blocks|key|1313834|text|创建一个与最终图像中的单元格相对应的二维数组，称为heatmap_cells，并将其实例化为全零。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1313835|为每个维度选择两个比例因子，它们以实际单位定义每个数组元素之间的差异，例如x_scale和y_scale。选择这些选项，这样您的所有数据点都将落入heatmap数组的范围内。|1313836|对于每个具有x_value和y_value的原始数据点|1313837|heatmap_cells[floor(x_value/x_scale),floor(y_value/y_scale)]%2B=1|1313838|entityMap^0|P|D|0|11|7|19|7|0|6|7|E|7|0|0|1R|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|M|$]]

Make a 2-dimensional array that corresponds to the cells in your final image, called say <code>heatmap_cells</code> and instantiate it as all zeroes.

Choose two scaling factors that define the difference between each array element in real units, for each dimension, say <code>x_scale</code> and <code>y_scale</code>. Choose these such that all your datapoints will fall within the bounds of the heatmap array.

For each raw datapoint with <code>x_value</code> and <code>y_value</code>:

<code>heatmap_cells[floor(x_value/x_scale),floor(y_value/y_scale)]+=1</code>

blocks|key|1313992|text|如果你不想要六边形，你可以使用numpy的histogram2d函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1313993|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()|code-block|syntax|javascript|1313994|这就产生了一个50x50的热图。比方说，如果你想要512x384，你可以在histogram2d的调用中放入bins=(512,+384)。|1313995|示例：|1313996|📷|atomic|1313997|​|1313998|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/60b530d5eb73e081d454d39af3ae1c0f.png|imageAlt^0|L|B|0|0|11|B|1I|F|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1B|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|R|3|S|5|6|7|1G|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10|11|-4]]]]

If you don't want hexagons, you can use numpy's <code>histogram2d</code> function:

<pre><code>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()
</code></pre>

This makes a 50x50 heatmap. If you want, say, 512x384, you can put <code>bins=(512, 384)</code> in the call to <code>histogram2d</code>.

Example: <img src="https://i.stack.imgur.com/YcXBC.png" alt="Matplotlib heat map example">

blocks|key|1314204|text|如果您使用的是1.2.x|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314205|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()|code-block|syntax|javascript|1314206|​|1314207|📷|atomic|offset|length|1314208|1314209|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/c644f5e7e21b6a38ab65a0eba3a433f8.png|imageAlt^0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|10|8|@]|9|@$L|11|M|12|1|13]]|A|$]]|$1|N|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V|W|-4]]]]

If you are using 1.2.x

<pre><code>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()
</code></pre>

<img src="https://i.stack.imgur.com/yj9ku.png" alt="gaussian_2d_heat_map">

blocks|key|1314348|text|Seaborn现在有了jointplot+function，它应该可以在这里很好地工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1314349|import+numpy+as+np
import+seaborn+as+sns
import+matplotlib.pyplot+as+plt

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

sns.jointplot(x=x,+y=y,+kind='hex')
plt.show()|code-block|syntax|javascript|1314350|​|1314351|📷|atomic|1314352|1314353|entityMap|0|LINK|mutability|MUTABLE|url|http://seaborn.pydata.org/generated/seaborn.jointplot.html|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/b2782e80c46b0b38f96cb3bb70d773f4.png|imageAlt^0|B|I|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|N|3|J|5|6|7|1C|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|X|S|Y|C|$Z|10|11|-4]]]]

Seaborn now has the <a href="http://seaborn.pydata.org/generated/seaborn.jointplot.html" rel="noreferrer">jointplot function</a> which should work nicely here:

