创建一个与最终图像中的单元格对应的二维数组，称为say heatmap_cells，并将其实例化为全零。

选择两个比例因子来定义每个数组元素在实际单位中的差异，对于每个维度，例如x_scale和y_scale。选择这些，使所有数据点都在热图数组的范围内。

对于每个带x_value和y_value的原始数据点:

heatmap_cells[地板(x_value / x_scale),地板(y_value / y_scale)] + = 1

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()

下面是我在100万个点集上做的一个，有3个类别(红色、绿色和蓝色)。如果您想尝试这个功能，这里有一个到存储库的链接。Github回购

histplot(
    X,
    Y,
    labels,
    bins=2000,
    range=((-3,3),(-3,3)),
    normalize_each_label=True,
    colors = [
        [1,0,0],
        [0,1,0],
        [0,0,1]],
    gain=50)

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:

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: