Question

我制作一个简单的python脚本来处理表格。我正在使用数组来存储单元格值。下面是代码：

table =[]
hlen = input("Please enter the amount of columns \n")
vlen = input("Please enter the amount of rows \n")
curcol = 1
currow = 1
totcell = hlen*vlen
while (totcell >= curcol * currow):
   str = input("Please input "+ str(curcol) +"," + str(currow))
   table.append(str)
   if (curcol >= hlen):
       currow =+ 1

//Process Table

程序运行甜蜜，要求第一个单元格为1,1。一切都很好，直到代码在重新加载时停止。 Heres pythons错误输出

Traceback (most recent call last):
  File "./Evacuation.py", line 13, in <module>
    str = input("Please input "+ str(curcol) +"," + str(currow))
TypeError: 'int' object is not callable

感谢您的帮助。

Answer 1

您使用变量名隐藏了内置str：

str = input("Please input "+ str(curcol) +"," + str(currow))

第二次，str(currow)您尝试调用 str，现在是int

将其命名为以外的任何内容！

另外，您使用的是Python 2，因此最好使用raw_input代替input

Answer 2

  m = input("Please input "+ str(curcol) +"," + str(currow))
  please use different name of variable not use 'str' because it is python default function for type  casting

  table =[]
  hlen = input("Please enter the amount of columns \n")
  vlen = input("Please enter the amount of rows \n")
  curcol = 1
  currow = 1
  totcell = hlen*vlen
  while (totcell >= curcol * currow):
       m = input("Please input "+ str(curcol) +"," + str(currow))
  table.append(m)
  if (curcol >= hlen):
    currow += 1



   Please enter the amount of columns 
   5
  Please enter the amount of rows 
   1
  Please input 1,11
  Please input 1,11
  Please input 1,1
  >>> ================================ RESTART ================================
  >>> 
  >>>Please enter the amount of columns 
  >>>1
  Please enter the amount of rows 
  1
 Please input 1,12


 see this program and run it .

Answer 3

您正在使用str作为您从输入中返回的int的变量名称。 Python指的是当你使用str（curcol）和str（currow）时，而不是Python字符串函数。

'int'对象不可调用

3 个答案: