矩阵中的每个值.x == y评估True对角线和False其他地方.在Python,True == 1和False == 0,这样你可以比较val == (x == y).括号很重要: val == x == y将是一个相当于的链式比较val == x and x == y

Answer 2

我用的是numpy:

(np.array(matrix1) == np.identity(len(matrix1))).all()

当然,如果你首先存储matrix1为一个numpy数组以避免转换,那会更好.