我想绘制一些数据。第一列包含x数据。但matplotlib没有绘制这个。我的错误在哪里?
import numpy as np
from numpy import cos
from scipy import *
from pylab import plot, show, ylim, yticks
from matplotlib import *
from pprint import pprint
n1 = 1.0
n2 = 1.5
#alpha, beta, intensity
data = [
[10, 22, 4.3],
[20, 42, 4.2],
[30, 62, 3.6],
[40, 83, 1.3],
[45, 102, 2.8],
[50, 123, 3.0],
[60, 143, 3.2],
[70, 163, 3.8],
]
for i in range(len(data)):
rhotang1 = (n1 * cos(data[i][0]) - n2 * cos(data[i][1]))
rhotang2 = (n1 * cos(data[i][0]) + n2 * cos(data[i][1]))
rhotang = rhotang1 / rhotang2
data[i].append(rhotang) #append 4th value
pprint(data)
x = data[:][0]
y1 = data[:][2]
y3 = data[:][3]
plot(x, y1, x, y3)
show()
编辑:http://paste.pocoo.org/show/205534/ 但它不起作用。
答案 0 :(得分:5)
您可以通过将数据转换为numpy数组来完成此操作:
data = np.array(data) # insert this new line after your appends
pprint(data)
x = data[:,0] # use the multidimensional slicing notation
y1 = data[:,2]
y3 = data[:,3]
plot(x, y1, x, y3)
还有几点:
您可以使用numpy以更清晰和矢量化的方式进行计算,例如
data = np.array(data)
rhotang1 = n1*cos(data[:,0]) - n2*cos(data[:,1])
rhotang2 = n1*cos(data[:,0]) + n2*cos(data[:,1])
y3 = rhotang1 / rhotang2
正如你所写的那样,你的计算可能无法提供你想要的东西,因为cos
等以弧度作为输入,你的数字看起来像度数。
答案 1 :(得分:2)
x = data[:][0]
y1 = data[:][2]
y3 = data[:][3]
这些行不符合您的想法。
首先,他们取一个整个数组的数组(即只是一个副本),然后从该数组中取出第0,第2或第3行,而不是列。
你可以尝试
x = [row[0] for row in x]
等
答案 2 :(得分:0)
试试这个:
#fresnel formula
import numpy as np
from numpy import cos
from scipy import *
from pylab import plot, show, ylim, yticks
from matplotlib import *
from pprint import pprint
n1 = 1.0
n2 = 1.5
#alpha, beta, intensity
data = np.array([
[10, 22, 4.3],
[20, 42, 4.2],
[30, 62, 3.6],
[40, 83, 1.3],
[45, 102, 2.8],
[50, 123, 3.0],
[60, 143, 3.2],
[70, 163, 3.8],
])
# Populate arrays
x = np.array([row[0] for row in data])
y1 = np.array([row[1] for row in data])
rhotang1 = n1*cos(data[:,0]) - n2*cos(data[:,1])
rhotang2 = n1*cos(data[:,0]) + n2*cos(data[:,1])
y3 = rhotang1 / rhotang2
plot(x, y1, 'r--', x, y3, 'g--')
show()