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当数据限制不同时,是否创建具有多个轴的等宽(正方形)绘图?

matplotlib python

bbengfort · 技术社区 · 7 年前

我想创建一个正方形的绘图使用多个轴使用 make_axes_locateable matplotlib documentation . 但是,虽然这在x和y数据具有相同范围的绘图上有效,但在范围相差几个数量级时不起作用。

import numpy as np
import matplotlib.pyplot as plt

from mpl_toolkits.axes_grid1 import make_axes_locatable

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)

_, ax = plt.subplots()
divider = make_axes_locatable(ax)

xhax = divider.append_axes("top", size=1, pad=0.1, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.1, sharey=ax)

ax.scatter(x, y)
xhax.hist(x)
yhax.hist(y, orientation="horizontal")

x0,x1 = ax.get_xlim()
y0,y1 = ax.get_ylim()
ax.set_aspect(abs(x1-x0)/abs(y1-y0))

plt.show()

set_aspect 回答 How do I make a matplotlib scatter plot square?

我试图用以下方法修复此问题:

ax.set_aspect(abs(x1-x0)/abs(y1-y0), share=True)

但这导致了以下情况:

使现代化 :此问题的关键限制之一是使用 使轴可定位 而不是 GridSpec

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.gridspec as grid

from mpl_toolkits.axes_grid1 import make_axes_locatable, axes_size


def joint_plot(x, y, ax=None):
    """
    Create a square joint plot of x and y.
    """
    if ax is None:
        ax = plt.gca()

    divider = make_axes_locatable(ax)
    xhax = divider.append_axes("top", size=1, pad=0.1, sharex=ax)
    yhax = divider.append_axes("right", size=1, pad=0.1, sharey=ax)

    ax.scatter(x, y)
    xhax.hist(x)
    yhax.hist(y, orientation="horizontal")

    x0,x1 = ax.get_xlim()
    y0,y1 = ax.get_ylim()
    ax.set_aspect(abs(x1-x0)/abs(y1-y0))

    plt.sca(ax)
    return ax, xhax, yhax


def color_plot(x, y, colors, ax=None):
    if ax is None:
        ax = plt.gca()

    divider = make_axes_locatable(ax)
    cbax = divider.append_axes("right", size="5%", pad=0.1)

    sc = ax.scatter(x, y, marker='o', c=colors, cmap='RdBu')
    plt.colorbar(sc, cax=cbax)

    ax.set_aspect("equal")

    plt.sca(ax)
    return ax, cbax


if __name__ == "__main__":
    _, axes = plt.subplots(nrows=2, ncols=2, figsize=(9,6))

    # Plot 1
    x = np.random.normal(100, 17, 120)
    y = np.random.normal(0.5, 0.1, 120)
    joint_plot(x, y, axes[0,0])

    # Plot 2
    x = np.random.normal(100, 17, 120)
    y = np.random.normal(100, 17, 120)
    c = np.random.normal(100, 17, 120)
    color_plot(x, y, c, axes[0,1])

    # Plot 3
    x = np.random.normal(100, 17, 120)
    y = np.random.normal(0.5, 0.1, 120)
    c = np.random.uniform(0.0, 1.0, 120)
    color_plot(x, y, c, axes[1,0])

    # Plot 4
    x = np.random.normal(0.5, 0.1, 120)
    y = np.random.normal(0.5, 0.1, 120)
    joint_plot(x, y, axes[1,1])

    plt.tight_layout()
    plt.show()

这个问题扩展了以下问题: Set equal aspect in plot with colorbar python interplay between axis('square') and set_xlim 因为只有轴约束。

2 回复 | 直到 7 年前

Thomas Kühn 7 年前

ax.set_aspect('equal') 一切正常。当然,如果只执行此操作,则记号标签的范围仅为0到1,因此必须应用一点matplotlib魔术将记号标签调整为原始数据范围。答案 here 显示了如何使用 FuncFormatter . 但是,由于原始记号是相对于间隔[0,1]选择的,因此使用 函数格式化程序 AutoLocator tick_values() 作用然后,这些刻度可以再次缩放到间隔[0,1],然后 函数格式化程序

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.ticker as mticker

from mpl_toolkits.axes_grid1 import make_axes_locatable

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)


fig,ax=plt.subplots()

divider = make_axes_locatable(ax)


##increased pad from 0.1 to 0.2 so that tick labels don't overlap
xhax = divider.append_axes("top", size=1, pad=0.2, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.2, sharey=ax)

##'normalizing' x and y values to be between 0 and 1:
xn = (x-min(x))/(max(x)-min(x))
yn = (y-min(y))/(max(y)-min(y))

