matplotlib 散点图的帕累托前沿答案

【问题标题】：Pareto front for matplotlib scatter plotmatplotlib 散点图的帕累托前沿
【发布时间】：2021-07-07 18:54:22
【问题描述】：

我正在尝试将帕累托前面添加到我拥有的散点图中。散点图数据为：

array([[1.44100000e+04, 3.31808987e+07],
       [1.21250000e+04, 3.22901074e+07],
       [6.03000000e+03, 2.84933900e+07],
       [8.32500000e+03, 2.83091317e+07],
       [6.68000000e+03, 2.56373373e+07],
       [5.33500000e+03, 1.89331461e+07],
       [3.87500000e+03, 1.84107940e+07],
       [3.12500000e+03, 1.60416570e+07],
       [6.18000000e+03, 1.48054565e+07],
       [4.62500000e+03, 1.33395341e+07],
       [5.22500000e+03, 1.23150492e+07],
       [3.14500000e+03, 1.20244820e+07],
       [6.79500000e+03, 1.19525083e+07],
       [2.92000000e+03, 9.18176770e+06],
       [5.45000000e+02, 5.66882578e+06]])

散点图如下所示：

我用这个tutorial 来绘制帕累托图，但由于某种原因，结果很奇怪，我得到了一条小红线：

这是我使用的代码：

def identify_pareto(scores):
    # Count number of items
    population_size = scores.shape[0]

    # Create a NumPy index for scores on the pareto front (zero indexed)
    population_ids = np.arange(population_size)

    # Create a starting list of items on the Pareto front
    # All items start off as being labelled as on the Parteo front
    pareto_front = np.ones(population_size, dtype=bool)
    print(pareto_front)
    # Loop through each item. This will then be compared with all other items
    for i in range(population_size):
        
        # Loop through all other items
        for j in range(population_size):
            
            # Check if our 'i' pint is dominated by out 'j' point
            if all(scores[j] >= scores[i]) and any(scores[j] > scores[i]):
               
                # j dominates i. Label 'i' point as not on Pareto front
                pareto_front[i] = 0
                # Stop further comparisons with 'i' (no more comparisons needed)
                break
    # Return ids of scenarios on pareto front
    return population_ids[pareto_front]


pareto = identify_pareto(scores)

pareto_front_df = pd.DataFrame(pareto_front)
pareto_front_df.sort_values(0, inplace=True)
pareto_front = pareto_front_df.values

#here I get as output weird results:
>>>
array([[ 5, 81],
       [15, 80],
       [30, 79],
       [55, 77],
       [70, 65],
       [80, 60],
       [90, 40],
       [97, 23],
       [99,  4]])

x_all = scores[:, 0]
y_all = scores[:, 1]
x_pareto = pareto_front[:, 0]
y_pareto = pareto_front[:, 1]

plt.scatter(x_all, y_all)
plt.plot(x_pareto, y_pareto, color='r')
plt.xlabel('Objective A')
plt.ylabel('Objective B')
plt.show()

结果是小红线。

我的问题是，我的错误在哪里？我怎样才能回到帕累托线？

【问题讨论】：

标签： python numpy matplotlib pareto-chart

【解决方案1】：

我认为您的代码没有任何问题，而是您的数据由分数表示的方式（如果分数是您呈现的第一个数组）。

数组[1.44100000e+04, 3.31808987e+07] 的第一个元素与其他值相比确实很大，因此它是函数内部唯一的外部迭代，其中if all(scores[j] >= scores[i]) and any(scores[j] > scores[i]): 条件不满足且未减少为零。所有其他点都减少到零。

我相信这是唯一用红点绘制的点。

【讨论】：