Question

我写了下面的代码，先创建一个向量列表，然后创建一个不带零的向量字典。

class SparseVec:     

    def __init__(self, n):
        self.val = [0] * n
        self.len = len(self.val)

    def __len__(self):
        return self.len

    def __getitem__(self,*i):
        return self.val

    def __setitem__(self, i, n):
        self.val[i] = n
        self.len = max(i, self.len)

         print 'self.len = ', self.len
    def nonzeros(self):
        nonzeroDict = {}
        for i in range(len(self.val)):
            if self.val[i] != 0:
                nonzeroDict[i] = self.val[i]
        return 'Sparse Vector{}'.format(nonzeroDict)
    def __add__(self, other):     
        length = self.len
        result = SparseVec(length)
        if self.len < len(other.val):
            length = len(other.val)
            result = SparseVec(length)        
        for i in range(self.len):
            result[i] = self.val[i] + other.val[i]
        for j in other.val:
            if j not in self.val:
                result[j] = other.val[j]
        return result 

    def __str__(self):
        return '{}'.format(self.val)

if __name__ == '__main__':

      a = SparseVec(5)
      a[2] = 9.2
      a[0] = -1
      a[3] = 0
      print a
      print a.nonzeros()

      b = SparseVec(5)
      b[1] = 1
      print b
      print b.nonzeros()

      c = a+b
      print c
      print c.nonzeros()

我如何重写__str__方法以得到类似这样的结果？

print a
[0] = 2  [1] = 5  etc...

我也想要这个：

for ai, i in a:
    print 'a[%d]=%g' % (i, ai)

提供与上面类似的结果，并在列表索引前面加上a：

a[0] = 2 a[1] = 5

Answer 1

>>> val = [2, 5, 7, 9]
>>> '  '.join('[{}] = {}'.format(index, elem)
...           for index, elem in enumerate(val))
'[0] = 2  [1] = 5  [2] = 7  [3] = 9'

在每个元素前面都带有一个“ a”：

>>> '  '.join('a[{}] = {}'.format(index, elem)
...           for index, elem in enumerate(val))
'a[0] = 2  a[1] = 5  a[2] = 7  a[3] = 9'

打印班级的列举清单

1 个答案: