如何在游戏（列表）中找到元素的位置？

Question

In this game, there is only one element at the position (0, 0) represented by 'o'. The rest are occupied. Write a function empty(game) that takes in a game and gives the list of the positions of all empty spaces. Each position is specified by an (i,j) . For example,

game =  [['o', 'x', 'x'],
         ['x', 'x', 'x'],
         ['x', 'x', 'x']]

empty(game) gives [(0,0)]

def empty_spaces(game):
    result = [0]
    num_of_rows = len(game)
    num_of_columns = len(game[0])
    for i in range(num_of_columns):
        for j in range(num_of_rows):
            if not "o":
                result += i[j]
    return result

However, the result I get is [0] . I would appreciate some help and thank you very much!

Answer 1

You have iterate over the game list and see if you find 'o' if there's a hit then append that elements co-ordinates to result . The errors in your I would like to point out are if not "o" is always true, result=[0] you wouldn't want to do it. Instead simply write result= [] or result= list() . result+= i[j] i in your code is not iterable it is a integer. Since we checking if every element is equal to "o" the time complexity is O(rows*columns) .

game =  [['o', 'o', 'x'],
         ['x', 'x', 'x'],
         ['x', 'x', 'o']]
result=[]
for i in range(len(game)):
    for j in range(len(game[i])):
        if game[i][j] == 'o':
            result.append((i,j))
print(result)

>>> [(0, 0), (0, 1), (2, 2)]

Answer 2

Something like this should work:

game =  [['x', 'x', 'o'],
         ['x', 'o', 'x'],
         ['x', 'x', 'o']]

def empty_spaces(board):
    result = []

    for r_index, row in enumerate(board):
        for c_index, column in enumerate(row):
            if 'o' in column:
                result.append((r_index, c_index))

    return result

print(empty_spaces)

>>> [(0, 2), (1, 1), (2, 2)]

Answer 3

The problem is with the statement if not 'o': .

For us humans it's obvious, within the context of the program, that we want to check whether or not the element at index (i, j) is the string 'o' .

Unfortunately, the nature and limitations of Python mean that we must be explicit and specific: if game[i][j] == 'o': .

This particular mistake is quite annoying since, despite the fact that part of the expression is missing, the code runs perfectly fine, because of something called truthyness. Truthyness (and its counterpart, falsyness), is the idea that something which isn't a boolean can be converted to/evaluated as one. In the case of Python strings, empty strings are falsy (they evaluate to False) nonempty strings are truthy (you get it). The if statement in your code is therefore evaluated like so: if not 'o': —> if not True: —> if False: . As you can tell, the contents of that if statement would never be executed.

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Here is how I would rewrite your code:

game =  [['o', 'x', 'x'],
         ['x', 'x', 'x'],
         ['x', 'o', 'x']]


def empty_indexes(board):
    index_list = []
    for row_idx, row in enumerate(board):
        for col_idx, elem in enumerate(row):
            if elem == 'o': 
                index_list.append((row_idx, col_idx))
    return index_list

enumerate() is a simple and extremely useful function which returns pairs composed of one of the elements in the input, and a counter, which is incremented each time. When used on a list, like in the code above, it returns what corresponds to (index, element) pairs.