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2 数学程序开胃菜
在上一章中( https://mp.weixin.qq.com/s/kKenXcEXIeLd_u_2kymF8A ),我们介绍了python的IDE;用numpy实现向量计算;用Matplotlib绘图;用sympy实现微积分和求导;用SciPy实现积分;用VPython实现弹跳球动画。
在本章中,您将了解 Python 命令式编程风格的线性程序结构以及分支和重复结构。面向对象编程和函数式编程的示例描述了使用 Python 编程的更多方法。
使用计算机解决的问题可以通过编程语言以多种方式建模和结构化。在应用计算机科学领域,命令式(过程式)编程、面向对象编程 (OOP) 和函数式编程风格已经确立。Python 支持所有这三种编程风格。
问题的解决方案也与逻辑条件相关联: 预期情况是否适用?此外,根据问题的不同,必须重复执行相同的任务,例如计算数学函数的值表。与其他过程式编程语言一样,Python 支持线性程序结构以及分支和重复结构。
2.1 线性程序结构
2.1.1 线性程序
- U=230
- R=11.8
- I=U/R
- b="The current is:"
- print(b, I, " A")
复制代码 执行:- The current is: 19.491525423728813 A
复制代码 python的整数是没有大小限制的,浮点数的限制如下:- >>> import sys
- >>> print(sys.float_info)
- sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsilon=2.220446049250313e-16, radix=2, rounds=1)
复制代码- #02_linear2.py
- U = 230
- R1, R2, R3=0.12, 0.52, 228
- Rg = R1 + R2 + R3
- I= U/Rg
- P1 = R1 * I**2
- P2 = R2 * I**2
- P3 = R3 * I**2
- print("Current I={0:6.3f} A " .format(I))
- print("P1={0:3.2f} W, P2={1:3.2f} W, P3={2:3.2f} W".format(P1,P2,P3))
复制代码 执行:- Current I= 1.006 A
- P1=0.12 W, P2=0.53 W, P3=230.72 W
复制代码- #02_linear2.py
- U = 230
- R1, R2, R3=0.12, 0.52, 228
- Rg = R1 + R2 + R3
- I= U/Rg
- P1 = R1 * I**2
- P2 = R2 * I**2
- P3 = R3 * I**2
- print("Current I={0:6.3f} A " .format(I))
- print("P1={0:3.2f} W, P2={1:3.2f} W, P3={2:3.2f} W".format(P1,P2,P3))
复制代码 执行:- ---Input---
- Voltage: 220
- Resistance: 34
- ---Output---
- Current 6.47 A
- Power 1423.53 W
- ---Input---
- Voltage:
复制代码 参考资料
2.2 函数
- 实例:使用函数来计算电流、电功率、电功和电能成本:
- #04_function1.py
- U, R = 230, 460
- t = 8
- price = 0.3
- def current():
- I = U / R
- print("Current: ", I, " A")
- def power():
- P = U**2 / R
- print("Power : ", P, " W")
- def work():
- P = U**2 /R
- W = P * t
- print("Work: ", W, " Wh")
- def cost():
- I = U / R
- W = U * I * t
- c = W * price / 1000.0
- print("Cost: ", c, " Euro")
- current()
- power()
- work()
- cost()
复制代码 执行:- Current: 0.5 A
- Power : 115.0 W
- Work: 920.0 Wh
- Cost: 0.276 Euro
复制代码 argument是调用函数时传递的值。该值被分配给函数中的指定参数。parameter是函数中使用的名称。
- #05_function2.py
- def current(U, R):
- return U / R
- def power(U, R):
- return U**2/R
- def work(U, R, t):
- P = U**2 / R
- W = P*t
- return W
- def cost(U, R, t, price):
- I = U / R
- W = U * I * t
- c =W * price / 1000.0
- return c
- Uq = 230 #V
- RLoad = 23 #ohms
- tn = 8 #h, hours
- price_actual = 0.3 #euro
- print("Current: ", current(Uq, RLoad), " A")
- print("Power : ", power(Uq, RLoad), " W")
- print("Work : ", work(Uq, RLoad,tn), " Wh")
- print("Cost : ", cost(Uq, RLoad,tn,price_actual), " euros")
复制代码
执行:- Current: 10.0 A
- Power : 2300.0 W
- Work : 18400.0 Wh
- Cost : 5.52 euros
复制代码
- 实例:带有多个返回值的函数:计算出体积、质量、惯性矩和加速力矩
与其他编程语言不同,Python 也允许返回多个值。我们来看一个直径为 1 分米、长度为 10 分米的实心钢圆柱体的例子。只需一个函数,就可以计算出体积、质量、惯性矩和加速力矩。所有四个值都应通过返回语句返回。
加速力矩 Mb 与角加速度 α 和惯性矩 J 成比例增加:
圆柱体的惯性矩 J 与质量m成正比,与半径 r 的平方成正比:
质量m由圆柱体的体积 V 和密度计算得出:
要计算体积 V,需要圆柱体的直径 d 和长度 l:
要完成这项任务,必须在 Python 开发环境的文本编辑器中按照语法规则以相反的顺序输入公式。- #06_function3.py
- rho = 7.85 #kg/dm^3, density for steel
- alpha = 1.2 #1/s^2, angular acceleration
- g = 3 #accuracy
- def cylinder(d,l):
- V=round(0.785*d**2*l,g)
- m=round(rho*V,g)
- J=round(0.5*m*(d/2/10)**2,g)
- Mb=round(alpha*J,g)
- return (V,m,J,Mb)
- #return V,m,J,Mb
- #return [V,m,J,Mb]
- d1=1 #dm
- l1=10 #dm
- T=cylinder(d1, l1)
- print("Cylinder data: ", T)
- print("Volume: ", T[0]," dm^3")
- print("Mass: ", T[1]," kg")
- print("Moment of inertia: ", T[2]," kgm^2")
- print("Acceleration torque:", T[3]," Nm")
复制代码 执行:- Cylinder data: (7.85, 61.622, 0.077, 0.092)
- Volume: 7.85 dm^3
- Mass: 61.622 kg
- Moment of inertia: 0.077 kgm^2
- Acceleration torque: 0.092 Nm
复制代码- #07_function4.py
- rho=7.85 #kg/dm^3, density of steel
- def volume(d,l):
- return 0.785*d**2*l
- def mass(d,l):
- return rho*volume(d,l)
- def moment_of_inertia(d,l):
- return 0.5*mass(d,l)*(d/2/10)**2
- def acceleration_torque(d,l,alpha):
- return alpha*moment_of_inertia(d,l)
- d1=1 #dm
- l1=10 #dm
- alpha1=1.2 #1/s^2, angular acceleration
- V=volume(d1,l1)
- m=mass(d1,l1)
- J=moment_of_inertia(d1,l1)
- Mb=acceleration_torque(d1,l1,alpha1)
- print("Volume: ", V, " dm^3")
- print("Mass: ", m, " kg")
- print("moment of inertia: ", J, " kgm^2")
- print("Acceleration torque: ", Mb, " Nm")
复制代码 执行:- Volume: 7.8500000000000005 dm^3
- Mass: 61.6225 kg
- moment of inertia: 0.07702812500000002 kgm^2
- Acceleration torque: 0.09243375000000002 Nm
复制代码 2.3 分支结构
根下的表达式也可以取负值。当出现这种情况时,方程将无法在实数空间内求解。因此,程序必须通过检查 D 是否≥ 0 来捕捉这种情况。对于要解决的问题,可以创建如图 2.3 所示的结构图。- #08_branch1.py
- import math as m
- p=-8.
- q=7.
- D=(p/2)**2 - q
- if D >= 0:
- x1 = -p/2 + m.sqrt(D)
- x2 = -p/2 - m.sqrt(D)
- print("x1 =",x1,"\nx2 =",x2)
- print("p =",-(x1+x2),"\nq =",x1*x2)
- else:
- print("The equation cannot be solved!")
复制代码 执行:- x1 = 7.0
- x2 = 1.0
- p = -8.0
- q = 7.0
复制代码- #09_multiple_selection1.py
- color=["black", "brown", "red", "orange", "yellow",
- "\ngreen","blue","purple","gray","white"]
- code="yellow" #input
- if code==color[0]:
- print("The color black is coded as 0.")
- elif code==color[1]:
- print("The color brown is coded as 1.")
- elif code==color[2]:
- print("The color red is coded as 2.")
- elif code==color[3]:
- print("The color orange is coded as 3.")
- elif code==color[4]:
- print("The color yellow is coded as 4.")
- elif code==color[5]:
- print("The color green is coded as 5.")
- elif code==color[6]:
- print("The color blue is coded as 6.")
- elif code==color[7]:
- print("The color purple is coded as 7.")
- elif code==color[8]:
- print("The color gray is coded as 8.")
- elif code==color[9]:
- print("The color white is coded as 9.")
复制代码 执行:- The color yellow is coded as 4.
复制代码 [code]#10_multiple_selection2.pyrate1,rate2,rate3=0.3,0.25,0.2 #eurosconsumption=5500 #kWhif 0 < consumption |
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