<pre><code>import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt

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

sns.jointplot(x=x, y=y, kind='hex')
plt.show()
</code></pre>

<a href="https://i.stack.imgur.com/9ageD.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9ageD.png" alt="demo image"></a>

blocks|key|1314576|text|最初的问题是...如何将散点值转换为网格值，对吗？histogram2d确实计算每个单元格的频率，但是，如果您有每个单元格的其他数据，而不仅仅是频率，则需要做一些额外的工作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1314577|x+=+data_x+#+between+-10+and+4,+log-gamma+of+an+svc
y+=+data_y+#+between+-4+and+11,+log-C+of+an+svc
z+=+data_z+#between+0+and+0.78,+f1-values+from+a+difficult+dataset|code-block|syntax|javascript|1314578|因此，我有一个数据集，其中包含X和Y坐标的Z结果。然而，我在计算感兴趣的区域之外的几个点(很大的差距)，而在一个小的感兴趣的区域中有成堆的点。|1314579|是的，在这里，它变得更加困难，但也更加有趣。一些库(对不起)：|1314580|from+matplotlib+import+pyplot+as+plt
from+matplotlib+import+cm
import+numpy+as+np
from+scipy.interpolate+import+griddata|1314581|pyplot是我今天的图形引擎，cm是一系列带有一些有趣选择的彩色地图。numpy用于计算，griddata用于将值附加到固定网格。|1314582|最后一个很重要，因为xy点的频率在我的数据中不是均匀分布的。首先，让我们从一些适合我的数据和任意网格大小的边界开始。原始数据的数据点也在这些x和y边界之外。|1314583|#determine+grid+boundaries
gridsize+=+500
x_min+=+-8
x_max+=+2.5
y_min+=+-2
y_max+=+7|1314584|因此，我们定义了一个具有500像素的网格，它介于x和y的最小值和最大值之间。|1314585|在我的数据中，高兴趣区中的可用值远远超过500个；而在低兴趣区，整个网格中甚至没有200个值；在x_min和x_max的图形边界之间，甚至更少。|1314586|因此，为了获得一张好的图片，任务是获得高息值的平均值，并填补其他地方的空白。|1314587|我现在定义了我的网格。对于每一对xx-yy，我想要一种颜色。|1314588|xx+=+np.linspace(x_min,+x_max,+gridsize)+#+array+of+x+values
yy+=+np.linspace(y_min,+y_max,+gridsize)+#+array+of+y+values
grid+=+np.array(np.meshgrid(xx,+yy.T))
grid+=+grid.reshape(2,+grid.shape[1]*grid.shape[2]).T|1314589|为什么是奇怪的形状？scipy.griddata想要一个(n，D)的形状。|1314590|Griddata通过预定义的方法计算网格中每个点的一个值。我选择“最近的”-空的网格点将填充来自最近邻居的值。这看起来好像信息较少的区域具有较大的单元格(即使不是这样)。人们可以选择“线性”插值，这样信息较少的区域看起来就不那么清晰了。品味问题，真的。|1314591|points+=+np.array([x,+y]).T+#+because+griddata+wants+it+that+way
z_grid2+=+griddata(points,+z,+grid,+method='nearest')
#+you+get+a+1D+vector+as+result.+Reshape+to+picture+format!
z_grid2+=+z_grid2.reshape(xx.shape[0],+yy.shape[0])|1314592|然后跳转到matplotlib来显示图|1314593|fig+=+plt.figure(1,+figsize=(10,+10))
ax1+=+fig.add_subplot(111)
ax1.imshow(z_grid2,+extent=[x_min,+x_max,y_min,+y_max,++],
++++++++++++origin='lower',+cmap=cm.magma)
ax1.set_title("SVC:+empty+spots+filled+by+nearest+neighbours")
ax1.set_xlabel('log+gamma')
ax1.set_ylabel('log+C')
plt.show()|1314594|在V型的尖端部分，你看，我在寻找最佳位置的过程中做了很多计算，而几乎所有其他地方的不太有趣的部分都有较低的分辨率。|1314595|​|1314596|📷|atomic|1314597|1314598|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.interpolate.griddata.html|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/fbbc21ab9b2d9472ebbb9020c6548de9.png|imageAlt^0|P|B|0|0|0|0|0|0|0|0|0|1C|5|1I|5|0|0|0|0|A|E|0|0|0|0|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|22|8|@$9|23|A|24|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|25|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|26|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|27|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|28|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|29|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|2A|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|2B|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|2C|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|2D|8|@$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|2I|8|@]|D|@]|E|$]]|$1|12|3|13|5|6|7|2J|8|@]|D|@]|E|$]]|$1|14|3|15|5|H|7|2K|8|@]|D|@]|E|$I|J]]|$1|16|3|17|5|6|7|2L|8|@]|D|@$9|2M|A|2N|1|2O]]|E|$]]|$1|18|3|19|5|6|7|2P|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|H|7|2Q|8|@]|D|@]|E|$I|J]]|$1|1C|3|1D|5|6|7|2R|8|@]|D|@]|E|$]]|$1|1E|3|1F|5|H|7|2S|8|@]|D|@]|E|$I|J]]|$1|1G|3|1H|5|6|7|2T|8|@]|D|@]|E|$]]|$1|1I|3|1J|5|6|7|2U|8|@]|D|@]|E|$]]|$1|1K|3|1L|5|1M|7|2V|8|@]|D|@$9|2W|A|2X|1|2Y]]|E|$]]|$1|1N|3|1J|5|6|7|2Z|8|@]|D|@]|E|$]]|$1|1O|3|-4|5|6|7|30|8|@]|D|@]|E|$]]]|1P|$1Q|$5|1R|1S|1T|E|$1U|1V]]|1W|$5|1X|1S|1Y|E|$1Z|20|21|-4]]]]

and the initial question was... how to convert scatter values to grid values, right?
<code>histogram2d</code> does count the frequency per cell, however, if you have other data per cell than just the frequency, you'd need some additional work to do.

<pre><code>x = data_x # between -10 and 4, log-gamma of an svc
y = data_y # between -4 and 11, log-C of an svc
z = data_z #between 0 and 0.78, f1-values from a difficult dataset
</code></pre>

So, I have a dataset with Z-results for X and Y coordinates. However, I was calculating few points outside the area of interest (large gaps), and heaps of points in a small area of interest.

Yes here it becomes more difficult but also more fun. Some libraries (sorry):

<pre><code>from matplotlib import pyplot as plt
from matplotlib import cm
import numpy as np
from scipy.interpolate import griddata
</code></pre>

pyplot is my graphic engine today,
cm is a range of color maps with some initeresting choice.
numpy for the calculations,
and griddata for attaching values to a fixed grid.

The last one is important especially because the frequency of xy points is not equally distributed in my data. First, let's start with some boundaries fitting to my data and an arbitrary grid size. The original data has datapoints also outside those x and y boundaries.

<pre><code>#determine grid boundaries
gridsize = 500
x_min = -8
x_max = 2.5
y_min = -2
y_max = 7
</code></pre>

So we have defined a grid with 500 pixels between the min and max values of x and y.

In my data, there are lots more than the 500 values available in the area of high interest; whereas in the low-interest-area, there are not even 200 values in the total grid; between the graphic boundaries of <code>x_min</code> and <code>x_max</code> there are even less.

So for getting a nice picture, the task is to get an average for the high interest values and to fill the gaps elsewhere.

I define my grid now. For each xx-yy pair, i want to have a color.

<pre><code>xx = np.linspace(x_min, x_max, gridsize) # array of x values
yy = np.linspace(y_min, y_max, gridsize) # array of y values
grid = np.array(np.meshgrid(xx, yy.T))
grid = grid.reshape(2, grid.shape[1]*grid.shape[2]).T
</code></pre>

Why the strange shape? <a href="https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.interpolate.griddata.html" rel="nofollow noreferrer">scipy.griddata</a> wants a shape of (n, D).

Griddata calculates one value per point in the grid, by a predefined method.
I choose "nearest" - empty grid points will be filled with values from the nearest neighbor. This looks as if the areas with less information have bigger cells (even if it is not the case). One could choose to interpolate "linear", then areas with less information look less sharp. Matter of taste, really.

<pre><code>points = np.array([x, y]).T # because griddata wants it that way
z_grid2 = griddata(points, z, grid, method='nearest')
# you get a 1D vector as result. Reshape to picture format!
z_grid2 = z_grid2.reshape(xx.shape[0], yy.shape[0])
</code></pre>

And hop, we hand over to matplotlib to display the plot

<pre><code>fig = plt.figure(1, figsize=(10, 10))
ax1 = fig.add_subplot(111)
ax1.imshow(z_grid2, extent=[x_min, x_max,y_min, y_max, ],
 origin='lower', cmap=cm.magma)
ax1.set_title("SVC: empty spots filled by nearest neighbours")
ax1.set_xlabel('log gamma')
ax1.set_ylabel('log C')
plt.show()
</code></pre>

Around the pointy part of the V-Shape, you see I did a lot of calculations during my search for the sweet spot, whereas the less interesting parts almost everywhere else have a lower resolution.

