具有两个缩放共享原点的Matplotlib轴
我需要在Matplotlib中用不同的Y轴比例覆盖两个数据集。数据包含正值和负值。我希望两个轴共享一个原点,但默认情况下Matplotlib不会对齐两个比例具有两个缩放共享原点的Matplotlib轴,matplotlib,scale,axis,Matplotlib,Scale,Axis,我需要在Matplotlib中用不同的Y轴比例覆盖两个数据集。数据包含正值和负值。我希望两个轴共享一个原点,但默认情况下Matplotlib不会对齐两个比例 import numpy as np import matplotlib.pyplot as plt fig = plt.figure() ax1 = fig.add_subplot(111) ax2 = ax1.twinx() ax1.bar(range(6), (2, -2, 1, 0, 0, 0)) ax2.plot(range(
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax2 = ax1.twinx()
ax1.bar(range(6), (2, -2, 1, 0, 0, 0))
ax2.plot(range(6), (0, 2, 8, -2, 0, 0))
plt.show()
我想可以使用执行一些计算。get_ylim()
和将两个刻度对齐。有更简单的解决办法吗
使用align_yaxis()函数:
import numpy as np
import matplotlib.pyplot as plt
def align_yaxis(ax1, v1, ax2, v2):
"""adjust ax2 ylimit so that v2 in ax2 is aligned to v1 in ax1"""
_, y1 = ax1.transData.transform((0, v1))
_, y2 = ax2.transData.transform((0, v2))
inv = ax2.transData.inverted()
_, dy = inv.transform((0, 0)) - inv.transform((0, y1-y2))
miny, maxy = ax2.get_ylim()
ax2.set_ylim(miny+dy, maxy+dy)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax2 = ax1.twinx()
ax1.bar(range(6), (2, -2, 1, 0, 0, 0))
ax2.plot(range(6), (0, 2, 8, -2, 0, 0))
align_yaxis(ax1, 0, ax2, 0)
plt.show()
为了确保y边界保持不变(因此不会将数据点移出绘图),并平衡两个y轴的调整,我对@HYRY的答案做了一些补充:
def align_yaxis(ax1, v1, ax2, v2):
"""adjust ax2 ylimit so that v2 in ax2 is aligned to v1 in ax1"""
_, y1 = ax1.transData.transform((0, v1))
_, y2 = ax2.transData.transform((0, v2))
adjust_yaxis(ax2,(y1-y2)/2,v2)
adjust_yaxis(ax1,(y2-y1)/2,v1)
def adjust_yaxis(ax,ydif,v):
"""shift axis ax by ydiff, maintaining point v at the same location"""
inv = ax.transData.inverted()
_, dy = inv.transform((0, 0)) - inv.transform((0, ydif))
miny, maxy = ax.get_ylim()
miny, maxy = miny - v, maxy - v
if -miny>maxy or (-miny==maxy and dy > 0):
nminy = miny
nmaxy = miny*(maxy+dy)/(miny+dy)
else:
nmaxy = maxy
nminy = maxy*(miny+dy)/(maxy+dy)
ax.set_ylim(nminy+v, nmaxy+v)
@在绘制以下两个点序列时,drevicko的回答对我来说是失败的:
l1 = [0.03, -0.6, 1, 0.05]
l2 = [0.8, 0.9, 1, 1.1]
fig, ax1 = plt.subplots()
ax1.plot(l1)
ax2 = ax1.twinx()
ax2.plot(l2, color='r')
align_yaxis(ax1, 0, ax2, 0)
。。。以下是我的版本:
def align_yaxis(ax1, ax2):
"""Align zeros of the two axes, zooming them out by same ratio"""
axes = (ax1, ax2)
extrema = [ax.get_ylim() for ax in axes]
tops = [extr[1] / (extr[1] - extr[0]) for extr in extrema]
# Ensure that plots (intervals) are ordered bottom to top:
if tops[0] > tops[1]:
axes, extrema, tops = [list(reversed(l)) for l in (axes, extrema, tops)]
