如何使用matplotlib创建相同数量的主要和次要y轴刻度？

通过手动设置辅助轴限制,我得到了这个图,这是我想要的输出：

我已经包含了可重现的代码：

import numpy as np
import matplotlib.pyplot as plt
data = np.loadtxt('data.dat',skiprows=2,delimiter=',',unpack=True).transpose()
time = data[:,0]
pressure = data[:,1]
lift = data[:,2]
figure_pressure_trace = plt.figure(figsize=(5.15,5.15))
figure_pressure_trace.clf()
P_vs_t = plt.subplot(111)
P_vs_t.plot(time,pressure,linewidth=1.0)
P_vs_t.set_ylabel(r'textit{Pressure (bar)}',labelpad=6)
P_vs_t.set_xlabel(r'textit{Time (ms)}',labelpad=6)
lift_vs_t = P_vs_t.twinx()
lift_vs_t.plot(time,lift,color='#4DAF4A')
lift_vs_t.set_ylabel(r'textit{Lift(mm)}',labelpad=6)
plt.show()
plt.close()

数据可用here.

更新：

我创建了一个函数来创建相同数量的刻度,整个代码是：

import numpy as np
import matplotlib.pyplot as plt

def equal_y_ticks(primary,secondary):
    y_min_primary,y_max_primary = primary.get_ybound()
    y_min_secondary,y_max_secondary = secondary.get_ybound()
    primary_ticks = len(primary.yaxis.get_major_ticks())
    secondary_ticks = len(secondary.yaxis.get_major_ticks())
    primary_spacing = (y_max_primary - y_min_primary) / (primary_ticks - 1)
    secondary_spacing = (y_max_secondary - y_min_secondary) / (secondary_ticks - 1)
    ticks = max(primary_ticks,secondary_ticks)
    if secondary_ticks < primary_ticks:
        y_max_secondary = y_min_secondary + (primary_ticks * secondary_spacing)
        secondary.yaxis.set_ticks(np.arange(y_min_secondary,y_max_secondary,secondary_spacing))
    else:
        y_max_primary = y_min_primary + (secondary_ticks * primary_spacing)
        primary.yaxis.set_ticks(np.arange(y_min_primary,y_max_primary,primary_spacing))

data = np.loadtxt('data.dat',color='#4DAF4A')
equal_y_ticks(P_vs_t,lift_vs_t)
lift_vs_t.set_ylabel(r'textit{Lift(mm)}',labelpad=6)
plt.show()
plt.close()

但是这个函数给了我这样的图(对于一些数据)：