##producinc the plots
ax.scatter(xn, yn)
xhax.hist(xn)
yhax.hist(yn, orientation="horizontal")

##turning off duplicate ticks (if needed):
plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

ax.set_aspect('equal')


##setting up ticks and labels to simulate real data:
locator = mticker.AutoLocator()

xticks = (locator.tick_values(min(x),max(x))-min(x))/(max(x)-min(x))
ax.set_xticks(xticks)
ax.xaxis.set_major_formatter(mticker.FuncFormatter(
    lambda t, pos: '{0:g}'.format(t*(max(x)-min(x))+min(x))
))

yticks = (locator.tick_values(min(y),max(y))-min(y))/(max(y)-min(y))
ax.set_yticks(yticks)
ax.yaxis.set_major_formatter(mticker.FuncFormatter(
    lambda t, pos: '{0:g}'.format(t*(max(y)-min(y))+min(y))
))

fig.tight_layout()
plt.show()

生成的图片看起来与预期的一样,并且在调整图像大小时也保持方形。

这与其说是解决方案,不如说是一种变通方法:

而不是使用 ax.set_aspect() figsize=(n,n) 到 plt.subplots 哪里 n xhax 以及 yhax 都是1英寸,这意味着 ax 也变成正方形。

import numpy as np
import matplotlib.pyplot as plt

from mpl_toolkits.axes_grid1 import make_axes_locatable

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)

fig, ax = plt.subplots(figsize=(5,5))

divider = make_axes_locatable(ax)

xhax = divider.append_axes("top", size=1, pad=0.1, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.1, sharey=ax)

ax.scatter(x, y)
xhax.hist(x)
yhax.hist(y, orientation="horizontal")

##turning off duplicate ticks:
plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

plt.show()

当然,一旦你调整你的身材,正方形的外观就会消失。但是,如果你已经知道你的身材的最终尺寸,并且只想保存它以供进一步使用,那么这应该是一个足够好的快速修复方法。

ImportanceOfBeingErnest 7 年前

这个 axes_grid1 Divider 或绝对坐标。

如果需要,可以在绝对坐标中手动指定轴的大小,以获得方形子地块。

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable, axes_size

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)

_, ax = plt.subplots()
divider = make_axes_locatable(ax)

xhax = divider.append_axes("top", size=1, pad=0.1, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.1, sharey=ax)

horiz = [axes_size.Fixed(2.8), axes_size.Fixed(.1), axes_size.Fixed(1)]
vert = [axes_size.Fixed(2.8), axes_size.Fixed(.1), axes_size.Fixed(1)]
divider.set_horizontal(horiz)
divider.set_vertical(vert)

ax.scatter(x, y)
xhax.hist(x)
yhax.hist(y, orientation="horizontal")

plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

plt.show()

要有一个仍然可以随体形大小缩放的自适应解决方案,可以定义一个自定义 Size ,它获取绝对尺寸填充轴和边缘轴的剩余值,并返回绝对坐标(英寸)中的最小值,用于两个方向,以便轴始终为正方形。

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.transforms import Bbox
from mpl_toolkits.axes_grid1 import make_axes_locatable, axes_size

class RemainderFixed(axes_size.Scaled):
    def __init__(self, xsizes, ysizes, divider):
        self.xsizes =xsizes
        self.ysizes =ysizes
        self.div = divider

    def get_size(self, renderer):
        xrel, xabs = axes_size.AddList(self.xsizes).get_size(renderer)
        yrel, yabs = axes_size.AddList(self.ysizes).get_size(renderer)
        bb = Bbox.from_bounds(*self.div.get_position()).transformed(self.div._fig.transFigure)
        w = bb.width/self.div._fig.dpi - xabs
        h = bb.height/self.div._fig.dpi - yabs
        return 0, min([w,h])

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)

fig, ax = plt.subplots()
divider = make_axes_locatable(ax)

margin_size = axes_size.Fixed(1)
pad_size = axes_size.Fixed(.1)
xsizes = [pad_size, margin_size]
ysizes = xsizes

xhax = divider.append_axes("top", size=margin_size, pad=pad_size, sharex=ax)
yhax = divider.append_axes("right", size=margin_size, pad=pad_size, sharey=ax)

divider.set_horizontal([RemainderFixed(xsizes, ysizes, divider)] + xsizes)
divider.set_vertical([RemainderFixed(xsizes, ysizes, divider)] + ysizes)

ax.scatter(x, y)
xhax.hist(x)
yhax.hist(y, orientation="horizontal")

plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

plt.show()