为什么散景比matplotlib慢得多

Question

我在Bokeh中绘制了一个箱形图，在matplotlib中绘制了另一个。对于相同的数据，Bokeh中的绘图速度大约慢100倍。为什么散景需要这么长时间？ 这是代码，我在Jupyter笔记本中运行它：

import pandas as pd
import numpy as np

import matplotlib.pyplot as plt
import matplotlib as mpl

from bokeh.charts import BoxPlot, output_notebook, show

from time import time

%matplotlib inline


# Generate data
N = 100000
x1 = 2 + np.random.randn(N)
y1 = ['a'] * N

x2 = -2 + np.random.randn(N)
y2 = ['b'] * N

X = list(x1) + list(x2)
Y = y1 + y2

data = pd.DataFrame()
data['Vals'] = X
data['Class'] = Y

df = data.apply(np.random.permutation)


# Time the bokeh plot
start_time = time()

p = BoxPlot(data, values='Vals', label='Class',\
            title="MPG Summary (grouped by CYL, ORIGIN)")
output_notebook()
show(p)

end_time = time()
print("Total time taken for Bokeh is {0}".format(end_time - start_time))


# time the matplotlib plot
start_time = time()

data.boxplot(column='Vals', by='Class', sym = 'o')

end_time = time()
print("Total time taken for matplotlib is {0}".format(end_time - start_time))

print语句产生以下输出：

  

Bokeh的总时间为11.8056321144104

     

matplotlib的总时间为0.1586170196533203

Answer 1

bokeh.charts.BoxPlot特别存在问题。不幸的是，bokeh.charts目前没有维护者，因此我无法说明何时可以修复或改进。

但是，如果它对您有用，我将在下面演示您可以使用完善且稳定的bokeh.plotting API来“手动”执行操作，然后时间可比较，如果不是比MPL更快：

from time import time

import pandas as pd
import numpy as np

from bokeh.io import output_notebook, show
from bokeh.plotting import figure

output_notebook()

# Generate data
N = 100000
x1 = 2 + np.random.randn(N)
y1 = ['a'] * N

x2 = -2 + np.random.randn(N)
y2 = ['b'] * N

X = list(x1) + list(x2)
Y = y1 + y2

df = pd.DataFrame()
df['Vals'] = X
df['Class'] = Y

# Time the bokeh plot
start_time = time()

# find the quartiles and IQR for each category
groups = df.groupby('Class')
q1 = groups.quantile(q=0.25)
q2 = groups.quantile(q=0.5)
q3 = groups.quantile(q=0.75)
iqr = q3 - q1
upper = q3 + 1.5*iqr
lower = q1 - 1.5*iqr

cats = ['a', 'b']

p = figure(x_range=cats)

# if no outliers, shrink lengths of stems to be no longer than the minimums or maximums
qmin = groups.quantile(q=0.00)
qmax = groups.quantile(q=1.00)
upper.score = [min([x,y]) for (x,y) in zip(list(qmax.loc[:,'Vals']),upper.Vals)]
lower.score = [max([x,y]) for (x,y) in zip(list(qmin.loc[:,'Vals']),lower.Vals)]

# stems
p.segment(cats, upper.Vals, cats, q3.Vals, line_color="black")
p.segment(cats, lower.Vals, cats, q1.Vals, line_color="black")

# boxes
p.vbar(cats, 0.7, q2.Vals, q3.Vals, fill_color="#E08E79", line_color="black")
p.vbar(cats, 0.7, q1.Vals, q2.Vals, fill_color="#3B8686", line_color="black")

# whiskers (almost-0 height rects simpler than segments)
p.rect(cats, lower.Vals, 0.2, 0.01, line_color="black")
p.rect(cats, upper.Vals, 0.2, 0.01, line_color="black")

p.xgrid.grid_line_color = None
p.ygrid.grid_line_color = "white"
p.grid.grid_line_width = 2
p.xaxis.major_label_text_font_size="12pt"

show(p)

end_time = time()
print("Total time taken for Bokeh is {0}".format(end_time - start_time))

这是一大块代码，但它很简单，可以包含一个可重用的函数。对我来说，上面的结果是：

Answer 2

我在Jupyter实验室中绘制了一些背景虚化的人物，这真的很慢。 bigreddot中的示例花费了7秒！重新启动Jupyter Lab后，一切恢复正常。