是否有任何python库可以让我绘制z = f(x,y),其中z表示为密集光栅化图像中的颜色(与一堆散点图的颜色相对)?如果是这样,我会使用什么功能?
看起来matplotlib.pyplot中的一些轮廓函数接近我想要的,但它们绘制轮廓线,我不希望这样。
答案 0 :(得分:9)
查看pcolor
中imshow
或matplotlib
的文档。
另一个好的开始是看看matplotlib画廊,看看是否有一个符合你要求的情节类型,然后使用示例代码作为你自己工作的起点:
答案 1 :(得分:9)
这是一个具体的简单示例(也适用于不能为x
和y
提取矩阵参数的函数):
# the function to be plotted
def func(x,y):
# gives vertical color bars if x is horizontal axis
return x
import pylab
# define the grid over which the function should be plotted (xx and yy are matrices)
xx, yy = pylab.meshgrid(
pylab.linspace(-3,3, 101),
pylab.linspace(-3,3, 111))
# indexing of xx and yy (with the default value for the
# 'indexing' parameter of meshgrid(..) ) is as follows:
#
# first index (row index) is y coordinate index
# second index (column index) is x coordinate index
#
# as required by pcolor(..)
# fill a matrix with the function values
zz = pylab.zeros(xx.shape)
for i in range(xx.shape[0]):
for j in range(xx.shape[1]):
zz[i,j] = func(xx[i,j], yy[i,j])
# plot the calculated function values
pylab.pcolor(xx,yy,zz)
# and a color bar to show the correspondence between function value and color
pylab.colorbar()
pylab.show()
答案 2 :(得分:1)
给予应有的信用:这只是Andre Holzner回答的一个微小变化。如果你必须,请向他投票!
import pylab
def f(x, y):
return pylab.cos(x) + pylab.sin(y)
xx = pylab.linspace(-5, 5, 100)
yy = pylab.linspace(-5, 5, 100)
zz = pylab.zeros([len(xx), len(yy)])
for i in xrange(len(xx)):
for j in xrange(len(yy)):
zz[j, i] = f(xx[i], yy[j])
pylab.pcolor(xx, yy, zz)
pylab.show()
使用严格的最小数组维度和索引可能更容易阅读语法。它依赖于以下几点(引自文档)。
If either or both of X and Y are 1-D arrays or column vectors, they will be expanded as needed into the appropriate 2-D arrays, making a rectangular grid.
答案 3 :(得分:1)
要扩展我上面的评论,这里有一些在网格上计算函数的可能方法
boffi@debian:~/Documents/tmp$ cat grid.py
import numpy as np
def z(x,y):
return np.sin(np.sqrt(x*x+y*y))
x = np.linspace(-1,1,11)
y = np.linspace(-2,2,21)
# naive
Z0 = np.zeros((len(y), len(x)))
for i, X in enumerate(x):
for j, Y in enumerate(y):
Z0[j,i] = z(X,Y)
# trampoline on a double list comprehension,
# it is possibly faster, sure it uses more memory
Z1 = np.array([[z(X,Y) for X in x] for Y in y])
# numpy has meshgrid,
# meshgrid uses twice memory as the result matrix but
# if used _correctly_ it's FAST
X, Y = np.meshgrid(x, y)
# numpy can avoid you explicit looping,
# but if you are so inclined...
Z2 = np.zeros((len(y), len(x)))
for r in range(len(y)):
for c in range(len(x)):
Z2[r, c] = z(X[r, c], Y[r, c])
# numpy has ufuncs, and
# t h i s i s t h e w a y t o g o
Z3 = z(X, Y)
# numpy has broadcasting (it's slower than Z = z(X, Y), less memory)
Z4 = z(x, y[:,None])
# note that x is still a _row_ of numbers, indexed by _columns_,
# while y[:,None] is now a _column_ of numbers, indexed by _rows_,
# so that Z4[row,column] <-- z(x[column], y[row])
# a bit of testing
# in previous answers, Z2 (i.e., explicit loops)
# is the preferred method --- here we show that the other four
# possible methods give you exactly the same result
print np.all(Z2==Z0)
print np.all(Z2==Z1)
print np.all(Z2==Z3)
print np.all(Z2==Z4)
boffi@debian:~/Documents/tmp$ python2 grid.py
True
True
True
True
boffi@debian:~/Documents/tmp$