- 随机数和蒙特卡洛模拟
- 求解单一变量非线性方程
- 求解线性系统方程
- 函数的数学积分
- 常微分方程的数值解
- 等势线绘图和曲线
- 等势线
-
-
import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D x_vals = np.linspace(-5,5,20) y_vals = np.linspace(0,10,20) X,Y = np.meshgrid(x_vals,y_vals) Z = X**2 * Y**0.5 line_count = 15 ax = Axes3D(plt.figure()) ax.plot_surface(X,Y,Z,rstride=1,cstride=1) plt.show()
- 等势线
-
非线性方程的数学解
-
一般实函数 Scipy.optimize
-
fsolve函数求零点（限定只给实数解）
-
import scipy.optimize as so from scipy.optimize import fsolve f = lambda x:x**2-1 fsolve(f,0.5) fsolve(f,-0.5) fsolve(f,[-0.5,0.5]) >>>fsolve(f,-0.5,full_output=True) >>>(array([-1.]), {'nfev': 9, 'fjac': array([[-1.]]), 'r': array([1.99999875]), 'qtf': array([3.82396337e-10]), 'fvec': array([4.4408921e-16])}, 1, 'The solution converged.') >>>help(fsolve) >>>Help on function fsolve in module scipy.optimize._minpack_py: fsolve(func, x0, args=(), fprime=None, full_output=0, col_deriv=0, xtol=1.49012e-08, maxfev=0, band=None, epsfcn=None, factor=100, diag=None) Find the roots of a function. Return the roots of the (non-linear) equations defined by ``func(x) = 0`` given a starting estimate. Parameters ---------- func : callable ``f(x, *args)`` A function that takes at least one (possibly vector) argument, and returns a value of the same length. x0 : ndarray The starting estimate for the roots of ``func(x) = 0``. args : tuple, optional Any extra arguments to `func`. fprime : callable ``f(x, *args)``, optional A function to compute the Jacobian of `func` with derivatives across the rows. By default, the Jacobian will be estimated. full_output : bool, optional If True, return optional outputs. col_deriv : bool, optional Specify whether the Jacobian function computes derivatives down the columns (faster, because there is no transpose operation). xtol : float, optional The calculation will terminate if the relative error between two consecutive iterates is at most `xtol`. maxfev : int, optional The maximum number of calls to the function. If zero, then ``100*(N+1)`` is the maximum where N is the number of elements in `x0`. band : tuple, optional If set to a two-sequence containing the number of sub- and super-diagonals within the band of the Jacobi matrix, the Jacobi matrix is considered banded (only for ``fprime=None``). epsfcn : float, optional A suitable step length for the forward-difference approximation of the Jacobian (for ``fprime=None``). If `epsfcn` is less than the machine precision, it is assumed that the relative errors in the functions are of the order of the machine precision. factor : float, optional A parameter determining the initial step bound (``factor * || diag * x||``). Should be in the interval ``(0.1, 100)``. diag : sequence, optional N positive entries that serve as a scale factors for the variables. Returns ------- x : ndarray The solution (or the result of the last iteration for an unsuccessful call). infodict : dict A dictionary of optional outputs with the keys: ``nfev`` number of function calls ``njev`` number of Jacobian calls ``fvec`` function evaluated at the output ``fjac`` the orthogonal matrix, q, produced by the QR factorization of the final approximate Jacobian matrix, stored column wise ``r`` upper triangular matrix produced by QR factorization of the same matrix ``qtf`` the vector ``(transpose(q) * fvec)`` ier : int An integer flag. Set to 1 if a solution was found, otherwise refer to `mesg` for more information. mesg : str If no solution is found, `mesg` details the cause of failure. See Also -------- root : Interface to root finding algorithms for multivariate functions. See the ``method=='hybr'`` in particular. Notes ----- ``fsolve`` is a wrapper around MINPACK's hybrd and hybrj algorithms. Examples -------- Find a solution to the system of equations: ``x0*cos(x1) = 4, x1*x0 - x1 = 5``. >>> from scipy.optimize import fsolve >>> def func(x): ... return [x[0] * np.cos(x[1]) - 4, ... x[1] * x[0] - x[1] - 5] >>> root = fsolve(func, [1, 1]) >>> root array([6.50409711, 0.90841421]) >>> np.isclose(func(root), [0.0, 0.0]) # func(root) should be almost 0.0. array([ True, True]) - 关键字参数 full_output=True
-
-
- 多项式的复数根
- np.roots([最高位系数，次高位系数，... ... x项系数，常数项])
-
>>>f = lambda x:x**4 + x -1 >>>np.roots([1,0,0,1,-1]) >>>array([-1.22074408+0.j , 0.24812606+1.03398206j, 0.24812606-1.03398206j, 0.72449196+0.j ])
-
- 求解线性等式 scipy.linalg
- 利用dir()获取常用函数
-
-
import numpy as np import scipy.linalg as sla from scipy.linalg import inv a = np.array([-1,5]) c = np.array([[1,3],[3,4]]) x = np.dot(inv(c),a) >>>x >>>array([ 3.8, -1.6])
- 利用dir()获取常用函数
- 数值积分 scipy.integrate
- 利用dir()获取你需要的信息
- 对自定义函数做积分
-
-
import scipy.integrate as si from scipy.integrate import quad import numpy as np import matplotlib.pyplot as plt f = lambda x:x**1.05*0.001 interval = 100 xmax = np.linspace(1,5,interval) integral,error = np.zeros(xmax.size),np.zeros(xmax.size) for i in range(interval): integral[i],error[i] = quad(f,0,xmax[i]) plt.plot(xmax,integral,label="integral") plt.plot(xmax,error,label="error") plt.show()
-
- 对震荡函数做积分
- quad 函数允许 调整他使用的网格
-
>>>(-0.4677718053224297, 2.5318630220102742e-05) >>>quad(np.cos,-1,1,limit=100) >>>(1.6829419696157932, 1.8684409237754643e-14) >>>quad(np.cos,-1,1,limit=1000) >>>(1.6829419696157932, 1.8684409237754643e-14) >>>quad(np.cos,-1,1,limit=10) >>>(1.6829419696157932, 1.8684409237754643e-14)
- 微分方程的数值解
- 向量场与流线图
- vector(x,y) = (y,-x)
-
import numpy as np import matplotlib.pyplot as plt coords = np.linspace(-1,1,30) X,Y = np.meshgrid(coords,coords) Vx,Vy = Y,-X plt.quiver(X,Y,Vx,Vy) plt.show() ------------------ import numpy as np import matplotlib.pyplot as plt coords = np.linspace(-2,2,10) X,Y = np.meshgrid(coords,coords) Z = np.exp(np.exp(X+Y)) ds = 4/6 dX,dY = np.gradient(Z,ds) plt.contourf(X,Y,Z,25) plt.quiver(X,Y,dX.transpose(),dY.transpose(),scale=25) plt.show()
-
-
