我有一组X,Y数据点(大约10k),很容易绘制成散点图,但我想用热图来表示。

我查看了Matplotlib中的示例,它们似乎都已经从热图单元值开始生成图像。

有没有一种方法可以将一堆不同的x, y转换为热图(其中x, y频率较高的区域会“更温暖”)?


当前回答

这些解决方案都不适用于我的应用程序,所以我想出了这个解决方案。本质上,我在每个点上都放置了一个二维高斯分布:

import cv2
import numpy as np
import matplotlib.pyplot as plt

def getGaussian2D(ksize, sigma, norm=True):
    oneD = cv2.getGaussianKernel(ksize=ksize, sigma=sigma)
    twoD = np.outer(oneD.T, oneD)
    return twoD / np.sum(twoD) if norm else twoD

def pt2heat(pts, shape, kernel=16, sigma=5):
    heat = np.zeros(shape)
    k = getGaussian2D(kernel, sigma)
    for y,x in pts:
        x, y = int(x), int(y)
        for i in range(-kernel//2, kernel//2):
            for j in range(-kernel//2, kernel//2):
                if 0 <= x+i < shape[0] and 0 <= y+j < shape[1]:
                    heat[x+i, y+j] = heat[x+i, y+j] + k[i+kernel//2, j+kernel//2]
    return heat


heat = pts2heat(pts, img.shape[:2])
plt.imshow(heat, cmap='heat')

以下是在相关图像上叠加的点,以及生成的热图:

其他回答

如果您正在使用1.2.x

import numpy as np
import matplotlib.pyplot as plt

x = np.random.randn(100000)
y = np.random.randn(100000)
plt.hist2d(x,y,bins=100)
plt.show()

而不是用np。我想回收py-sphviewer,这是一个使用自适应平滑内核渲染粒子模拟的python包,可以很容易地从pip安装(见网页文档)。考虑以下基于示例的代码:

import numpy as np
import numpy.random
import matplotlib.pyplot as plt
import sphviewer as sph

def myplot(x, y, nb=32, xsize=500, ysize=500):   
    xmin = np.min(x)
    xmax = np.max(x)
    ymin = np.min(y)
    ymax = np.max(y)

    x0 = (xmin+xmax)/2.
    y0 = (ymin+ymax)/2.

    pos = np.zeros([len(x),3])
    pos[:,0] = x
    pos[:,1] = y
    w = np.ones(len(x))

    P = sph.Particles(pos, w, nb=nb)
    S = sph.Scene(P)
    S.update_camera(r='infinity', x=x0, y=y0, z=0, 
                    xsize=xsize, ysize=ysize)
    R = sph.Render(S)
    R.set_logscale()
    img = R.get_image()
    extent = R.get_extent()
    for i, j in zip(xrange(4), [x0,x0,y0,y0]):
        extent[i] += j
    print extent
    return img, extent
    
fig = plt.figure(1, figsize=(10,10))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222)
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224)


# Generate some test data
x = np.random.randn(1000)
y = np.random.randn(1000)

#Plotting a regular scatter plot
ax1.plot(x,y,'k.', markersize=5)
ax1.set_xlim(-3,3)
ax1.set_ylim(-3,3)

heatmap_16, extent_16 = myplot(x,y, nb=16)
heatmap_32, extent_32 = myplot(x,y, nb=32)
heatmap_64, extent_64 = myplot(x,y, nb=64)

ax2.imshow(heatmap_16, extent=extent_16, origin='lower', aspect='auto')
ax2.set_title("Smoothing over 16 neighbors")

ax3.imshow(heatmap_32, extent=extent_32, origin='lower', aspect='auto')
ax3.set_title("Smoothing over 32 neighbors")

#Make the heatmap using a smoothing over 64 neighbors
ax4.imshow(heatmap_64, extent=extent_64, origin='lower', aspect='auto')
ax4.set_title("Smoothing over 64 neighbors")

plt.show()

产生如下图像:

如你所见,这些图像看起来非常漂亮,我们能够识别出它上面不同的子结构。这些图像是在一个特定的域内为每个点扩展一个给定的权重,由平滑长度定义,而平滑长度又由到更近的nb邻居的距离给出(我选择了16,32和64作为示例)。因此,高密度区域通常分布在较小的区域,与低密度区域相比。

myplot函数是我写的一个非常简单的函数它是为了将x y数据交给py-sphviewer来完成这个魔术。

下面是Jurgy使用scipy.cKDTree实现的最近邻方法。在我的测试中,它快了大约100倍。

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.spatial import cKDTree


def data_coord2view_coord(p, resolution, pmin, pmax):
    dp = pmax - pmin
    dv = (p - pmin) / dp * resolution
    return dv


n = 1000
xs = np.random.randn(n)
ys = np.random.randn(n)

resolution = 250

extent = [np.min(xs), np.max(xs), np.min(ys), np.max(ys)]
xv = data_coord2view_coord(xs, resolution, extent[0], extent[1])
yv = data_coord2view_coord(ys, resolution, extent[2], extent[3])


def kNN2DDens(xv, yv, resolution, neighbours, dim=2):
    """
    """
    # Create the tree
    tree = cKDTree(np.array([xv, yv]).T)
    # Find the closest nnmax-1 neighbors (first entry is the point itself)
    grid = np.mgrid[0:resolution, 0:resolution].T.reshape(resolution**2, dim)
    dists = tree.query(grid, neighbours)
    # Inverse of the sum of distances to each grid point.
    inv_sum_dists = 1. / dists[0].sum(1)

    # Reshape
    im = inv_sum_dists.reshape(resolution, resolution)
    return im


fig, axes = plt.subplots(2, 2, figsize=(15, 15))
for ax, neighbours in zip(axes.flatten(), [0, 16, 32, 63]):

    if neighbours == 0:
        ax.plot(xs, ys, 'k.', markersize=5)
        ax.set_aspect('equal')
        ax.set_title("Scatter Plot")
    else:

        im = kNN2DDens(xv, yv, resolution, neighbours)

        ax.imshow(im, origin='lower', extent=extent, cmap=cm.Blues)
        ax.set_title("Smoothing over %d neighbours" % neighbours)
        ax.set_xlim(extent[0], extent[1])
        ax.set_ylim(extent[2], extent[3])

plt.savefig('new.png', dpi=150, bbox_inches='tight')

非常类似于@Piti的答案,但使用1次调用而不是2次调用来生成点:

import numpy as np
import matplotlib.pyplot as plt

pts = 1000000
mean = [0.0, 0.0]
cov = [[1.0,0.0],[0.0,1.0]]

x,y = np.random.multivariate_normal(mean, cov, pts).T
plt.hist2d(x, y, bins=50, cmap=plt.cm.jet)
plt.show()

输出:

如果你不想要六边形,你可以使用numpy的histogram2d函数:

import numpy as np
import numpy.random
import matplotlib.pyplot as plt

# Generate some test data
x = np.random.randn(8873)
y = np.random.randn(8873)

heatmap, xedges, yedges = np.histogram2d(x, y, bins=50)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]

plt.clf()
plt.imshow(heatmap.T, extent=extent, origin='lower')
plt.show()

这是一个50x50的热图。如果你想要,比如说512x384,你可以在调用histogram2d时放入bins=(512,384)。

例子: