创建一个修改了一个元素的新元组

def item(i, v):
  if i != 1: return v
  return strangestuff(v)

for row in rows:
  t = tuple(item(i, c.InnerText)
            for i, c in enumerate(row.Descendants[TableCell]())
           )

我会这样做:

temp_list = [c.InnerText for c in row.Descendants[TableCell]()]
temp_list[2] = "Something different"
t = tuple(temp_list)

它的工作原理如下:

>>> temp_list = [i for i in range(4)]
>>> temp_list[2] = "Something different"
>>> t = tuple(temp_list)
>>> t
(0, 1, 'Something different', 3)

for row in rows:
    t1, t2, t3, t4 = tuple([c.InnerText for c in row.Descendants[TableCell]()])
    t1 = ...
    t = (t1, t2, t3, t4)

l=列表(旧元组) l[2]=foo t=元组(l)

newtuple=update(oldtuple,(无,无,val,无))

或许正确的结构是拉链

newtuple=update(oldtuple,((2,val),(3,val)))