如何在python中將3D功能繪制為2D色彩圖？

Question

是否有任何python庫可以讓我繪制z = f（x，y），其中z表示為密集光柵化圖像中的顏色（與一堆散點圖的顏色相對）？ 如果是這樣，我會使用什么功能？

看起來matplotlib.pyplot中的一些輪廓函數接近我想要的，但它們繪制輪廓線，我不希望這樣。

Answer 1

這是一個具體的簡單示例（也適用於不能為x和y采用矩陣參數的x ）：

# 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() 

Answer 2

在matplotlib查看pcolor或imshow的文檔。

另一個好的開始是看看matplotlib畫廊，看看是否有一個符合你要求的情節類型，然后使用示例代碼作為你自己工作的起點：

http://matplotlib.sourceforge.net/gallery.html

Answer 3

給予應有的信用：這只是Andre Holzner答案的微小變化。 如果必須，請向他投票！

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()

使用嚴格的最小數組維度和索引可能更容易閱讀語法。 它依賴於以下幾點（引自文檔）。

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.

Answer 4

為了擴展我上面的評論，這里有一些計算網格上的函數的可能方法

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$