<a href="https://i.stack.imgur.com/w5HYZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w5HYZ.png" alt="Heatmap of a SVC in high resolution"></a>

blocks|key|1314683|text|非常类似于@Piti's+answer，但使用1次调用而不是2次来生成点：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1314684|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()|code-block|syntax|javascript|1314685|输出：|1314686|​|1314687|📷|atomic|1314688|1314689|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/13233972/900394|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/78dc8dcb538b48ae2489765a5c1b731c.png|imageAlt^0|5|E|0|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|19|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|C|$]]|$1|M|3|N|5|O|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|P|3|L|5|6|7|1F|8|@]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|Z|U|10|C|$11|12|13|-4]]]]

Very similar to <a href="https://stackoverflow.com/a/13233972/900394">@Piti's answer</a>, but using 1 call instead of 2 to generate the points:

<pre><code>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()
</code></pre>

Output:

<a href="https://i.stack.imgur.com/76BQz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/76BQz.png" alt="2d_gaussian_heatmap"></a>

blocks|key|1099805|text|很抱歉我来晚了，但我刚才也有类似的问题。公认的答案(+@ptomato)帮了我很大的忙，但我也想把这篇文章贴出来，以防对某人有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099806|'''+I+wanted+to+create+a+heatmap+resembling+a+football+pitch+which+would+show+the+different+actions+performed+'''

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

#fixing+random+state+for+reproducibility
np.random.seed(1234324)

fig+=+plt.figure(12)
ax1+=+fig.add_subplot(121)
ax2+=+fig.add_subplot(122)

#Ratio+of+the+pitch+with+respect+to+UEFA+standards+
hmap=+np.full((6,+10),+0)
#print(hmap)

xlist+=+np.random.uniform(low=0.0,+high=100.0,+size=(20))
ylist+=+np.random.uniform(low=0.0,+high+=100.0,+size+=(20))

#UEFA+Pitch+Standards+are+105m+x+68m
xlist+=+(xlist/100)*10.5
ylist+=+(ylist/100)*6.5

ax1.scatter(xlist,ylist)

#int+of+the+co-ordinates+to+populate+the+array
xlist_int+=+xlist.astype+(int)
ylist_int+=+ylist.astype+(int)

#print(xlist_int,+ylist_int)

for+i,+j+in+zip(xlist_int,+ylist_int):
++++#this+populates+the+array+according+to+the+x,y+co-ordinate+values+it+encounters+
++++hmap[j][i]=+hmap[j][i]+%2B+1+++

#Reversing+the+rows+is+necessary+
hmap+=+hmap[::-1]

#print(hmap)
im+=+ax2.imshow(hmap)|code-block|syntax|javascript|1099807|结果如下所示|1099808|📷|atomic|offset|length|1099809|​|1099810|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/9b774829d6fe5063100e62cc0f422f69.png|imageAlt^0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|11|8|@]|9|@$L|12|M|13|1|14]]|A|$]]|$1|N|3|O|5|6|7|15|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W|X|-4]]]]

I'm afraid I'm a little late to the party but I had a similar question a while ago. The accepted answer (by @ptomato) helped me out but I'd also want to post this in case it's of use to someone. 

<pre><code>
''' I wanted to create a heatmap resembling a football pitch which would show the different actions performed '''

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

#fixing random state for reproducibility
np.random.seed(1234324)

fig = plt.figure(12)
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)

#Ratio of the pitch with respect to UEFA standards 
hmap= np.full((6, 10), 0)
#print(hmap)

xlist = np.random.uniform(low=0.0, high=100.0, size=(20))
ylist = np.random.uniform(low=0.0, high =100.0, size =(20))

#UEFA Pitch Standards are 105m x 68m
xlist = (xlist/100)*10.5
ylist = (ylist/100)*6.5

ax1.scatter(xlist,ylist)

