有没有办法在python中绘制方向字段?
我的尝试是修改http://www.compdigitec.com/labs/files/slopefields.py给予
#!/usr/bin/python
import math
from subprocess import CalledProcessError, call, check_call
def dy_dx(x, y):
try:
# declare your dy/dx here:
return x**2-x-2
except ZeroDivisionError:
return 1000.0
# Adjust window parameters
XMIN = -5.0
XMAX = 5.0
YMIN = -10.0
YMAX = 10.0
XSCL = 0.5
YSCL = 0.5
DISTANCE = 0.1
def main():
fileobj = open("data.txt", "w")
for x1 in xrange(int(XMIN / XSCL), int(XMAX / XSCL)):
for y1 in xrange(int(YMIN / YSCL), int(YMAX / YSCL)):
x= float(x1 * XSCL)
y= float(y1 * YSCL)
slope = dy_dx(x,y)
dx = math.sqrt( DISTANCE/( 1+math.pow(slope,2) ) )
dy = slope*dx
fileobj.write(str(x) + " " + str(y) + " " + str(dx) + " " + str(dy) + "\n")
fileobj.close()
try:
check_call(["gnuplot","-e","set terminal png size 800,600 enhanced font \"Arial,12\"; set xrange [" + str(XMIN) + ":" + str(XMAX) + "]; set yrange [" + str(YMIN) + ":" + str(YMAX) + "]; set output 'output.png'; plot 'data.txt' using 1:2:3:4 with vectors"])
except (CalledProcessError, OSError):
print "Error: gnuplot not found on system!"
exit(1)
print "Saved image to output.png"
call(["xdg-open","output.png"])
return 0
if __name__ == '__main__':
main()
然而,我从中获得的最佳图像是。 如何获得看起来更像第一张图像的输出?另外,如何添加三条实线?
答案 0 :(得分:11)
您可以使用此matplotlib代码作为基础。根据您的需要进行修改。 我已更新代码以显示相同长度的箭头。
也可以将轴形式“框”更改为“箭头”。如果您需要更改,请告诉我,我可以添加它。
import matplotlib.pyplot as plt
from scipy import *
from scipy import integrate
from scipy.integrate import ode
import numpy as np
fig = plt.figure(num=1)
ax=fig.add_subplot(111)
## Vector field function
def vf(t,x):
dx=np.zeros(2)
dx[0]=1
dx[1]=x[0]**2-x[0]-2
return dx
#Solution curves
t0=0; tEnd=10; dt=0.01;
r = ode(vf).set_integrator('vode', method='bdf',max_step=dt)
ic=[[-3.5,-10], [-3,-10], [-2.5,-10]]
color=['r','b','g']
for k in range(len(ic)):
Y=[];T=[];S=[];
r.set_initial_value(ic[k], t0).set_f_params()
while r.successful() and r.t +dt < tEnd:
r.integrate(r.t+dt)
Y.append(r.y)
S=np.array(np.real(Y))
ax.plot(S[:,0],S[:,1], color = color[k], lw = 1.25)
#Vector field
X,Y = np.meshgrid( np.linspace(-5,5,20),np.linspace(-10,10,20) )
U = 1
V = X**2-X-2
#Normalize arrows
N = np.sqrt(U**2+V**2)
U2, V2 = U/N, V/N
ax.quiver( X,Y,U2, V2)
plt.xlim([-5,5])
plt.ylim([-10,10])
plt.xlabel(r"$x$")
plt.ylabel(r"$y$")
plt.show()
答案 1 :(得分:1)
我很高兴使用pygame将其中一个作为业余爱好项目。我绘制了每个像素的斜率,使用蓝色阴影表示正值,使用红色阴影表示阴性。黑色是未定义的。这是dy/dx = log(sin(x/y)+cos(y/x))
:
dy/dx=log(sin(x/y)+cos(y/x) http://oi43.tinypic.com/90umn4.jpg
你可以放大&amp; out - 这里放大了中间的上半部分:
并点击一个点来绘制通过该点的直线:
这只是440行代码,所以here is the .zip of all the files。我想我会在这里摘录相关内容。
等式本身作为字符串中的有效Python表达式输入,例如"log(sin(x/y)+cos(y/x))"
。然后是compile
d。此函数在此处绘制颜色字段,其中self.func.eval()
给出给定点的dy/dx
。这里的代码有点复杂,因为我让它分阶段渲染 - 先是32x32块,然后是16x16等,以使用户更加快捷。
def graphcolorfield(self, sqsizes=[32,16,8,4,2,1]):
su = ScreenUpdater(50)
lastskip = self.xscreensize
quitit = False
for squaresize in sqsizes:
xsquaresize = squaresize
ysquaresize = squaresize
if squaresize == 1:
self.screen.lock()
y = 0
while y <= self.yscreensize:
x = 0
skiprow = y%lastskip == 0
while x <= self.xscreensize:
if skiprow and x%lastskip==0:
x += squaresize
continue
color = (255,255,255)
try:
m = self.func.eval(*self.ct.untranscoord(x, y))
if m >= 0:
if m < 1:
c = 255 * m
color = (0, 0, c)
else:
#c = 255 - 255 * (1.0/m)
#color = (c, c, 255)
c = 255 - 255 * (1.0/m)
color = (c/2.0, c/2.0, 255)
else:
pm = -m
if pm < 1:
c = 255 * pm
color = (c, 0, 0)
else:
c = 255 - 255 * (1.0/pm)
color = (255, c/2.0, c/2.0)
except:
color = (0, 0, 0)
if squaresize > 1:
self.screen.fill(color, (x, y, squaresize, squaresize))
else:
self.screen.set_at((x, y), color)
if su.update():
quitit = True
break
x += xsquaresize
if quitit:
break
y += ysquaresize
if squaresize == 1:
self.screen.unlock()
lastskip = squaresize
if quitit:
break
这是通过点绘制直线的代码:
def _grapheqhelp(self, sx, sy, stepsize, numsteps, color):
x = sx
y = sy
i = 0
pygame.draw.line(self.screen, color, (x, y), (x, y), 2)
while i < numsteps:
lastx = x
lasty = y
try:
m = self.func.eval(x, y)
except:
return
x += stepsize
y = y + m * stepsize
screenx1, screeny1 = self.ct.transcoord(lastx, lasty)
screenx2, screeny2 = self.ct.transcoord(x, y)
#print "(%f, %f)-(%f, %f)" % (screenx1, screeny1, screenx2, screeny2)
try:
pygame.draw.line(self.screen, color,
(screenx1, screeny1),
(screenx2, screeny2), 2)
except:
return
i += 1
stx, sty = self.ct.transcoord(sx, sy)
pygame.draw.circle(self.screen, color, (int(stx), int(sty)), 3, 0)
它向后运行&amp;从那一点开始转发:
def graphequation(self, sx, sy, stepsize=.01, color=(255, 255, 127)):
"""Graph the differential equation, given the starting point sx and sy, for length
length using stepsize stepsize."""
numstepsf = (self.xrange[1] - sx) / stepsize
numstepsb = (sx - self.xrange[0]) / stepsize
self._grapheqhelp(sx, sy, stepsize, numstepsf, color)
self._grapheqhelp(sx, sy, -stepsize, numstepsb, color)
我从来没有画过实际线条,因为像素方法看起来太酷了。
答案 2 :(得分:0)
尝试将参数值更改为:
XSCL = .2
YSCL = .2
这些参数决定轴上采样的点数。
根据您的评论,您还需要绘制派生dy_dx(x,y)适用的函数。
目前,您只计算和绘制由函数dy_dx(x,y)计算的斜率线。除了斜率之外,你还需要找到(在这种情况下是3个)绘制函数。
首先定义一个函数:
def f1_x(x):
return x**3-x**2-2x;
然后,在循环中,您还必须将函数的所需值写入fileobj文件。