Python 翻转和旋转等高线图的numpy阵列

短版：

我有一个10x10 numpy数组，其等高线图（用

pyplot.contourf

绘制）如下所示

现在，我想让它看起来像这样——假设曲线在X轴和Y轴上是对称的

长版本

我有一个10x10 numpy数组

z

作为

x

和

y

的函数。其中

x=y=np.arange（0.002,0.022,0.002）

。这是我试过的

import numpy as np
import matplotlib.pyplot as plt
z=np.array([[ 2.08273679, -0.06591932, -1.14525488, -1.49923222, -1.74361248,
    -1.81418446, -1.90115591, -1.94329043, -1.93130228, -1.96064259],
   [ 0.20180514, -0.94522815, -1.34635828, -1.58844515, -1.7528935 ,
    -1.84438752, -1.86257547, -1.9439332 , -1.99009407, -1.94829146],
   [-1.09749238, -1.48234452, -1.64234357, -1.75344742, -1.83019763,
    -1.88547473, -1.92958533, -1.940775  , -1.95535063, -1.9629588 ],
   [-1.62892483, -1.70176401, -1.76263555, -1.84966414, -1.87139241,
    -1.91879916, -1.90796703, -1.96632612, -1.95794984, -1.94585536],
   [-1.71551518, -1.91806287, -1.86999609, -1.90800839, -1.92515012,
    -1.93386969, -1.96487487, -1.95405297, -1.97032435, -1.96087146],
   [-1.81904322, -1.94790171, -2.        , -1.96932249, -1.91842475,
    -1.98101775, -1.98521938, -1.97618539, -1.95892852, -2.01410874],
   [-1.8138236 , -1.90877811, -1.93966404, -1.98406259, -1.95253807,
    -1.95867436, -1.96679456, -2.01126218, -1.99885932, -1.99369292],
   [-1.9927308 , -1.97658099, -1.91586737, -1.96813381, -1.98416011,
    -1.98639893, -1.99997964, -1.99746813, -1.98126505, -1.97767361],
   [-1.96406473, -1.92609437, -1.99171257, -1.94687523, -1.9823819 ,
    -1.97786533, -2.02323228, -1.98559114, -1.99172681, -2.00881064],
   [-1.92470024, -1.99537152, -1.99419303, -1.97261023, -1.9673841 ,
    -1.98801505, -2.02412735, -2.01394008, -2.01956817, -2.04963448]])
x=y=np.arange(0.002,0.022,0.002)
#The following gives the plot I currently have
plt.figure()
plt.contourf(x,y,z)
plt.show()
#Tried to flip the matrix z using np.flipud and np.fliplr
plt.figure()
plt.contourf(x,y,z)
plt.contourf(-x,y,np.fliplr(z))
plt.contourf(x,-y,np.flipud(z))
plt.contourf(-x,-y,np.flipud(np.fliplr(z)))
plt.show()
#Also tried to rotate the matrix z using np.rot90
plt.figure()
plt.contourf(x,y,z)
plt.contourf(x,-y,np.rot90(z))
plt.contourf(-x,-y,np.rot90(z,2))
plt.contourf(-x,y,np.rot90(z,3))
plt.show()

