Python:x-y-plot与matplotlib

Question

我想绘制一些数据。第一列包含x数据。但是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/ 但没用。

Answer 1

x = data[:][0]
y1 = data[:][2]
y3 = data[:][3]

这些台词不符合你的想法。

首先从数组中提取一个片段，这是整个数组（也就是说，只是一个副本），然后从该数组中提取第0、第2或第3行，而不是列。

你可以试试

x = [row[0] for row in x]

等等

Answer 2

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()

Answer 3

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，如下所示

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

正如您所写，您的计算可能不会给出您想要的结果，因为cosetc以弧度作为输入，您的数字看起来像度数。