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如何用Python绘出三值等值线图?

plotly python

Martin Drozdik rlbond · 技术社区 · 6 年前

但我不知道如何使用pythonapi实现这一点。引用有一个名为 Ternary Contour Plots in Python 但它不绘制数值函数,而是将单纯形映射到给定离散集的函数。

import plotly as py

A = [0, .2, .2, .2, 0, .6, .75, .9, 0, 1, .8, .3]
B = [1, .2, .4, .1, 0, .4, .05, 0, .8, 0, .05, .3]
C = [0, .6, .4, .7, 1, 0, .2, .1, .2, 0, .15, .4]
Z = [.1, .5, .1, .2, 1, .8, .4, 0, .1, .6, 1, .7]

trace = {
    "type": 'scatterternary',
    "carpet": 'scattercontour',
    "a": A,
    "b": B,
    "c": C,
    "z": Z
}

layout = {
    'title': 'Simple Ternary Contour Plot with Python'
}

figure = dict(data=[trace], layout=layout)
py.offline.plot(figure, validate=False)

但我得到的却是这样的情节:

2 回复 | 直到 6 年前

petem 6 年前

如果有人还感兴趣, here 是一个Jupyter笔记本,看看如何生成一个绘出的三元等高线图。

Martin Drozdik rlbond 6 年前

如果可以生成要绘制任意比例轮廓的数据,则有一种解决方法。在这种情况下,你可以用三元坐标来计算曲面坐标。

import plotly as py
import plotly.graph_objs as go
import numpy as np


def compute_in_barycentric_coordinates(a, b, c):
    epsilon = 1e-6
    if 0.0 < a < 1.0 and 0.0 < b < 1.0 and 0.0 < c < 1.0 and abs(a + b + c - 1.0) < epsilon:
        return pow(a, 2) * pow(b, 1.5) * pow(c, 1.8)
    return None


x_data = np.linspace(0, 1, 101)
y_data = np.linspace(0, 1, 101)
z_data = []
f = 1.0 / pow(3.0, 0.5)
for y in y_data:
    next_slice = []
    for x in x_data:
        a = 1 - x - f*y
        b = x - f*y
        c = 2*f*y
        z = compute_in_barycentric_coordinates(a, b, c)
        next_slice.append(z)
    z_data.append(next_slice)


data = [
    go.Contour(
        x=x_data,
        y=y_data,
        z=z_data,
    )
]


py.offline.plot(data)