我用上面的代码得到了下面的图

及

---

理想情况下，我也希望通过插值图来填充原点处的不连续性。但对于初学者来说，他们希望获得正确的方向。非常感谢您的帮助

将数组

z

的

fliplr

和

flipud

的结果合并到一个新的双尺寸数组

zz

，然后进行绘图。根据第一个图，必须使用nan值跳过间隔（-0.002；+0.002）中的x和y：

import numpy as np
import matplotlib.pyplot as plt
z=np.array([[ 2.08273679, -0.06591932, -1.14525488, -1.49923222, -1.74361248,
    -1.81418446, -1.90115591, -1.94329043, -1.93130228, -1.96064259],
   [ 0.20180514, -0.94522815, -1.34635828, -1.58844515, -1.7528935 ,
    -1.84438752, -1.86257547, -1.9439332 , -1.99009407, -1.94829146],
   [-1.09749238, -1.48234452, -1.64234357, -1.75344742, -1.83019763,
    -1.88547473, -1.92958533, -1.940775  , -1.95535063, -1.9629588 ],
   [-1.62892483, -1.70176401, -1.76263555, -1.84966414, -1.87139241,
    -1.91879916, -1.90796703, -1.96632612, -1.95794984, -1.94585536],
   [-1.71551518, -1.91806287, -1.86999609, -1.90800839, -1.92515012,
    -1.93386969, -1.96487487, -1.95405297, -1.97032435, -1.96087146],
   [-1.81904322, -1.94790171, -2.        , -1.96932249, -1.91842475,
    -1.98101775, -1.98521938, -1.97618539, -1.95892852, -2.01410874],
   [-1.8138236 , -1.90877811, -1.93966404, -1.98406259, -1.95253807,
    -1.95867436, -1.96679456, -2.01126218, -1.99885932, -1.99369292],
   [-1.9927308 , -1.97658099, -1.91586737, -1.96813381, -1.98416011,
    -1.98639893, -1.99997964, -1.99746813, -1.98126505, -1.97767361],
   [-1.96406473, -1.92609437, -1.99171257, -1.94687523, -1.9823819 ,
    -1.97786533, -2.02323228, -1.98559114, -1.99172681, -2.00881064],
   [-1.92470024, -1.99537152, -1.99419303, -1.97261023, -1.9673841 ,
    -1.98801505, -2.02412735, -2.01394008, -2.01956817, -2.04963448]])
x=y=np.linspace(-0.020,0.020,21)

zz = np.empty((21,21)); zz[:,:] = np.nan
zz[11:,11:] = z
zz[11:,:10] = np.fliplr(z)
zz[:10,:] = np.flipud(zz[11:,:])

plt.figure()
plt.contourf(x,y,zz)
plt.show()

要填充间隙，请跳过坐标阵列的一个点：

...
x=y=np.linspace(-0.020,0.020,20)
zz = np.empty((20,20)); zz[:,:] = np.nan
zz[10:,10:] = z
zz[10:,:10] = np.fliplr(z)
zz[:10,:] = np.flipud(zz[10:,:])
...

---

您的问题是，即使您否定

x

和

y

，它们的顺序保持不变，因此对于否定的

x

，您从

-0.002

转到

-0.022

，这意味着在绘图过程中翻转的

z

会被翻转回来。要实现您的目标，您可以执行以下操作：

#either don't flip z
plt.figure()
plt.contourf(x,y,z)
plt.contourf(-x,y,z)
plt.contourf(x,-y,z)
plt.contourf(-x,-y,z)
plt.show()

#or reverse also -x and -y:
plt.figure()
plt.contourf(x,y,z)
plt.contourf(-x[::-1],y,np.fliplr(z))
plt.contourf(x,-y[::-1],np.flipud(z))
plt.contourf(-x[::-1],-y[::-1],np.flipud(np.fliplr(z)))
plt.show()

如果您只是将

z

和翻转的

z

连接起来，那么一切都会按预期工作

plt.contourf

负责插值本身

ztotal = np.concatenate([np.fliplr(z),z],axis=1)
ztotal = np.concatenate([np.flipud(ztotal),ztotal],axis=0)
xtotal = np.concatenate([-x[::-1],x],axis=0)
ytotal = np.concatenate([-y[::-1],y],axis=0)

plt.figure()
plt.contourf(xtotal,ytotal,ztotal)
plt.show()

---

我一直认为顺序在负轴上变化。这很有帮助。谢谢你的

nan

hack。你的答案和另一个一样好。我接受了这一点，因为它让我意识到了问题所在。我不认为标题编辑是必要的，因为问题和答案都适用于

plt.contour

。你的标题编辑不是限制了范围吗？