Question

好的，所以我有下面的代码，用于实时绘制来自串行接收的嵌入式设备的一些数据。它不是一个生产工具，而是一个内部工具，因此它不是非常用户友好。问题在于，无论我做什么，我都无法显示小的网格线，即使这里它们设置为True, which=both。我可以做任何我想要的主要网格线，但未成年人不会出现。有任何想法吗？这是代码：

import numpy as np
from matplotlib import pyplot as plt
from matplotlib import animation
import serial

SERIAL_PORT_NUM=9

...a bunch of constants...
#windows starts serial port numbers at 1, python starts at 0
SERIAL_PORT_NUM = SERIAL_PORT_NUM - 1
"""
Open the serial port
"""
ser =serial.Serial(port=SERIAL_PORT_NUM,baudrate=115200,bytesize=8,parity='N',stopbits=1,timeout=None,xonxoff=0,rtscts=0)

# First set up the figure, the axis, and the plot element we want to animate
raw_adc_fig = plt.figure()
raw_adc_ax = plt.axes(xlim=(0, 200), ylim=(0, 2047))
raw_adc_ax.grid(True, which='both')
raw_adc_fig.suptitle("Raw ADC data")
plt.ylabel("ADC values (hex)")
plt.xlabel("time (sec)")
raw_adc_line, = raw_adc_ax.plot([], [], lw=2)

def read_serial(serial_port):
    tmp = ''
    same_line = True
    while same_line:
        tmp += serial_port.read(1)
        if tmp != '':
            if tmp[-1] == '*':
                same_line = False
    tmp = tmp.rstrip()
    tmp = tmp.lstrip()
    return tmp

def process_serial(input_data):
    output_data = 0
    intermediate_data = input_data[A_TYPE_START_POS:A_TYPE_STOP_POS + 1]
    if( intermediate_data != ''):
        output_data =  int(intermediate_data , 16 )
    else:
        print "bad data"
        output_data = -100

    return output_data

def get_sound_value(serial_port):
    cur_line = ''

    get_next_line = True
    # read in the next line until a sound packet of type A is found
    while( get_next_line ):
        cur_line = read_serial(serial_port)
        if( (cur_line != '') and (cur_line[0:3] == ROUTER_SOUND_DATA) and (len(cur_line) == D_TYPE_STOP_POS + 2) ):
          get_next_line = False

    sound_value = process_serial(cur_line)
    return sound_value

# initialization function: plot the background of each frame
def raw_adc_init():
    raw_adc_line.set_data([], [])
    return raw_adc_line,

# animation function.  This is called sequentially
def raw_adc_animate(i):
    sound_data_list.append( get_sound_value(ser) )
    y = sound_data_list
    if( len(y) == 190 ):
        del y[0]
    x = np.linspace(0, len(y), len(y))
    raw_adc_line.set_data(x, y)
    return raw_adc_line,

# call the animator.  blit=True means only re-draw the parts that have changed.
raw_adc_anim = animation.FuncAnimation(raw_adc_fig, raw_adc_animate, init_func=raw_adc_init, frames=200, interval=1000, blit=True)

编辑：修正了打开串口的错误。已将timeout=0更改为timeout=None。

Answer 1

不幸的是，ax.grid在这方面有点令人困惑。（这是一个设计错误/普通问题。）它会打开次要网格，但是次要标记仍然关闭。

除了致电plt.minorticks_on外，您还需要致电ax.minorticks_on或ax.grid(True, which='both')。

Answer 2

您应该使用plt.minorticks_on()。

import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(1)
ax = fig.add_subplot(111)

x = np.linspace(0,10,41)
y = np.sin(x)

plt.plot(x,y)
plt.grid(b=True, which='major', color='k', linestyle='-')
plt.grid(b=True, which='minor', color='r', linestyle='-', alpha=0.2)
plt.minorticks_on()
plt.show()

无法在matplotlib图中显示较小的网格线

2 个答案: