如何设置散点图的 bin 坐标
How to set bin coordinates for a scatter plot
我要return占据特定区域的散点数。通常，我会使用 2dhistogram 和 pcolormesh.
但是如果我想设置代表不代表网格的不规则尺寸的 bin 坐标，我该怎么做？
下面是我的数据集的示例。
import matplotlib.pyplot as plt
import matplotlib as mpl
import math
import numpy as np
x1 = np.random.randint(80, size=(400, 10))
y1 = np.random.randint(80, size=(400, 10))
x2 = np.random.randint(80, size=(400, 10))
y2 = np.random.randint(80, size=(400, 10))
fig, ax = plt.subplots()
ax.grid(False)
plt.scatter(x1[0],y1[0], c = 'r', zorder = 2)
plt.scatter(x2[0],y2[0], c = 'b', zorder = 2)
ang1 = 0, 50
ang2 = 100, 50
angle = math.degrees(math.acos(5.5/9.15))
xy = 50, 50
Halfway = mpl.lines.Line2D((50,50), (0,100), c = 'white')
arc1 = mpl.patches.Arc(ang1, 65, 100, angle = 0, theta2 = angle, theta1 = 360-angle, lw = 2)
arc2 = mpl.patches.Arc(ang2, 65, 100, angle = 0, theta2 = 180+angle, theta1 = 180-angle, lw = 2)
Oval = mpl.patches.Ellipse(xy, 100, 100, lw = 3, alpha = 0.1)
ax.add_line(Halfway)
ax.add_patch(arc1)
ax.add_patch(arc2)
ax.add_patch(Oval)
plt.text(15, 75, '1', fontsize = 8)
plt.text(35, 90, '2', fontsize = 8)
plt.text(65, 90, '3', fontsize = 8)
plt.text(85, 75, '4', fontsize = 8)
ax.autoscale()
plt.draw()
我要设置的 bin 标记为 1-4。是否可以设置 return 这些垃圾箱的坐标？
如果我可以设置这些坐标，那么我想return每个散点占据的bin。
输出：
更新：
如果我想要在散点图中的每一行的每个箱中显示 xy 的导出，我会写出 (x1[0], y1[0]) 并将数据转置为 return:
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
然后我将(x1[0], y1[0])更改为(x1[1], y1[1])以获得第二行数据。
1 2 3 4
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
然后我会结合这些来创建：
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
我有 1000 行，所以我正在尝试创建一种方法来使用整个 (x1, y1) 为每行数据生成每个 bin 中的坐标。
预期输出：
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
2 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
3 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
4 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
5....
6....
如果我尝试 (x1, y1) 我得到错误：
err = (arc_vertices[:,0] - x)**2 + (arc_vertices[:,1] - y)**2 ValueError: operands could not be broadcast together with shapes (70,) (10,)
我真的不喜欢这种方法。使用数据的 x 坐标计算点的 y 坐标落在曲线上的位置似乎更好。
此方法的工作原理类似，但使用弧的有限顶点：
arc1v = ax.transData.inverted().transform(arc1.get_verts())
arc2v = ax.transData.inverted().transform(arc2.get_verts())
for (x,y) in zip(x1[0], y1[0]):
err = (arc1v[:,0] - x)**2 + (arc1v[:,1] - y)**2
nearest = (arc1v[err == min(err)])[0]
line_x = (x, nearest[0])
line_y = (y, nearest[1])
ax.add_line(mpl.lines.Line2D(line_x, line_y))
if x > nearest[0]:
ax.scatter(x, y, marker='^', s=100, c='k', zorder=1)
else:
ax.scatter(x, y, marker='v', s=100, c='k', zorder=1)
此 "labels" 指向（左）曲线的左侧，带有一个向下的三角形，并指向它的右侧，带有一个向上的三角形。图表上的线条指向曲线上最近定义的顶点，仅供说明之用。
您也可以对另一条曲线执行此操作，bin 2/3 划分很简单。
这是一个示例输出图：
更新：
这里有一个更完整的答案：
import matplotlib.pyplot as plt
import matplotlib as mpl
import math
import numpy as np
BIN_23_X = 50 # The separator between bin 2 and 3
x1 = np.random.randint(80, size=(400, 10))
y1 = np.random.randint(80, size=(400, 10))
x2 = np.random.randint(80, size=(400, 10))
y2 = np.random.randint(80, size=(400, 10))
fig, ax = plt.subplots()
ax.grid(False)
plt.scatter(x1[0],y1[0], c = 'r', zorder = 2)
plt.scatter(x2[0],y2[0], c = 'b', zorder = 2)
ang1 = 0, 50
ang2 = 100, 50
angle = math.degrees(math.acos(5.5/9.15))
xy = 50, 50
Halfway = mpl.lines.Line2D((BIN_23_X,BIN_23_X), (0,100), c = 'white')
arc1 = mpl.patches.Arc(ang1, 65, 100, angle = 0, theta2 = angle, theta1 = 360-angle, lw = 2)
arc2 = mpl.patches.Arc(ang2, 65, 100, angle = 0, theta2 = 180+angle, theta1 = 180-angle, lw = 2)
Oval = mpl.patches.Ellipse(xy, 100, 100, lw = 3, alpha = 0.1)
ax.add_line(Halfway)
ax.add_patch(arc1)
ax.add_patch(arc2)
ax.add_patch(Oval)
plt.text(15, 75, '1', fontsize = 8)
plt.text(35, 90, '2', fontsize = 8)
plt.text(65, 90, '3', fontsize = 8)
plt.text(85, 75, '4', fontsize = 8)
# Classification helpers
def get_nearest_arc_vert(x, y, arc_vertices):
err = (arc_vertices[:,0] - x)**2 + (arc_vertices[:,1] - y)**2
nearest = (arc_vertices[err == min(err)])[0]
return nearest
arc1v = ax.transData.inverted().transform(arc1.get_verts())
arc2v = ax.transData.inverted().transform(arc2.get_verts())
def classify_pointset(vx, vy):
bins = {(k+1):[] for k in range(4)}
for (x,y) in zip(vx, vy):
nx1, ny1 = get_nearest_arc_vert(x, y, arc1v)
nx2, ny2 = get_nearest_arc_vert(x, y, arc2v)
if x < nx1: # Is this point in bin 1? To the left of arc1?
bins[1].append((x,y))
elif x > nx2: # Is this point in bin 4? To the right of arc2?
bins[4].append((x,y))
else:
# If we get here, the point is in either bin 2 or 3. We'll consider points
# that fall on the line to be in bin 3.
if x < BIN_23_X: # Is this point to the left BIN_23_X? => Bin 2
bins[2].append((x,y))
else: # Otherwise, the point is in Bin 3
bins[3].append((x,y))
return bins
# Classify points
bins_red = classify_pointset(x1[0], y1[0])
bins_blue = classify_pointset(x2[0], y2[0])
# Display classifications
print("Red:")
for x in bins_red.items():
print(" ", x)
print("Blue:")
for x in bins_blue.items():
print(" ", x)
# "Annotate" classifications
for (x,y) in (bins_red[1] + bins_blue[1]):
ax.scatter(x, y, marker='^', s=100, c='k', zorder=1)
for (x,y) in (bins_red[2] + bins_blue[2]):
ax.scatter(x, y, marker='v', s=100, c='k', zorder=1)
for (x,y) in (bins_red[3] + bins_blue[3]):
ax.scatter(x, y, marker='^', s=100, c='y', zorder=1)
for (x,y) in (bins_red[4] + bins_blue[4]):
ax.scatter(x, y, marker='v', s=100, c='y', zorder=1)
ax.autoscale()
plt.draw()
plt.show()
产生：
在这里，点是 "annotated"，它们后面的形状对应于它们被分类到的 bins：
Bin Anno. Color Triangle Pointing
-------------------------------------------
Bin 1 Black Up
Bin 2 Black Down
Bin 3 Yellow Up
Bin 4 Yellow Down
该代码还显示了分类结果（classify_pointset 的输出是一个字典，以 bin 编号 (1-4) 为键，值是在 bin 中找到的点的点坐标：
Red:
(1, [(14, 30), (4, 18), (12, 48)])
(2, [(49, 41)])
(3, [(62, 79), (50, 7), (68, 19), (71, 1), (59, 27), (77, 0)])
(4, [])
Blue:
(1, [(20, 74), (11, 17), (12, 75)])
(2, [(41, 19), (30, 15)])
(3, [(61, 75)])
(4, [(79, 73), (69, 58), (76, 34), (78, 65)])
您不必以图形方式注释图形，它只是用于说明，您可以只使用 classify_pointset（bins_red 和 bins_blue）返回的字典。
更新 2
以下代码生成一个列表列表（仍为 1 索引），因此您可以通过访问 all_points[1] 找到 bin 1 中的所有点（红色和蓝色）。 all_points 列表中的第一个元素（索引 0）是 None，因为我们将列表保持为 1 索引。
# Generate a list of lists, the outer index corresponds to the bin number (1-indexed)
all_points = [None] * 5
for bin_key in [1,2,3,4]:
all_points[bin_key] = bins_red[bin_key] + bins_blue[bin_key]
# Just for display.
for bin_key, bin_points in enumerate(all_points):
print(bin_key, bin_points)
输出：
0 None
1 [(1, 8), (16, 72), (23, 67), (12, 19), (24, 51), (24, 47), (15, 23), (18, 51)]
2 [(39, 75), (35, 27), (48, 55), (45, 53), (45, 22)]
3 [(66, 58), (55, 64), (70, 1), (71, 15), (73, 3), (71, 75)]
4 [(74, 62)]
我要return占据特定区域的散点数。通常，我会使用 2dhistogram 和 pcolormesh.
但是如果我想设置代表不代表网格的不规则尺寸的 bin 坐标，我该怎么做？
下面是我的数据集的示例。
import matplotlib.pyplot as plt
import matplotlib as mpl
import math
import numpy as np
x1 = np.random.randint(80, size=(400, 10))
y1 = np.random.randint(80, size=(400, 10))
x2 = np.random.randint(80, size=(400, 10))
y2 = np.random.randint(80, size=(400, 10))
fig, ax = plt.subplots()
ax.grid(False)
plt.scatter(x1[0],y1[0], c = 'r', zorder = 2)
plt.scatter(x2[0],y2[0], c = 'b', zorder = 2)
ang1 = 0, 50
ang2 = 100, 50
angle = math.degrees(math.acos(5.5/9.15))
xy = 50, 50
Halfway = mpl.lines.Line2D((50,50), (0,100), c = 'white')
arc1 = mpl.patches.Arc(ang1, 65, 100, angle = 0, theta2 = angle, theta1 = 360-angle, lw = 2)
arc2 = mpl.patches.Arc(ang2, 65, 100, angle = 0, theta2 = 180+angle, theta1 = 180-angle, lw = 2)
Oval = mpl.patches.Ellipse(xy, 100, 100, lw = 3, alpha = 0.1)
ax.add_line(Halfway)
ax.add_patch(arc1)
ax.add_patch(arc2)
ax.add_patch(Oval)
plt.text(15, 75, '1', fontsize = 8)
plt.text(35, 90, '2', fontsize = 8)
plt.text(65, 90, '3', fontsize = 8)
plt.text(85, 75, '4', fontsize = 8)
ax.autoscale()
plt.draw()
我要设置的 bin 标记为 1-4。是否可以设置 return 这些垃圾箱的坐标？
如果我可以设置这些坐标，那么我想return每个散点占据的bin。 输出：
更新：
如果我想要在散点图中的每一行的每个箱中显示 xy 的导出，我会写出 (x1[0], y1[0]) 并将数据转置为 return:
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
然后我将(x1[0], y1[0])更改为(x1[1], y1[1])以获得第二行数据。
1 2 3 4
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
然后我会结合这些来创建：
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
我有 1000 行，所以我正在尝试创建一种方法来使用整个 (x1, y1) 为每行数据生成每个 bin 中的坐标。
预期输出：
1 2 3 4
0 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
1 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
2 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
3 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
4 [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)] [(x,y),(x,y)]
5....
6....
如果我尝试 (x1, y1) 我得到错误：
err = (arc_vertices[:,0] - x)**2 + (arc_vertices[:,1] - y)**2 ValueError: operands could not be broadcast together with shapes (70,) (10,)
我真的不喜欢这种方法。使用数据的 x 坐标计算点的 y 坐标落在曲线上的位置似乎更好。
此方法的工作原理类似，但使用弧的有限顶点：
arc1v = ax.transData.inverted().transform(arc1.get_verts())
arc2v = ax.transData.inverted().transform(arc2.get_verts())
for (x,y) in zip(x1[0], y1[0]):
err = (arc1v[:,0] - x)**2 + (arc1v[:,1] - y)**2
nearest = (arc1v[err == min(err)])[0]
line_x = (x, nearest[0])
line_y = (y, nearest[1])
ax.add_line(mpl.lines.Line2D(line_x, line_y))
if x > nearest[0]:
ax.scatter(x, y, marker='^', s=100, c='k', zorder=1)
else:
ax.scatter(x, y, marker='v', s=100, c='k', zorder=1)
此 "labels" 指向（左）曲线的左侧，带有一个向下的三角形，并指向它的右侧，带有一个向上的三角形。图表上的线条指向曲线上最近定义的顶点，仅供说明之用。
您也可以对另一条曲线执行此操作，bin 2/3 划分很简单。
这是一个示例输出图：
更新：
这里有一个更完整的答案：
import matplotlib.pyplot as plt
import matplotlib as mpl
import math
import numpy as np
BIN_23_X = 50 # The separator between bin 2 and 3
x1 = np.random.randint(80, size=(400, 10))
y1 = np.random.randint(80, size=(400, 10))
x2 = np.random.randint(80, size=(400, 10))
y2 = np.random.randint(80, size=(400, 10))
fig, ax = plt.subplots()
ax.grid(False)
plt.scatter(x1[0],y1[0], c = 'r', zorder = 2)
plt.scatter(x2[0],y2[0], c = 'b', zorder = 2)
ang1 = 0, 50
ang2 = 100, 50
angle = math.degrees(math.acos(5.5/9.15))
xy = 50, 50
Halfway = mpl.lines.Line2D((BIN_23_X,BIN_23_X), (0,100), c = 'white')
arc1 = mpl.patches.Arc(ang1, 65, 100, angle = 0, theta2 = angle, theta1 = 360-angle, lw = 2)
arc2 = mpl.patches.Arc(ang2, 65, 100, angle = 0, theta2 = 180+angle, theta1 = 180-angle, lw = 2)
Oval = mpl.patches.Ellipse(xy, 100, 100, lw = 3, alpha = 0.1)
ax.add_line(Halfway)
ax.add_patch(arc1)
ax.add_patch(arc2)
ax.add_patch(Oval)
plt.text(15, 75, '1', fontsize = 8)
plt.text(35, 90, '2', fontsize = 8)
plt.text(65, 90, '3', fontsize = 8)
plt.text(85, 75, '4', fontsize = 8)
# Classification helpers
def get_nearest_arc_vert(x, y, arc_vertices):
err = (arc_vertices[:,0] - x)**2 + (arc_vertices[:,1] - y)**2
nearest = (arc_vertices[err == min(err)])[0]
return nearest
arc1v = ax.transData.inverted().transform(arc1.get_verts())
arc2v = ax.transData.inverted().transform(arc2.get_verts())
def classify_pointset(vx, vy):
bins = {(k+1):[] for k in range(4)}
for (x,y) in zip(vx, vy):
nx1, ny1 = get_nearest_arc_vert(x, y, arc1v)
nx2, ny2 = get_nearest_arc_vert(x, y, arc2v)
if x < nx1: # Is this point in bin 1? To the left of arc1?
