对数刻度中的Seaborn条形图

Question

我正在使用pyplot.boxplots在logscale中绘制boxplot，然后我想使用seaborn在其上进行swarmplot / strip绘图。 但是，水手们把我们整个范围弄得一团糟，并且阴谋诡计。知道如何在seaborn中自定义职位吗？

att1= np.sort(np.unique(df1.att1))

w = 0.085
width = lambda p, w: 10 ** (np.log10(p) + w / 2.) - 10 ** (np.log10(p) - w / 2.)
custom_widths = width(freqns, w)


ax.boxplot([df1[df1.att1== xi].att2 for xi in att1], positions=att1,
           boxprops={'facecolor': 'none'}, medianprops={'color': 'black'}, patch_artist=True,
           widths=custom_widths)
ax.set_xscale("log")
fig.set_size_inches(10.5, 8)
means = [np.median(df1[df1.Frequency == xi].CapDensity) for xi in freqs]
plt.plot(freqns, means, '--k*', lw=1.2)

这是不含条形图的图像：

sns.stripplot(x="Frequency", y="CapDensity",data=df1, edgecolor="black", linewidth=.3, jitter=0.1, zorder=0.5, ax=ax)

这是我在boxplot顶部剥离图的时候。

Answer 1

问题是您将0,1,2,3,...用作箱线图的位置，而seaborn始终会在内部位置from matplotlib import pyplot as plt import numpy as np import pandas as pd df = pd.DataFrame({'x': np.random.choice([1, 3, 5, 8, 10, 30, 50, 100], 500), 'y': np.random.normal(750, 20, 500)}) xvals = np.unique(df.x) w = 0.085 width = lambda p, w: 10 ** (np.log10(p) + w / 2.) - 10 ** (np.log10(p) - w / 2.) custom_widths = width(xvals, w) fig, ax = plt.subplots(figsize=(12, 4)) ax.set_xscale('log') ax.boxplot([df[df.x == xi].y for xi in xvals], positions=xvals, showfliers=False, boxprops={'facecolor': 'none'}, medianprops={'color': 'black'}, patch_artist=True, widths=custom_widths) medians = [np.median(df[df.x == xi].y) for xi in xvals] ax.plot(xvals, medians, '--k*', lw=2) ax.set_xticks(xvals) for xi, wi in zip(xvals, custom_widths): yis = df[df.x == xi].y ax.scatter(xi + np.random.uniform(-wi / 2, wi / 2, yis.size), yis) plt.show() 上放置一个简易绘图。最简单的解决方案是通过matplotlib创建剥离图。 （一个小样创建起来要复杂得多。）

假设您的数据与上一个问题类似，则可以将这样的图创建为：

def rows_generator(df):
    i = 0
    while (i+3) <= df.shape[0]:
        yield df.iloc[i:(i+3):1, :]
        i += 1

i = 1
for df in rows_generator(df):
    print(f'Time #{i}')
    print(df)
    i += 1