python之检查元素是否在列表中

我有一个矩阵或列表列表:

x = [[1,2,3],  
     [2,3,1],  
     [3,1,2]] 

我的目标是检查是否

1)矩阵的每一列包含，从1到n的每个整数恰好出现一次。

2) 矩阵的每一行包含从 1 到 n 的每个整数恰好一次。

这是我在解决Udacity的intro to programming类(class)时遇到的问题练习。这是我解决问题的方法。我知道这是漫长而低效的。那么解决这个问题的简短有效方法是什么？？

def check(p): 
    j = 0 
    for e in p: 
        i = 1 + j 
    s = str(p) 
 
    if s.find('.')!= -1: 
        return False 
 
    while i < len(p): 
        if p[i] == e: 
            return False 
        if p[i] > len(p) or p[i] < 1: 
            return False 
        i += 1 
        j += 1 
    return True 
 
def check_sudoku(p): 
    i = 0 
    z = [] 
    a = [] 
    x = 0 
    for e in p: 
        r = check(e) 
        if r == False: 
            return r 
 
    #Below here is to transpose the list 
    while x < len(p): 
        z.append(1) 
        x += 1 
    while i < len(p): 
        for e in p: 
            a.append(e.pop()) 
        z[i] = a 
        i +=  1 
        a = [] 
 
    #Below here is to check the transpose 
    for g in z: 
        r = check(g) 
        if r == False: 
            return r 
    return True 

请您参考如下方法:

您可以像这样简化检查程序:

def check_valid(matrix): 
    n = len(matrix[0]) 
    valid_num = range(1, n+1)     # make a list of valid number from 1 to n 
 
    # sort rows and column of matrix in ascending order and compare with valid_num  
    for line in matrix: 
        if sorted(line) != valid_num: 
            return False 
    for column in zip(*matrix): 
        if sorted(column) != valid_num: 
            return False 
 
    # If all rows and column is valid, then  
    return True  

编辑:遵循@vocalno KISS 规则但速度更快并且适用于数字和字符

def check_valid2(matrix): 
    n = len(matrix) 
 
    for line in matrix+zip(*matrix): 
        if len(set(line)) != n: 
            return False 
 
    return True 
 
In[9]: %timeit for m in (x, y): check_valid_mine(m) 
100000 loops, best of 3: 8.55 µs per loop 
 
In[10]: %timeit for m in [x,y]: check_valid2(m) 
100000 loops, best of 3: 5.77 µs per loop 

编辑 2:关于速度

def check_valid3(matrix): 
    n = len(matrix) 
 
    for line in matrix+zip(*matrix): 
        if not len(set(line)) - n: 
            return False 
 
    return True 
 
In[19]: %timeit for m in [x,y]: check_valid3(m) 
100000 loops, best of 3: 2.29 µs per loop