bins[1].append((x,y))
elif x > nx2: # Is this point in bin 4? To the right of arc2?
bins[4].append((x,y))
else:
# If we get here, the point is in either bin 2 or 3. We'll consider points
# that fall on the line to be in bin 3.
if x < BIN_23_X: # Is this point to the left BIN_23_X? => Bin 2
bins[2].append((x,y))
else: # Otherwise, the point is in Bin 3
bins[3].append((x,y))
return bins
# Classify points
bins_red = classify_pointset(x1[0], y1[0])
bins_blue = classify_pointset(x2[0], y2[0])
# Display classifications
print("Red:")
for x in bins_red.items():
print(" ", x)
print("Blue:")
for x in bins_blue.items():
print(" ", x)
# "Annotate" classifications
for (x,y) in (bins_red[1] + bins_blue[1]):
ax.scatter(x, y, marker='^', s=100, c='k', zorder=1)
for (x,y) in (bins_red[2] + bins_blue[2]):
ax.scatter(x, y, marker='v', s=100, c='k', zorder=1)
for (x,y) in (bins_red[3] + bins_blue[3]):
ax.scatter(x, y, marker='^', s=100, c='y', zorder=1)
for (x,y) in (bins_red[4] + bins_blue[4]):
ax.scatter(x, y, marker='v', s=100, c='y', zorder=1)
ax.autoscale()
plt.draw()
plt.show()
产生：
在这里，点是 "annotated"，它们后面的形状对应于它们被分类到的 bins：
Bin Anno. Color Triangle Pointing ------------------------------------------- Bin 1 Black Up Bin 2 Black Down Bin 3 Yellow Up Bin 4 Yellow Down
该代码还显示了分类结果（classify_pointset 的输出是一个字典，以 bin 编号 (1-4) 为键，值是在 bin 中找到的点的点坐标：
Red: (1, [(14, 30), (4, 18), (12, 48)]) (2, [(49, 41)]) (3, [(62, 79), (50, 7), (68, 19), (71, 1), (59, 27), (77, 0)]) (4, []) Blue: (1, [(20, 74), (11, 17), (12, 75)]) (2, [(41, 19), (30, 15)]) (3, [(61, 75)]) (4, [(79, 73), (69, 58), (76, 34), (78, 65)])
您不必以图形方式注释图形，它只是用于说明，您可以只使用 classify_pointset（bins_red 和 bins_blue）返回的字典。
更新 2
以下代码生成一个列表列表（仍为 1 索引），因此您可以通过访问 all_points[1] 找到 bin 1 中的所有点（红色和蓝色）。 all_points 列表中的第一个元素（索引 0）是 None，因为我们将列表保持为 1 索引。
# Generate a list of lists, the outer index corresponds to the bin number (1-indexed)
all_points = [None] * 5
for bin_key in [1,2,3,4]:
all_points[bin_key] = bins_red[bin_key] + bins_blue[bin_key]
# Just for display.
for bin_key, bin_points in enumerate(all_points):
print(bin_key, bin_points)
输出：
0 None 1 [(1, 8), (16, 72), (23, 67), (12, 19), (24, 51), (24, 47), (15, 23), (18, 51)] 2 [(39, 75), (35, 27), (48, 55), (45, 53), (45, 22)] 3 [(66, 58), (55, 64), (70, 1), (71, 15), (73, 3), (71, 75)] 4 [(74, 62)]