在Python中交叉两个图,找到x值:

Shi*_*ash 23 python matplotlib

设0 <= x <= 1.我有两列f和g,长度分别为5000.现在我的情节:

  plt.plot(x, f, '-')
  plt.plot(x, g, '*')
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我想找到曲线相交的点'x'.我不想找到f和g的交集.我可以简单地用以下方法做到:

   set(f) & set(g)
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Mat*_*att 70

您可以np.sign结合使用np.diffnp.argwhere获取线交叉的点的索引(在这种情况下,点是[ 0, 149, 331, 448, 664, 743]):

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(0, 1000)
f = np.arange(0, 1000)
g = np.sin(np.arange(0, 10, 0.01) * 2) * 1000

plt.plot(x, f, '-')
plt.plot(x, g, '-')

idx = np.argwhere(np.diff(np.sign(f - g))).flatten()
plt.plot(x[idx], f[idx], 'ro')
plt.show()
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交点图

首先它使用计算f - g和相应的符号np.sign.应用np.diff揭示了符号变化的所有位置(例如,线交叉).使用np.argwhere给出了确切的指数.

  • 那可能是因为有 10 个交叉点@ShivaPrakash (5认同)
  • `x [idx]`会给你相应的值. (2认同)
  • 谢谢,今天它为我节省了一些时间。我不确定您是否需要`reshape(-1)`,并且可以解释`+ 0'在做什么吗?另外,如果有人希望它可以处理具有'nan`值的数组,则此变体也应起作用:`idx = np.argwhere((np.diff(np.sign(f-g))!= 0)&np.isfinite (np.diff(np.sign(f-g))))` (2认同)

Geo*_*rgy 17

对于那些正在使用或开放使用Shapely库进行几何相关计算的人来说,获得交集会容易得多。你只需LineString要从每一行构建并得到它们intersection如下:

import numpy as np
import matplotlib.pyplot as plt
from shapely.geometry import LineString

x = np.arange(0, 1000)
f = np.arange(0, 1000)
g = np.sin(np.arange(0, 10, 0.01) * 2) * 1000

plt.plot(x, f)
plt.plot(x, g)

first_line = LineString(np.column_stack((x, f)))
second_line = LineString(np.column_stack((x, g)))
intersection = first_line.intersection(second_line)

if intersection.geom_type == 'MultiPoint':
    plt.plot(*LineString(intersection).xy, 'o')
elif intersection.geom_type == 'Point':
    plt.plot(*intersection.xy, 'o')
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在此处输入图片说明

要将xy值作为 NumPy 数组获取,您只需编写:

x, y = LineString(intersection).xy
# x: array('d', [0.0, 149.5724669847373, 331.02906176584617, 448.01182730277833, 664.6733061190541, 743.4822641140581])
# y: array('d', [0.0, 149.5724669847373, 331.02906176584617, 448.01182730277833, 664.6733061190541, 743.4822641140581])
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或者如果一个交点只有一个点:

x, y = intersection.xy
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bun*_*kus 5

好吧,我正在为两条大小不同且 x 值不同的曲线寻找 matplotlib。这是我想出的:

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

fig = plt.figure()
ax = fig.add_subplot(111)

# x1 = [1,2,3,4,5,6,7,8]
# y1 = [20,100,50,120,55,240,50,25]
# x2 = [3,4,5,6,7,8,9]
# y2 = [25,200,14,67,88,44,120]

x1=[1.4,2.1,3,5.9,8,9,12,15]
y1=[2.3,3.1,1,3.9,8,9,11,9]
x2=[1,2,3,4,6,8,9,12,14]
y2=[4,12,7,1,6.3,7,5,6,11]

ax.plot(x1, y1, color='lightblue',linewidth=3, marker='s')
ax.plot(x2, y2, color='darkgreen', marker='^')

y_lists = y1[:]
y_lists.extend(y2)
y_dist = max(y_lists)/200.0

x_lists = x1[:]
x_lists.extend(x2)  
x_dist = max(x_lists)/900.0
division = 1000
x_begin = min(x1[0], x2[0])     # 3
x_end = max(x1[-1], x2[-1])     # 8

points1 = [t for t in zip(x1, y1) if x_begin<=t[0]<=x_end]  # [(3, 50), (4, 120), (5, 55), (6, 240), (7, 50), (8, 25)]
points2 = [t for t in zip(x2, y2) if x_begin<=t[0]<=x_end]  # [(3, 25), (4, 35), (5, 14), (6, 67), (7, 88), (8, 44)]
# print points1
# print points2

