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Matplotlib 数字分组（小数分隔符）

发布于 2024-08-27 02:24:21 字数 89 浏览 8 评论 0原文

基本上，当使用 matplotlib 生成绘图时，y 轴上的刻度会达到数百万。如何打开数字分组（即 1000000 显示为 1,000,000）或打开小数点分隔符？

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甜妞爱困 2024-09-03 02:24:21

我认为没有内置函数可以做到这一点。（这就是我读完你的问题后的想法；我刚刚检查过，但在文档中找不到）。

无论如何，推出自己的产品都很容易。

（下面是一个完整的示例 - 即，它将生成一个 mpl 绘图，其中一个轴具有已命名的刻度标签 - 尽管创建自定义刻度标签只需要五行代码 - 函数的三行代码（包括 import 语句）用于创建自定义标签，两行用于创建新标签并将它们放置在指定的轴上。）

# first code a function to generate the axis labels you want 
# ie, turn numbers greater than 1000 into commified strings (12549 => 12,549)

import locale
locale.setlocale(locale.LC_ALL, 'en_US')
fnx = lambda x : locale.format("%d", x, grouping=True)

from matplotlib import pyplot as PLT
import numpy as NP

data = NP.random.randint(15000, 85000, 50).reshape(25, 2)
x, y = data[:,0], data[:,1]

fig = PLT.figure()
ax1 = fig.add_subplot(111)
ax1.plot(x, y, "ro")
default_xtick = range(20000, 100000, 10000)

# these two lines are the crux:
# create the custom tick labels
new_xtick = map(fnx, default_xtick)
# set those labels on the axis
ax1.set_xticklabels(new_xtick)

PLT.show()

I don't think there's a built-in function to do this. (That's what i thought after i read your Q; i just checked and couldn't find one in the Documentation).

In any event, it's easy to roll your own.

(Below is a complete example--ie, it will generate an mpl plot with one axis having commified tick labels--although five lines of code are all you need to create custom tick labels--three (including import statement) for the function used to create the custom labels, and two lines to create the new labels and place them on the specified axis.)

# first code a function to generate the axis labels you want 
# ie, turn numbers greater than 1000 into commified strings (12549 => 12,549)

import locale
locale.setlocale(locale.LC_ALL, 'en_US')
fnx = lambda x : locale.format("%d", x, grouping=True)

from matplotlib import pyplot as PLT
import numpy as NP

data = NP.random.randint(15000, 85000, 50).reshape(25, 2)
x, y = data[:,0], data[:,1]

fig = PLT.figure()
ax1 = fig.add_subplot(111)
ax1.plot(x, y, "ro")
default_xtick = range(20000, 100000, 10000)

# these two lines are the crux:
# create the custom tick labels
new_xtick = map(fnx, default_xtick)
# set those labels on the axis
ax1.set_xticklabels(new_xtick)

PLT.show()

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不一样的天空 2024-09-03 02:24:21

我无法使用 doug 发布的答案，因为命令 locale.setlocale(locale.LC_ALL, ' en_US') 由于区域设置不受支持，在我的 WSL 环境中引发了错误。

幸运的是，从 Python 3.8 开始，您可以利用 f 字符串进行变量格式化，包括数字分组。我将 fnx lambda 函数定义为 fnx = lambda x : f'{x:,}' 并且代码按预期工作。

这是完整的工作代码，已经修改。

fnx = lambda x : f'{x:,}'

from matplotlib import pyplot as plt
import numpy as np

data = np.random.randint(15000, 85000, 50).reshape(25, 2)
x, y = data[:, 0], data[:, 1]

fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.plot(x, y, "ro")
default_xtick = range(20000, 100000, 10000)

# The crucial part:
# Create custom tick labels
new_xtick = map(fnx, default_xtick)
# Set these labels on the axis
ax1.set_xticklabels(new_xtick)

plt.show()

请注意，必须安装 Python 库 matplotlib 和 numpy 才能运行此代码。

I couldn't to use the answer posted by doug because the command locale.setlocale(locale.LC_ALL, 'en_US') raised an error in my WSL environment due to the unsupported locale.

Fortunately, starting from Python 3.8, you can utilize f-strings for variable formatting, including digit grouping. I defined the fnx lambda function as fnx = lambda x : f'{x:,}' and the code worked as intended.

Here is the the full working code, already modified.

fnx = lambda x : f'{x:,}'

from matplotlib import pyplot as plt
import numpy as np

data = np.random.randint(15000, 85000, 50).reshape(25, 2)
x, y = data[:, 0], data[:, 1]

fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.plot(x, y, "ro")
default_xtick = range(20000, 100000, 10000)

# The crucial part:
# Create custom tick labels
new_xtick = map(fnx, default_xtick)
# Set these labels on the axis
ax1.set_xticklabels(new_xtick)

plt.show()

Note that the Python libraries matplotlib and numpy must be installed for this code to run.

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