为什么散景比 matplotlib 慢这么多

Question

我在 Bokeh 中绘制了一个箱线图，在 matplotlib 中绘制了另一个。 对于相同的数据，在 Bokeh 中绘制大约要慢 100 倍。 为什么散景需要这么长时间？ 这是代码，我在 Jupyter notebook 中运行了它：

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))

打印语句产生以下输出：

散景所需的总时间为 11.8056321144104

matplotlib 花费的总时间为 0.1586170196533203

Answer 1

bokeh.charts.BoxPlot存在一些问题。 不幸的是， bokeh.charts目前没有维护者，所以我无法说明它何时可能得到修复或改进。

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”做事，然后时间与 MPL 相当，如果bokeh.plotting话：

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))

这是一段代码，但它很简单，可以包装成一个可重用的函数。 对我来说，上述结果是：

Answer 2

我在 Jupyter 实验室中绘制了一些散景图，它真的很慢。 bigreddot的例子用了 7 秒！ 重启 Jupyter 实验室后，一切恢复正常。