我正在编写使用循环(目前是for循环)的代码，利用Heron的方法，找到平方根的估计值，同时还显示迭代和相对变化。
到目前为止我写的是:
# Problem 1.
def square_root_for(a, x0, max_iter = 10, tol=1e-14):
""" (number, integer, number) -> float
Return an estimate of the square root of a number using the Heron's method.
>>> square_root_for(5, 5)
Iteration | Estimate | Relative Change
-------------------------------------------------
1 | 3.00000000000000 | 0.4000000000000000
2 | 2.33333333333333 | 0.2222222222222222
3 | 2.23809523809524 | 0.0408163265306123
4 | 2.23606889564336 | 0.0009053870529653
5 | 2.23606797749998 | 0.0000004106060359
6 | 2.23606797749979 | 0.0000000000000842
7 | 2.23606797749979 | 0.0000000000000000
2.23606797749979
"""
x = [x0]
x.append(a/x0)
print('Iteration | Estimate | Relative Change')
print('-------------------------------------------------')
for i in list(range(1,( max_iter + 1),1)):
#x[i+1] = (s1 + x[i]) / 2
change = (abs(x[i] - x[i-1]) / x[i-1])
if change > tol:
x.append(1/2 * (x[i] + (a / x[i])))
print('{} | {:.14f} | {:.16f}'.format(i, x[i], change))
i = i + 1
# Don't change or delete the 5 lines of code below.
a = 5
max_iter = 100
tol = 1e-15
x_final = square_root_for(a, a, max_iter, tol)
print('Final estimate using square_root_for is {0}'.format(x_final))
它大部分工作，但我的例子显示为:
Iteration | Estimate | Relative Change
-------------------------------------------------
1 | 1.00000000000000 | 0.8000000000000000
2 | 3.00000000000000 | 2.0000000000000000
3 | 2.33333333333333 | 0.2222222222222222
4 | 2.23809523809524 | 0.0408163265306123
5 | 2.23606889564336 | 0.0009053870529653
6 | 2.23606797749998 | 0.0000004106060359
7 | 2.23606797749979 | 0.0000000000000842
8 | 2.23606797749979 | 0.0000000000000000
这是不正确的。第一行的评估和相对变更不应该在那里，而应该是第二行的评估和相对变更，对应于迭代1。也应该只有7次迭代，但由于这个问题，有8次。
代码返回以下错误:
Traceback (most recent call last):
File "C:UsersagsmiDesktopst114homework6.py", line 61, in <module>
x_final = square_root_for(a, a, max_iter, tol)
File "C:UsersagsmiDesktopst114homework6.py", line 50, in square_root_for
change = (abs(x[i] - x[i-1]) / x[i-1])
IndexError: list index out of range
总的来说，我的问题是，我如何解决这个问题，让我的第一次迭代返回第二次返回的内容，这样就只有7次迭代而不是8次。
谢谢!
将list()从for循环中移除，然后进行测试。
for i in range(1,( max_iter + 1),1):
同时删除for循环中的i = i+1。
并修改if语句:
if change > tol:
x.append(1/2 * (x[i] + (a / x[i])))
print('{} | {:.14f} | {:.16f}'.format(i, x[i], change))
else:
break
# Problem 1.
def square_root_for(a, x0, max_iter = 10, tol=1e-14):
""" (number, integer, number) -> float
Return an estimate of the square root of a number using the Heron's method.
>>> square_root_for(5, 5)
Iteration | Estimate | Relative Change
-------------------------------------------------
1 | 3.00000000000000 | 0.4000000000000000
2 | 2.33333333333333 | 0.2222222222222222
3 | 2.23809523809524 | 0.0408163265306123
4 | 2.23606889564336 | 0.0009053870529653
5 | 2.23606797749998 | 0.0000004106060359
6 | 2.23606797749979 | 0.0000000000000842
7 | 2.23606797749979 | 0.0000000000000000
2.23606797749979
"""
x = [x0]
x.append(a/x0)
print('Iteration | Estimate | Relative Change')
print('-------------------------------------------------')
for i in range(1,max_iter):
#x[i+1] = (s1 + x[i]) / 2
change = (abs(x[i] - x[i-1]) / x[i-1])
if change > tol:
x.append(1/2 * (x[i] + (a / x[i])))
else:
break
print('{} | {:.14f} | {:.16f}'.format(i, x[i], change))
return(x[i])
# Don't change or delete the 5 lines of code below.
a = 5
max_iter = 100
tol = 1e-15
x_final = square_root_for(a, a, max_iter, tol)
print('Final estimate using square_root_for is {0}'.format(x_final))
我取出了I = I +1行，简化了AAR提到的范围语句，但是您得到的错误是因为它正在循环并期望找到一个数组值，所以当达到所需的精度时，您需要一个中断来逃离for循环。编辑-您还需要从函数返回值x[i]。
输出Iteration | Estimate | Relative Change
-------------------------------------------------
1 | 1.00000000000000 | 0.8000000000000000
2 | 3.00000000000000 | 2.0000000000000000
3 | 2.33333333333333 | 0.2222222222222222
4 | 2.23809523809524 | 0.0408163265306123
5 | 2.23606889564336 | 0.0009053870529653
6 | 2.23606797749998 | 0.0000004106060359
7 | 2.23606797749979 | 0.0000000000000842
Final estimate using square_root_for is 2.23606797749979