python - 绘制连接两个 swarmplots 的各个数据点的彩色线

Question

我有:

import numpy as np
import pandas as pd
import seaborn as sb
import matplotlib.pyplot as plt

# Generate random data
set1 = np.random.randint(0, 40, 24)
set2 = np.random.randint(0, 100, 24)

# Put into dataframe and plot
df = pd.DataFrame({'set1': set1, 'set2': set2})
data = pd.melt(df)
sb.swarmplot(data=data, x='variable', y='value')

用 seaborn 的 swarmplot 函数绘制的两个随机分布:

我希望两个分布的各个图都用彩色线连接，以便数据框中第 1 组的第一个数据点与第 2 组的第一个数据点连接。我意识到如果没有 seaborn，这可能会相对简单，但我想保留各个数据点不重叠的特性。有什么方法可以访问 seaborn swarmfunction 中的各个地 block 坐标吗？

Answer 1

编辑:感谢@Mead，他在 2021 年 8 月 23 日之前指出了我帖子中的一个错误(我忘了在之前的版本中对位置进行排序)。

我通过 Paul Brodersen 给出了很好的答案试一试，尽管他这么说

Madness lies this way

...实际上我认为它非常简单并且产生了不错的结果:

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import seaborn as sns

# Generate random data
rng = np.random.default_rng(42)
set1 = rng.integers(0, 40, 5)
set2 = rng.integers(0, 100, 5)

# Put into dataframe
df = pd.DataFrame({"set1": set1, "set2": set2})
print(df)
data = pd.melt(df)

# Plot
fig, ax = plt.subplots()
sns.swarmplot(data=data, x="variable", y="value", ax=ax)

# Now connect the dots
# Find idx0 and idx1 by inspecting the elements return from ax.get_children()
# ... or find a way to automate it
idx0 = 0
idx1 = 1
locs1 = ax.get_children()[idx0].get_offsets()
locs2 = ax.get_children()[idx1].get_offsets()

# before plotting, we need to sort so that the data points
# correspond to each other as they did in "set1" and "set2"
sort_idxs1 = np.argsort(set1)
sort_idxs2 = np.argsort(set2)

# revert "ascending sort" through sort_idxs2.argsort(),
# and then sort into order corresponding with set1
locs2_sorted = locs2[sort_idxs2.argsort()][sort_idxs1]

for i in range(locs1.shape[0]):
    x = [locs1[i, 0], locs2_sorted[i, 0]]
    y = [locs1[i, 1], locs2_sorted[i, 1]]
    ax.plot(x, y, color="black", alpha=0.1)

它打印:

   set1  set2
0     3    85
1    30     8
2    26    69
3    17    20
4    17     9

并且您可以看到数据在图中相应地链接。