Question

def set(seq, row, col, val):
    if seq[row-1][col-1] != val:
        x = seq.remove(seq[row-1]) #[[32,33]]
        y = seq.pop(row-1) #[50,92]
        z = y.remove(row-1) #[50]  It is wrong from here onwards
        a = y.insert(row-1,val) #[50,50]
        b = x.append(a) # [[32,33],[50,50]]
        return b

鉴于

test_case_100 = make_matrix([[32, 33], [50, 92]]) # [[32, 33], [50, 92]]

然后set函数应该改变矩阵如下

set(test_case_100, 2, 2, 50)  # [[32, 33], [50, 50]]
set(test_case_100, 2, 1, 30) #[[32, 33], [30, 50]]
set(test_case_100, 1, 2, 29) #[[32, 29], [30, 50]]
set(test_case_100, 1, 1, -20) #[[-20, 29], [30, 50]]
print(set(test_case_100, 2, 2, 50))  # [[-20, 29], [30, 50]]

Answer 1

正如@thefourtheye在评论中所指出的那样，你使这个过得更加艰难。

def set_matrix(matrix, row, col, value):
    try:
        matrix[row-1][col-1] = value
    except IndexError:
        raise IndexError("No value at x:{}, y:{}".format(row,col))

这可能是将OOP引入编码风格的好时机。

class Matrix(list):
    def __init__(self, *args):
        for arg in args:
            self.append(arg)
    def setval(self,row,col,value):
        try:
            self[row-1][col-1] = value
        except IndexError:
            raise IndexError("No value at x:{}, y:{}".format(row,col))

test_matrix = Matrix([32, 33], [50, 92])
test_matrix.setval(2,2,50)
print(test_matrix) # [[32, 33], [50, 50]]

你在这里使用类似新课程的原因是你可以做一些很酷的事情：

class Matrix(list):
    # code goes here
    def __str__(self):
        COLSIZE = max(map(len,(val for col in self for val in col)))+1
        NUMCOLS = len(self[0]) # all rows equal size I hope!
        formatstring = "{}{}{}".format("{:>",COLSIZE,"}")*NUMCOLS
        return "\n".join(formatstring.format(*row) for row in self)

>>> print(Matrix([1,2],[3,4],[5,6]))
1 2
3 4
5 6
>>> print(Matrix([226,1],[330,1000],[15,17]))
226 1
330 1000
15  17

更改嵌套列表中的整数

1 个答案: