是否可以调用另一个python脚本来只访问脚本中的定义而不访问其他内容?在
在我要导入的脚本中,有一些绘图我想取消,因为这个其他程序不需要。也就是说,我只想访问Stumpff函数的定义,而不需要绘制图形。在
我要导入的脚本是:
#!/usr/bin/env ipython
# This program plots the Stumpff functions C(z) and S(z)
import numpy as np
import pylab
from matplotlib.ticker import MaxNLocator
def C(z):
if z > 0:
return (1 - np.cos(z ** 0.5)) / z
elif z < 0:
return (np.cosh(np.sqrt(-z)) - 1) / -z
return 0.5
def S(z):
if z > 0:
return (np.sqrt(z) - np.sin(z ** 0.5)) / np.sqrt(z) ** 3
elif z < 0:
return (np.sinh(np.sqrt(-z)) - np.sqrt(-z)) / np.sqrt(-z) ** 3
return 1.0 / 6.0
vC = np.vectorize(C)
vS = np.vectorize(S)
z = np.linspace(-50.0, 500.0, 100000.0)
y = vC(z)
y2 = vS(z)
fig = pylab.figure()
ax = fig.add_subplot(111)
ax.plot(z, y, 'r')
ax.plot(z, y2, 'b')
pylab.legend(('$C(z)$', '$S(z)$'), loc = 0)
pylab.xlim((-50, 0))
pylab.ylim((0, 12))
pylab.xlabel('$z$')
pylab.gca().xaxis.set_major_locator(MaxNLocator(prune = 'lower'))
pylab.savefig('stumpffneg50to0.eps', format = 'eps')
fig2 = pylab.figure()
ax2 = fig2.add_subplot(111)
ax2.plot(z, y, 'r')
ax2.plot(z, y2, 'b')
pylab.legend(('$C(z)$', '$S(z)$'), loc = 1)
pylab.xlim((0, 30))
pylab.ylim((0, 0.5))
pylab.xlabel('$z$')
pylab.gca().xaxis.set_major_locator(MaxNLocator(prune = 'lower'))
pylab.savefig('stumpff0to30.eps', format = 'eps')
fig3 = pylab.figure()
ax3 = fig3.add_subplot(111)
ax3.plot(z, y, 'r')
ax3.plot(z, y2, 'b')
pylab.legend(('$C(z)$', '$S(z)$'), loc = 0)
pylab.xlim((0, 500))
pylab.ylim((0, 0.05))
pylab.xlabel('$z$')
pylab.gca().xaxis.set_major_locator(MaxNLocator(prune = 'lower'))
pylab.savefig('stumpff0to500.eps', format = 'eps')
pylab.show()
之后,我的新脚本能理解C(z)和{}?在