生成与您类似的数据后：
import numpy as np
import matplotlib.pyplot as plt
plt.ioff()
import matplotlib.dates as mdates
import datetime as dt
import time
# --- Build date blocks ---
now = time.time() # first time
delta_time = 60 # distance between points in seconds
n_jumps = 10 # number of pieces with no data
avg_jump_size = 60*60*24 # 86400 s = 1 day
jumps = abs(np.random.normal(avg_jump_size/2.,
avg_jump_size/2,
n_jumps+1)) + avg_jump_size/2.
# `abs` just to make sure the jump is positive, and ` + avg_jump_size/2.` to
# make sure it's larger than the time step.
avg_n_poins_per_block = 2*60*60/delta_time # 2 hours of acquisition per block
blocks_sizes = abs(np.random.normal(avg_n_poins_per_block/2.,
avg_n_poins_per_block/2.,
n_jumps+1)) + avg_n_poins_per_block/2.
times = np.array([]) # array to place all dates
for i in range(n_jumps):
block = np.arange(now, now+delta_time*blocks_sizes[i], delta_time)
times = np.concatenate((times, block))
now += jumps[i]
# last block
block = np.arange(now, now+delta_time*blocks_sizes[-1], delta_time)
times = np.concatenate((times, block))
def time2mdate_str(number=None):
"""
Convert a time given by `time.time()` to a `datetime` instance
from `matplotlib.mdate`.
"""
if number is None:
number = time.time()
# time.time() returns the number of seconds since the Epoch
# (1970-01-01 00:00:00).
# But mdate represents time in days since 0001-01-01 00:00:00, plus 1 day.
# (http://matplotlib.org/api/dates_api.html)
# So we convert to days:
number /= 60*60*24
# and then we add the 1969 years:
# http://www.rapidtables.com/calc/time/days-in-year.htm
number += 1969*365.2425
# and now it should be off by only (!) ~11h (close enough)
a = mdates.num2date(number)
return a
# list of time strings:
dates_test = [time2mdate_str(t).strftime("%Y-%m-%d %H:%M:%S") for t in times]
# some random data:
errors = np.random.normal(0.025, 0.01, len(times))
xaxis = np.arange(len(errors)) # omiter
# Original code:
x = [dt.datetime.strptime(d,'%Y-%m-%d %H:%M:%S') for d in dates_test]
x = [mdates.date2num(i) for i in x]
fig, ax = plt.subplots(nrows=3, figsize=(8, 6), sharex = True)
ax[0].plot(xaxis, errors)
ax4 = ax[0].twiny()
ax4.plot_date(x, errors, fmt="r-")
ax4.xaxis.set_major_formatter(mdates.DateFormatter('%d/%m \n%H:%M'))
fig.tight_layout()
fig.show()
我的第一个想法是分离每个数据块（您仍然可以减去初始小时，使第一个 x 值为 00:00）：
# break into blocks:
x = np.array(x)
deltas = x[1:] - x[:-1]
# assume there's no break right after the first value and
# find where the difference between consecutive times is larger
# than 1.1*deltas[0] (margin of 1.1* for float comparison)
break_indexes = np.where(deltas > deltas[0]*1.1)[0]+1
# add borders (will be useful for iterating over the list):
break_indexes = np.concatenate(([0],break_indexes,[-1]))
n_jumps = len(break_indexes) - 1
# offset to make sure each line does not overlap with another:
offset = 1.5*np.max(errors)
fig2, ax2 = plt.subplots(figsize=(8, 6))
for i in range(n_jumps):
i_0 = break_indexes[i]
slice_ = slice(i_0, break_indexes[i+1])
ax2.plot(x[slice_]-x[i_0]+x[0], errors[slice_]+offset*i, label=dates_test[i_0])
ax2.xaxis.set_major_formatter(mdates.DateFormatter('%H:%M'))
ax2.set_xlim(x[0])
ax2.legend()
fig2.tight_layout()
fig2.show()
然后我想你会发现使用每个时间块的开始作为刻度很有用，尽管这会导致标签重叠：
# use blocks as ticks:
fig3, ax3 = plt.subplots(nrows=3, figsize=(8, 6), sharex = True)
ax3[0].plot(xaxis, errors)
ax3 = ax3[0].twiny()
ax3.plot_date(x, errors, fmt="r-")
ax3.set_xticks(x[break_indexes][:-1])
ax3.xaxis.set_major_formatter(mdates.DateFormatter('%d/%m \n%H:%M'))
fig3.tight_layout()
fig3.show()
最后我认为您可能想要删除这些空白，但将标签保留在正确的位置：
# use blocks as ticks and eliminate the gaps:
# (could have used the inverse of this to create those gaps)
x_without_gaps = x.copy()
delta = x[1] - x[0]
for i in range(1,n_jumps):
i0 = break_indexes[i]
i1 = break_indexes[i+1]
if i1 == -1:
i1 = None
x_without_gaps[i0:i1] -= x_without_gaps[i0] - x_without_gaps[i0-1] - delta
#x_without_gaps += x[0]
fig4, ax4 = plt.subplots(nrows=3, figsize=(8, 6), sharex = True)
ax4[0].plot(xaxis, errors)
ax4[0].set_xlim(0,len(errors)-1)
ax5 = ax4[0].twiny()
ax5.plot_date(x_without_gaps, errors, fmt="r-")
ax5.set_xticks(x_without_gaps[break_indexes][:-1])
ax5.set_xticklabels([date.strftime('%d/%m \n%H:%M') for date in
mdates.num2date(x[break_indexes][:-1])])
# the following line would clear the values placed!
#ax5.xaxis.set_major_formatter(mdates.DateFormatter('%d/%m \n%H:%M'))
fig4.tight_layout()
fig4.show()
带有错误刻度标签的旧结果（每次运行时数据都是随机的，因此新示例中的标签明显不同，但我不想重新上传每张图片 - 相信或自己尝试并验证 =op ） ：
请注意，您无法真正看到红色后面的蓝线，因此匹配似乎很好 - 只要您设置 xlim 以确保这一点。