產生類似的數據到你的後：
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()
我首先想到的是分隔每個數據塊（你還可以減去初始小時，有第一x值是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()
然後，我還以爲你會發現它有用每次使用塊的開始爲蜱雖然這可能會導致標籤的疊加：
# 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()
最後我想你可能想REM奧雅納這些差距，但保持標籤在正確的地方：
# 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()
老結果和錯誤的刻度標記（此數據在每次運行隨機的，因此在新的例如標籤有明顯的不同，但我不想重新上傳每張圖片 - 信任或嘗試自己並驗證= op）： 
請注意，您無法真正看到紅色背後的藍線，因此匹配似乎沒有問題 - 只要您設置xlim以確保。 
你可以嘗試[編輯蜱（http://matplotlib.org/examples/ticks_and_spines/tick-formatters.html）到你想要的值，[這個例子]（ http://matplotlib.org/examples/axes_grid/demo_parasite_axes2.html）可能是有用的 – berna1111
此外，對日期的陰謀似乎在收集所有的數據，它只是沒有像對抗指數，但聚集在幾個日期。你確定時間戳是全部等間隔和正確轉換？ – berna1111
@ berna1111感謝您的回覆。我不認爲在這種情況下提到的例子有幫助。 ii）這是一個好點。時間戳不是等間隔的，但這只是它們的方式。這就是爲什麼我想用它們作爲刻度標籤，而不是實際的刻度。 – ninjaSurfer