如何格式化seaborn / matplotlib轴刻度标签从数字到成千上万？ （125436至125.4K）

Question

import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
import seaborn as sns
import pandas as pd
sns.set(style="darkgrid")    
fig, ax = plt.subplots(figsize=(8, 5))    
palette = sns.color_palette("bright", 6)
g = sns.scatterplot(ax=ax, x="Area", y="Rent/Sqft", hue="Region", marker='o', data=df, s=100, palette= palette)
g.legend(bbox_to_anchor=(1, 1), ncol=1)
g.set(xlim = (50000,250000))

如何将轴格式从数字更改为自定义格式？例如125000至125.00K

Answer 1

格式化标准单位中的刻度标签的规范方法是使用EngFormatter。 matplotlib文档中也有an example。

在这里看起来可能如下。

import numpy as np; np.random.seed(42)
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
import seaborn as sns
import pandas as pd

df = pd.DataFrame({"xaxs" : np.random.randint(50000,250000, size=20),
                   "yaxs" : np.random.randint(7,15, size=20),
                   "col"  : np.random.choice(list("ABC"), size=20)})

fig, ax = plt.subplots(figsize=(8, 5))    
palette = sns.color_palette("bright", 6)
sns.scatterplot(ax=ax, x="xaxs", y="yaxs", hue="col", data=df, 
                marker='o', s=100, palette="magma")
ax.legend(bbox_to_anchor=(1, 1), ncol=1)
ax.set(xlim = (50000,250000))

ax.xaxis.set_major_formatter(ticker.EngFormatter())

plt.show()

Answer 2

IIUC，您可以格式化xticks并进行设置：

In[60]:
#generate some psuedo data
df = pd.DataFrame({'num':[50000, 75000, 100000, 125000], 'Rent/Sqft':np.random.randn(4), 'Region':list('abcd')})
df

Out[60]: 
      num  Rent/Sqft Region
0   50000   0.109196      a
1   75000   0.566553      b
2  100000  -0.274064      c
3  125000  -0.636492      d

In[61]:
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
import seaborn as sns
import pandas as pd
sns.set(style="darkgrid")    
fig, ax = plt.subplots(figsize=(8, 5))    
palette = sns.color_palette("bright", 4)
g = sns.scatterplot(ax=ax, x="num", y="Rent/Sqft", hue="Region", marker='o', data=df, s=100, palette= palette)
g.legend(bbox_to_anchor=(1, 1), ncol=1)
g.set(xlim = (50000,250000))
xlabels = ['{:,.2f}'.format(x) + 'K' for x in g.get_xticks()/1000]
g.set_xticklabels(xlabels)

Out[61]: 

这里的关键是这一行：

xlabels = ['{:,.2f}'.format(x) + 'K' for x in g.get_xticks()/1000]
g.set_xticklabels(xlabels)

因此，这会将所有刻度线除以1000，然后格式化它们并设置xtick标签

更新 感谢@ScottBoston提出了一种更好的方法：

ax.xaxis.set_major_formatter(ticker.FuncFormatter(lambda x, pos: '{:,.2f}'.format(x/1000) + 'K'))

请参见docs

Answer 3

这是我要解决的方法：（类似于ScottBoston）

from matplotlib.ticker import FuncFormatter

f = lambda x, pos: f'{x/10**3:,.0f}K'
ax.xaxis.set_major_formatter(FuncFormatter(f))

Answer 4

使用 Seaborn 而没有导入 matplotlib ：

import seaborn as sns
sns.set()

chart = sns.relplot(x="x_val", y="y_val", kind="line", data=my_data)

ticks = chart.axes[0][0].get_xticks()

xlabels = ['$' + '{:,.0f}'.format(x) for x in ticks]

chart.set_xticklabels(xlabels)
chart.fig

谢谢EdChum的回答，让我有90％的经验。