我是Python新手,有任何编程背景。我试图为相同的x数据集绘制y的2个数据集,用get对它进行线性回归,得到R^2值。我就是这样走到现在的:
import matplotlib
import matplotlib.pyplot as pl
from scipy import stats
#first order
'''sin(Δθ)'''
y1 = [-0.040422445,-0.056402365,-0.060758191]
#second order
'''sin(Δθ)'''
y2 = [-0.083967708, -0.107420964, -0.117248521]
''''λ, theo (nm)'''
x= [404.66, 546.07, 579.06]
pl.title('Angular displacements vs. Theoretical wavelength')
pl.xlabel('theoretical λ (in nm)')
pl.y1label('sin(Δθ) of 1st order images')
pl.y2label('sin(Δθ) of 2nd order images')
plot1 = pl.plot(x, y1, 'r')
plot2 = pl.plot(x, y2, 'b')
pl.legend([plot1, plot2], ('1st order images', '2nd order images'), 'best', numpoints=1)
slope1, intercept1, r_value1, p_value1, std_err1 = stats.linregress(x,y1)
slope2, intercept2, r_value2, p_value2, std_err2 = stats.linregress(x,y2)
print "r-squared:", r_value1**2
print "r-squared:", r_value2**2
pl.show()
...i没有得到任何情节,我得到了错误:"UnicodeDecodeError:'ascii‘编解码器无法在第12位解码字节0xce :序号不在范围(128)“
但我不明白。有人能帮忙告诉我我的密码出了什么问题吗?谢谢你
发布于 2016-09-11 10:47:03
此代码中有多个错误。
y1label
和y2label
不是pl
模块的对象# blah blah
不同于'''blah blah'''
。第一个是注释,第二个是表达式。可以将第二个变量分配给变量(a = '''blah blah'''
),但不能将第一个变量分配给变量:a = # blah blah
生成一个SyntaxError。下面是一个应该有效的代码:
import matplotlib
import matplotlib.pyplot as pl
from scipy import stats
y1 = [-0.040422445,-0.056402365,-0.060758191]
y2 = [-0.083967708, -0.107420964, -0.117248521]
x= [404.66, 546.07, 579.06]
pl.title('Angular displacements vs. Theoretical wavelength')
pl.xlabel(r'theoretical $\lambda$ (in nm)')
pl.ylabel(r'sin($\Delta\theta$)')
y1label = '1st order images'
y2label = '2nd order images'
plot1 = pl.plot(x, y1, 'r', label=y1label)
plot2 = pl.plot(x, y2, 'b', label=y2label)
pl.legend()
slope1, intercept1, r_value1, p_value1, std_err1 = stats.linregress(x,y1)
slope2, intercept2, r_value2, p_value2, std_err2 = stats.linregress(x,y2)
print "r-squared:", r_value1**2
print "r-squared:", r_value2**2
pl.show()
https://stackoverflow.com/questions/39435177
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