当从一个变量中减去前一个变量的新值时，如何中断一段时间？

Question

当从一个变量中减去前一个变量的新值时，如何中断一段时间

代码是：

Calc = (PTO2.h - coefficients[1]) / coefficients[0]
while True:
    NewP = IAPWS97(P=PH5, s=Calc)
    Calc = (NewP.h - coefficients[1]) / coefficients[0]
    print(NewP.h)

结果如下：

3181.2423174700475, 3125.5329929699737, 3145.170908432667, 
3138.216970209225, 3140.675480138904, 3139.805801319479,
3140.1133819014494, 3140.0045917261796, 3140.043069467109, 
3140.029460245017, 3140.034273686281, 3140.032571219946,
3140.033173365131

这个想法是在值不再增加时停止，即3140，它应该是最终值。 这个问题可以通过5到6次迭代来解决

Answer 1

请检查这是否符合您的要求：

# Define required difference
difference = 0.1
solutions = []

# Add 1st solution or guess
# func is whatever function you are using, with the required arguments, arg.
solutions.append(func(arg))
delta = soltutions[-1]

# Check if delta is still positive and if the delta meets the requirements
while delta > 0 and delta > difference:
   solutions.append(func(arg))
   delta = solutions[-2] - solutions[-1]

print(f'Final result is: {solutions[-1]}')

此解决方案假定，如果变量开始变为负值，您希望完成执行。如果符号不重要，可以在delta中使用abs()