Python：检查一个数字是否已经在列表列表的网格中

Question

In Python, I'm trying to create a duplicate checker that can check if the number that I'm entering is already in the grid. If not present, it should return False . So for example, numbers 1, 8, 9 should return False and others should return True . Can someone help me?

grid = [[0, 2, 3],
        [4, 0, 5],
        [6, 7, 0]]


number_rows = len(grid)
def possible(n):
    global grid
    if number_rows == n:
        return False
    return True
    

possible(1)  # ---> True

Answer 1

If the grid is a list of lists you can use a for/else block :

def possible(n, grid):
    for x in grid:
        if n in x:
            return False
    else:
        return True

grid = [[0, 2, 3],
        [4, 0, 5],
        [6, 7, 0]]

print(possible(1, grid))  # ---> True
print(possible(2, grid))  # ---> False

Answer 2

Without using packages such as NumPy, the easiest way I can think of is to use Python's any with list comprehension. Consider something like:

def possible (number, grid):
    check = [number in sublist for sublist in grid]

    return not any(check)

In the function, check will loop through the sublists and return a list of Trues and Falses depending on if number exist in each sublist. For instance, if you pass 2 as the number , check will be [True, False, False] because 2 only exist in the first sublist.

Then, the any function, like the name suggests, if there are any True s within check . Finally, the not keyword inverts it, since you want the function to return True if there are not already in the "grid".

Answer 3

number_rows = len(grid) only counts the number of lists in the main list grid[a,b,c] . In your code, you're testing if the number you input is equal to the length of the main list. Use the n in list to test if it's true or false.

def possible(n):
    temp = [a for b in grid for a in b] ##put all the values from the grid into a single layer list
    if n in temp: 
        return False
    return True