x_axis = np.linspace(x_begin, x_end, division)
idx = 0
id_px1 = 0
id_px2 = 0
x1_line = []
y1_line = []
x2_line = []
y2_line = []
xpoints = len(x_axis)
intersection = []
while idx < xpoints:
    # Iterate over two line segments
    x = x_axis[idx]
    if id_px1>-1:
        if x >= points1[id_px1][0] and id_px1<len(points1)-1:
            y1_line = np.linspace(points1[id_px1][1], points1[id_px1+1][1], 1000) # 1.4 1.401 1.402 etc. bis 2.1
            x1_line = np.linspace(points1[id_px1][0], points1[id_px1+1][0], 1000)
            id_px1 = id_px1 + 1
            if id_px1 == len(points1):
                x1_line = []
                y1_line = []
                id_px1 = -1
    if id_px2>-1:
        if x >= points2[id_px2][0] and id_px2<len(points2)-1:
            y2_line = np.linspace(points2[id_px2][1], points2[id_px2+1][1], 1000)
            x2_line = np.linspace(points2[id_px2][0], points2[id_px2+1][0], 1000)
            id_px2 = id_px2 + 1
            if id_px2 == len(points2):
                x2_line = []
                y2_line = []
                id_px2 = -1
    if x1_line!=[] and y1_line!=[] and x2_line!=[] and y2_line!=[]:
        i = 0
        while abs(x-x1_line[i])>x_dist and i < len(x1_line)-1:
            i = i + 1
        y1_current = y1_line[i]
        j = 0
        while abs(x-x2_line[j])>x_dist and j < len(x2_line)-1:
            j = j + 1
        y2_current = y2_line[j]
        if abs(y2_current-y1_current)<y_dist and i != len(x1_line) and j != len(x2_line):
            ymax = max(y1_current, y2_current)
            ymin = min(y1_current, y2_current)
            xmax = max(x1_line[i], x2_line[j])
            xmin = min(x1_line[i], x2_line[j])
            intersection.append((x, ymin+(ymax-ymin)/2))
            ax.plot(x, y1_current, 'ro') # Plot the cross point
    idx += 1    
print "intersection points", intersection
plt.show()
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cry*_*ick 5

这是一个解决方案,它:

  • 处理N维数据
  • 使用欧几里得距离,而不仅仅是在y轴上找到交叉点
  • 在处理大量数据时效率更高(它查询KD-tree,它应该以对数时间而不是线性时间进行查询)。
  • 您可以distance_upper_bound在KD树查询中更改,以定义足够接近的程度。
  • 如果需要,您可以同时查询具有许多点的KD树。注意:如果您需要一次查询数千个点,则可以通过使用另一个KD-tree查询KD-tree来获得显着的性能提升。

在此处输入图片说明

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.spatial import cKDTree
from scipy import interpolate

fig = plt.figure()
ax = fig.add_axes([0, 0, 1, 1], projection='3d')
ax.axis('off')

def upsample_coords(coord_list):
    # s is smoothness, set to zero
    # k is degree of the spline. setting to 1 for linear spline
    tck, u = interpolate.splprep(coord_list, k=1, s=0.0)
    upsampled_coords = interpolate.splev(np.linspace(0, 1, 100), tck)
    return upsampled_coords

# target line
x_targ = [1, 2, 3, 4, 5, 6, 7, 8]
y_targ = [20, 100, 50, 120, 55, 240, 50, 25]
z_targ = [20, 100, 50, 120, 55, 240, 50, 25]
targ_upsampled = upsample_coords([x_targ, y_targ, z_targ])
targ_coords = np.column_stack(targ_upsampled)

# KD-tree for nearest neighbor search
targ_kdtree = cKDTree(targ_coords)

# line two
x2 = [3,4,5,6,7,8,9]
y2 = [25,35,14,67,88,44,120]
z2 = [25,35,14,67,88,44,120]
l2_upsampled = upsample_coords([x2, y2, z2])
l2_coords = np.column_stack(l2_upsampled)

# plot both lines
ax.plot(x_targ, y_targ, z_targ, color='black', linewidth=0.5)
ax.plot(x2, y2, z2, color='darkgreen', linewidth=0.5)

# find intersections
for i in range(len(l2_coords)):
    if i == 0:  # skip first, there is no previous point
        continue

    distance, close_index = targ_kdtree.query(l2_coords[i], distance_upper_bound=.5)

    # strangely, points infinitely far away are somehow within the upper bound
    if np.isinf(distance):
        continue

    # plot ground truth that was activated
    _x, _y, _z = targ_kdtree.data[close_index]
    ax.scatter(_x, _y, _z, 'gx')
    _x2, _y2, _z2 = l2_coords[i]
    ax.scatter(_x2, _y2, _z2, 'rx')  # Plot the cross point


plt.show()
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