索引错误:无法打印cost_function

Question

我试图运行一个for循环，通过定义一个函数来打印三个不同斜率和偏差=0的代价函数。数据集有5行，成本函数是根据考勤来预测分数。如果为每个斜率值定义三个独立的函数，我就可以打印成本函数。这是我的代码：

dataset = {"Attendance":[100, 87, 15, 63, 47], "Marks": [100, 95, 6, 73, 50]}
Marks = pd.DataFrame(dataset, columns = ["Attendance", "Marks"])
bias = 0
slope = {"values": [-1, 0, 3]}
slope = pd.DataFrame(slope)
def error():
  a = []
  sum_of_squared_error = 0
  for i in range(len(slope)):
    for j in range(0, len(Marks)):
      x = Marks.iloc[j, 0]
      y = Marks.iloc[j, 1]
      sum_of_squared_error += (y - (slope.iloc[0, i]*x + bias)) ** 2
      cost_function = sum_of_squared_error / (2 * len(Marks))
    a.append(cost_function)
  return a
error()

我得到了这个错误。

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-126-1bf54ede9ae1> in <module>()
     13       a.append(cost_function)
     14   return a
---> 15 error()

5 frames
/usr/local/lib/python3.6/dist-packages/pandas/core/indexing.py in _validate_integer(self, key, axis)
   2061         len_axis = len(self.obj._get_axis(axis))
   2062         if key >= len_axis or key < -len_axis:
-> 2063             raise IndexError("single positional indexer is out-of-bounds")
   2064 
   2065     def _getitem_tuple(self, tup: Tuple):

IndexError: single positional indexer is out-of-bounds

Answer 1

您没有正确地访问斜率df的元素。slope.shape返回(3, 1)，以便迭代行号，而不是列号。

sum_of_squared_error += (y - (slope.iloc[0, i]*x + bias)) ** 2应该是：sum_of_squared_error += (y - (slope.iloc[i, 0]*x + bias)) ** 2

此外，应该在内部循环和外部循环之间将sum_of_squared_error重置为0：

import pandas as pd

dataset = {"Attendance":[100, 87, 15, 63, 47], "Marks": [100, 95, 6, 73, 50]}
Marks = pd.DataFrame(dataset, columns = ["Attendance", "Marks"])
bias = 0
slope = {"values": [-1, 0, 3]}
slope = pd.DataFrame(slope)
def error():
  a = []
  sum_of_squared_error = 0
  for i in range(len(slope)):
    for j in range(0, len(Marks)):
      x = Marks.iloc[j, 0]
      y = Marks.iloc[j, 1]
      sum_of_squared_error += (y - (slope.iloc[i, 0]*x + bias)) ** 2
    cost_function = sum_of_squared_error / (2 * len(Marks))
    sum_of_squared_error = 0
    a.append(cost_function)
  return a
error()

输出：

>>> error()
[10147.0, 2689.0, 9081.4]

Answer 2

错误出现在网上-

    sum_of_squared_error += (y - (slope.iloc[0, i]*x + bias)) ** 2

之所以会发生这种情况，是因为iloc只能在轴的length-1之前接受值。

参考文献- https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.iloc.html

在你的例子0中，我取了以下值-

• 0,0 - OK
• 0,1 -不确定
• 0，2-不确定

因为，在slope中没有列1和2。