如何在 Bokeh 中使用自定义标签作为刻度？

Question

我了解如何指定在 Bokeh 中显示的特定刻度，但我的问题是是否有一种方法可以分配特定的标签来显示与位置。例如

plot.xaxis[0].ticker=FixedTicker(ticks=[0,1])

仅显示 0 和 1 处的 x 轴标签，但如果我不想显示 0 和 1，而是想显示 Apple 和 Orange，该怎么办？类似的东西

plot.xaxis[0].ticker=FixedTicker(ticks=[0,1], labels=['Apple', 'Orange'])

直方图不适用于我正在绘制的数据。有没有办法像这样在 Bokeh 中使用自定义标签？

Answer 1

固定刻度可以直接作为“ticker”值传递，并且可以提供主要标签覆盖来显式为特定值提供自定义标签：

from bokeh.plotting import figure, output_file, show
    
p = figure()
p.circle(x=[1,2,3], y=[4,6,5], size=20)
    
p.xaxis.ticker = [1, 2, 3]
p.xaxis.major_label_overrides = {1: 'A', 2: 'B', 3: 'C'}
    
output_file("test.html")
    
show(p)

Answer 2

编辑：更新了 Bokeh

0.12.5

，但也可以在其他答案中看到更简单的方法。

这对我有用：

import pandas as pd
from bokeh.charts import Bar, output_file, show
from bokeh.models import TickFormatter
from bokeh.core.properties import Dict, Int, String

class FixedTickFormatter(TickFormatter):
    """
    Class used to allow custom axis tick labels on a bokeh chart
    Extends bokeh.model.formatters.TickFormatte
    """

    JS_CODE =  """
        import {Model} from "model"
        import * as p from "core/properties"

        export class FixedTickFormatter extends Model
          type: 'FixedTickFormatter'
          doFormat: (ticks) ->
            labels = @get("labels")
            return (labels[tick] ? "" for tick in ticks)
          @define {
            labels: [ p.Any ]
          }
    """

    labels = Dict(Int, String, help="""
    A mapping of integer ticks values to their labels.
    """)

    __implementation__ = JS_CODE

skills_list = ['cheese making', 'squanching', 'leaving harsh criticisms']
pct_counts = [25, 40, 1]
df = pd.DataFrame({'skill':skills_list, 'pct jobs with skill':pct_counts})
p = Bar(df, 'index', values='pct jobs with skill', title="Top skills for ___ jobs", legend=False)
label_dict = {}
for i, s in enumerate(skills_list):
    label_dict[i] = s

p.xaxis[0].formatter = FixedTickFormatter(labels=label_dict)
output_file("bar.html")
show(p)

Answer 3

这可以作为分类数据处理，请参阅 bokeh 文档。

from bokeh.plotting import figure, show

categories = ['A', 'B','C' ]

p = figure(x_range=categories)
p.circle(x=categories, y=[4, 6, 5], size=20)

show(p)

Answer 4

x 刻度的标签如果不落在 10 的倍数上，则不会 可见。

肮脏的黑客是使用填充来以 10 的倍数扩展刻度标签，如图所示 -

label_dict = { 0: "g (1-7)", 1: "", 2: "", 3: "", 4: "", 5: "", 6: "", 7: "", 8: "", 9: "", 10: "O (8-13)", 11: "", 12: "", 13: "", 14: "", 15: "", 16: "", 17: "", 18: "", 19: "", 20: "E (14-18)            I (19-24)", 21: "", 22: "", 23: "", 24: "", 25: "", 26: "", 27: "", 28: "", 29: "", 30: "C (25-35)                ", 31: "", 32: "", 33: "", 34: "", 35: "", 36: "", 37: "", 38: "", 39: "", 40: "λ(36) k(37)      A / L (38-40)", 41: "", 42: "", 43: "", 44: "", 45: "", 46: "", 47: "", 48: "", 49: ""}