#int of the co-ordinates to populate the array
xlist_int = xlist.astype (int)
ylist_int = ylist.astype (int)

#print(xlist_int, ylist_int)

for i, j in zip(xlist_int, ylist_int):
 #this populates the array according to the x,y co-ordinate values it encounters 
 hmap[j][i]= hmap[j][i] + 1 

#Reversing the rows is necessary 
hmap = hmap[::-1]

#print(hmap)
im = ax2.imshow(hmap)


</code></pre>

Here's the result
<a href="https://i.stack.imgur.com/nZL1I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nZL1I.png" alt="enter image description here"></a>

blocks|key|1099903|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099904|📷|atomic|offset|length|1099905|1099906|这是我用3个类别(红色，绿色和蓝色)在一百万点集上做的一个。如果你想尝试这个功能，这里有一个到存储库的链接。Github+Repo|1099907|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)|code-block|syntax|javascript|1099908|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/3be514a0e711f590f051e3e1a8c1173e.png|imageAlt|1|LINK|MUTABLE|url|https://github.com/JoelStansbury/Histplot/blob/master/Histplot.ipynb^0|0|0|1|0|0|0|1I|B|1|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@$E|14|F|15|1|16]]|A|$]]|$1|G|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@$E|19|F|1A|1|1B]]|A|$]]|$1|J|3|K|5|L|7|1C|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V|W|-4]]|X|$5|Y|S|Z|A|$10|11]]]]

<a href="https://i.stack.imgur.com/Fo08O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fo08O.png" alt="enter image description here"></a>

Here's one I made on a 1 Million point set with 3 categories (colored Red, Green, and Blue). Here's a link to the repository if you'd like to try the function. <a href="https://github.com/JoelStansbury/Histplot/blob/master/Histplot.ipynb" rel="nofollow noreferrer">Github Repo</a>

<pre><code>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)
</code></pre>

blocks|key|1099987|text|下面是使用scipy.cKDTree实现的Jurgy's+great+nearest+neighbour+approach。在我的测试中，它大约快了100倍。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1099988|​|1099989|📷|atomic|1099990|1099991|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+%25d+neighbours"+%25+neighbours)
++++++++ax.set_xlim(extent[0],+extent[1])
++++++++ax.set_ylim(extent[2],+extent[3])

plt.savefig('new.png',+dpi=150,+bbox_inches='tight')|code-block|syntax|javascript|1099992|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.cKDTree.html|1|https://stackoverflow.com/a/47350350/1391441|2|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/3d9d193fbf16bce2c56b49d1da98903a.png|imageAlt^0|5|D|0|L|14|1|0|0|0|1|2|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]|$A|18|B|19|1|1A]]|C|$]]|$1|D|3|E|5|6|7|1B|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|I|3|E|5|6|7|1G|8|@]|9|@]|C|$]]|$1|J|3|K|5|L|7|1H|8|@]|9|@]|C|$M|N]]|$1|O|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|Z|S|10|C|$11|12|13|-4]]]]

Here's <a href="https://stackoverflow.com/a/47350350/1391441">Jurgy's great nearest neighbour approach</a> but implemented using <a href="https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.cKDTree.html" rel="noreferrer">scipy.cKDTree</a>. In my tests it's about 100x faster.

<a href="https://i.stack.imgur.com/nWGHW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/nWGHW.png" alt="enter image description here"></a>

<pre><code>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')
</code></pre>

I have a set of X,Y data points (about 10k) that are easy to plot as a scatter plot but that I would like to represent as a heatmap.

I looked through the examples in MatPlotLib and they all seem to already start with heatmap cell values to generate the image.

Is there a method that converts a bunch of x,y, all different, to a heatmap (where zones with higher frequency of x,y would be "warmer")?

Generate a heatmap in MatPlotLib using a scatter data set

我有一组X，Y数据点(大约10k)，这些数据点很容易绘制为散点图，但我想将其表示为热图。我浏览了MatPlotLib中的示例，它们似乎都已经从热图单元格值开始生成图像。有没有一种方法可以将一堆完全不同的x，y转换成热图(其中x，y频率较高的区域将“变暖”)？

问使用散点数据集在MatPlotLib中生成热图
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