# How much would the plot overflow if we kept current zoom levels?
tot_span = tops[1] + 1 - tops[0]
b_new_t = extrema[0][0] + tot_span * (extrema[0][1] - extrema[0][0])
t_new_b = extrema[1][1] - tot_span * (extrema[1][1] - extrema[1][0])
axes[0].set_ylim(extrema[0][0], b_new_t)
axes[1].set_ylim(t_new_b, extrema[1][1])
def align_yaxis_np(ax1, ax2):
"""Align zeros of the two axes, zooming them out by same ratio"""
axes = np.array([ax1, ax2])
extrema = np.array([ax.get_ylim() for ax in axes])
tops = extrema[:,1] / (extrema[:,1] - extrema[:,0])
# Ensure that plots (intervals) are ordered bottom to top:
if tops[0] > tops[1]:
axes, extrema, tops = [a[::-1] for a in (axes, extrema, tops)]
# How much would the plot overflow if we kept current zoom levels?
tot_span = tops[1] + 1 - tops[0]
extrema[0,1] = extrema[0,0] + tot_span * (extrema[0,1] - extrema[0,0])
extrema[1,0] = extrema[1,1] + tot_span * (extrema[1,0] - extrema[1,1])
[axes[i].set_ylim(*extrema[i]) for i in range(2)]
原则上,对齐零(或其他值,其他提供的解决方案可以接受)的可能性是无限的:无论将零放置在y轴的何处,都可以缩放两个系列中的每一个,以使其适合。我们只是选择这样的位置,在转换后,两个覆盖相同高度的垂直间隔。
或者换言之,我们将它们最小化,使其具有与不对齐图相同的因子。
(这并不意味着0位于绘图的一半:例如,如果一个绘图全部为负值,而另一个绘图全部为正值,则会发生这种情况。)
Numpy版本:
def align_yaxis(ax1, ax2):
"""Align zeros of the two axes, zooming them out by same ratio"""
axes = (ax1, ax2)
extrema = [ax.get_ylim() for ax in axes]
tops = [extr[1] / (extr[1] - extr[0]) for extr in extrema]
# Ensure that plots (intervals) are ordered bottom to top:
if tops[0] > tops[1]:
axes, extrema, tops = [list(reversed(l)) for l in (axes, extrema, tops)]
# How much would the plot overflow if we kept current zoom levels?
tot_span = tops[1] + 1 - tops[0]
b_new_t = extrema[0][0] + tot_span * (extrema[0][1] - extrema[0][0])
t_new_b = extrema[1][1] - tot_span * (extrema[1][1] - extrema[1][0])
axes[0].set_ylim(extrema[0][0], b_new_t)
axes[1].set_ylim(t_new_b, extrema[1][1])
def align_yaxis_np(ax1, ax2):
"""Align zeros of the two axes, zooming them out by same ratio"""
axes = np.array([ax1, ax2])
extrema = np.array([ax.get_ylim() for ax in axes])
tops = extrema[:,1] / (extrema[:,1] - extrema[:,0])
# Ensure that plots (intervals) are ordered bottom to top:
if tops[0] > tops[1]:
axes, extrema, tops = [a[::-1] for a in (axes, extrema, tops)]
# How much would the plot overflow if we kept current zoom levels?
tot_span = tops[1] + 1 - tops[0]
extrema[0,1] = extrema[0,0] + tot_span * (extrema[0,1] - extrema[0,0])
extrema[1,0] = extrema[1,1] + tot_span * (extrema[1,0] - extrema[1,1])
[axes[i].set_ylim(*extrema[i]) for i in range(2)]
我已经制定了一个解决方案,从上面开始,它将对齐任意数量的轴:
def align_yaxis_np(axes):
"""Align zeros of the two axes, zooming them out by same ratio"""
axes = np.array(axes)
extrema = np.array([ax.get_ylim() for ax in axes])
# reset for divide by zero issues
for i in range(len(extrema)):
if np.isclose(extrema[i, 0], 0.0):
extrema[i, 0] = -1
if np.isclose(extrema[i, 1], 0.0):
extrema[i, 1] = 1
# upper and lower limits
lowers = extrema[:, 0]
uppers = extrema[:, 1]
# if all pos or all neg, don't scale
all_positive = False
all_negative = False
if lowers.min() > 0.0:
all_positive = True
if uppers.max() < 0.0:
all_negative = True
if all_negative or all_positive:
# don't scale
return
# pick "most centered" axis
res = abs(uppers+lowers)
min_index = np.argmin(res)
# scale positive or negative part
multiplier1 = abs(uppers[min_index]/lowers[min_index])
multiplier2 = abs(lowers[min_index]/uppers[min_index])
for i in range(len(extrema)):
# scale positive or negative part based on which induces valid
if i != min_index:
lower_change = extrema[i, 1] * -1*multiplier2
upper_change = extrema[i, 0] * -1*multiplier1
if upper_change < extrema[i, 1]:
extrema[i, 0] = lower_change
else:
extrema[i, 1] = upper_change
# bump by 10% for a margin
extrema[i, 0] *= 1.1
extrema[i, 1] *= 1.1
# set axes limits
[axes[i].set_ylim(*extrema[i]) for i in range(len(extrema))]
def align\u yaxis\u np(轴):
“”“将两个轴的零点对齐,按相同的比例缩小”
轴=np.阵列(轴)
extrema=np.array([ax.get_ylim()表示轴中的ax])
#重置为零除问题
对于范围内的i(len(极值)):
如果np.isclose(极值[i,0],0.0):
极值[i,0]=-1
如果np.isclose(极值[i,1],0.0):
极值[i,1]=1
#上限和下限
下限=极值[:,0]
上限=极值[:,1]
#如果全部为pos或neg,则不要缩放
全阳性=假
全部否定=错误
如果降低.min()>0.0:
全部为正=真
如果uppers.max()小于0.0:
全部为负=真
如果全部为负或全部为正:
#不要缩放
返回
#选择“最中心”轴
res=abs(上部+下部)
最小索引=np.argmin(res)
#缩放正或负部分
乘数1=abs(上[最小指数]/下[最小指数])
乘数2=abs(降低[最小指数]/升高[最小指数])
对于范围内的i(len(极值)):
#缩放正或负的部分,根据该部分导出有效的
如果我min_索引:
下限变化=极值[i,1]*-1*乘法器2
上限变化=极值[i,0]*-1*乘法器1
如果上限变化<极值[i,1]:
极值[i,0]=下限变化
其他:
极值[i,1]=上限变化
#增加10%的利润
极值[i,0]*=1.1
极值[i,1]*=1.1
#设置轴限制
[轴[i]。为范围(len(extrema))中的i设置_ylim(*extrema[i])]
4个随机序列的示例(您可以在4组单独的y轴标签上看到离散范围):
此处的其他答案似乎过于复杂,不一定适用于所有场景(例如,ax1为负,ax2为正)。有两种简单的方法始终有效:
总是在图的中间放置0个y轴
有点花哨,并保留了正负比,见下文
的解决方案适用于两个以上的轴:
import numpy as np
def align_yaxis(axes):
y_lims = np.array([ax.get_ylim() for ax in axes])
# force 0 to appear on all axes, comment if don't need
y_lims[:, 0] = y_lims[:, 0].clip(None, 0)
y_lims[:, 1] = y_lims[:, 1].clip(0, None)
# normalize all axes
y_mags = (y_lims[:,1] - y_lims[:,0]).reshape(len(y_lims),1)
y_lims_normalized = y_lims / y_mags
# find combined range
y_new_lims_normalized = np.array([np.min(y_lims_normalized), np.max(y_lims_normalized)])
# denormalize combined range to get new axes
new_lims = y_new_lims_normalized * y_mags
for i, ax in enumerate(axes):
ax.set_ylim(new_lims[i])
我需要对齐两个子地块,但不在它们的零处。其他的解决方案对我来说不太管用
我的程序的主代码如下所示。子地块未对齐。此外,我只更改align_yaxis
功能,并保持所有其他代码不变
import matplotlib.pyplot as plt
def align_yaxis(ax1, v1, ax2, v2):
return 0
x = range(10)
y1 = [3.2, 1.3, -0.3, 0.4, 2.3, -0.9, 0.2, 0.1, 1.3, -3.4]
y2, s = [], 100
for i in y1:
s *= 1 + i/100
y2.append(s)
fig = plt.figure()
ax1 = fig.add_subplot()
ax2 = ax1.twinx()
ax1.axhline(y=0, color='k', linestyle='-', linewidth=0.5)
ax1.bar(x, y1, color='tab:blue')
ax2.plot(x, y2, color='tab:red')
fig.tight_layout()
align_yaxis(ax1, 0, ax2, 100)
plt.show()
使用@HYRY的解决方案,我得到了对齐的子图,但是第二个子图不在图中。你看不到它
def align_yaxis(ax1, v1, ax2, v2):
"""adjust ax2 ylimit so that v2 in ax2 is aligned to v1 in ax1"""
_, y1 = ax1.transData.transform((0, v1))
_, y2 = ax2.transData.transform((0, v2))
inv = ax2.transData.inverted()
_, dy = inv.transform((0, 0)) - inv.transform((0, y1-y2))
miny, maxy = ax2.get_ylim()
ax2.set_ylim(miny+dy, maxy+dy)
使用@drevicko的解决方案,我也得到了对齐的绘图。但是现在第一个子图不在图片中,第一个Y轴非常奇怪
def align_yaxis(ax1, v1, ax2, v2):
"""adjust ax2 ylimit so that v2 in ax2 is aligned to v1 in ax1"""
_, y1 = ax1.transData.transform((0, v1))
_, y2 = ax2.transData.transform((0, v2))
adjust_yaxis(ax2,(y1-y2)/2,v2)
adjust_yaxis(ax1,(y2-y1)/2,v1)
def adjust_yaxis(ax,ydif,v):
"""shift axis ax by ydiff, maintaining point v at the same location"""
inv = ax.transData.inverted()
_, dy = inv.transform((0, 0)) - inv.transform((0, ydif))
miny, maxy = ax.get_ylim()
miny, maxy = miny - v, maxy - v
if -miny>maxy or (-miny==maxy and dy > 0):
nminy = miny
nmaxy = miny*(maxy+dy)/(miny+dy)
else:
nmaxy = maxy
nminy = maxy*(miny+dy)/(maxy+dy)
ax.set_ylim(nminy+v, nmaxy+v)
所以我对@drevicko的解决方案进行了一些调整,得到了我想要的
def align_yaxis(ax1, v1, ax2, v2):
"""adjust ax2 ylimit so that v2 in ax2 is aligned to v1 in ax1"""
_, y1 = ax1.transData.transform((0, v1))
_, y2 = ax2.transData.transform((0, v2))
adjust_yaxis(ax1,(y2 - y1)/2,v1)
adjust_yaxis(ax2,(y1 - y2)/2,v2)
def adjust_yaxis(ax,ydif,v):
"""shift axis ax by ydiff, maintaining point v at the same location"""
inv = ax.transData.inverted()
_, dy = inv.transform((0, 0)) - inv.transform((0, ydif))
miny, maxy = ax.get_ylim()
nminy = miny - v + dy - abs(dy)
nmaxy = maxy - v + dy + abs(dy)
ax.set_ylim(nminy+v, nmaxy+v)
这可能不是您想要的,但这帮助我将整数排列在两个不同的垂直轴上:
ax1.set_ylim(0,4000)
ax2.set_ylim(0,120)
ax2.set_yticks(np.linspace(ax2.get_yticks()[0], ax2.get_yticks()[-1], len(ax1.get_yticks())))
请您对if/else声明添加评论。我发现这种方法仍然会切断数据。如果不查看您的数据,很难做到这一点(从而找出数据被切断的原因)。你能提供更多的信息吗?也许调试并建议编辑?(在这里ping我,以防你编辑,这样我就可以接受它-代码编辑通常不被接受!)谢谢今天稍后我将整理一个可复制的示例。如果您能解释if/else和rescalingok的逻辑,那就太好了,if
基本上决定了miny
或maxy
的绝对值是否更大(abs(miny)
只有当它为负值时才更大)。换句话说,它离0点更远(实际上,v
-点,因为你可以在某个值上对齐)。@devicko:这里很难给出一个可复制的例子,所以我提出了一个新问题