Python matplotlib，獲取xtick標簽的位置

Question

我如何獲得xtick主要標簽的位置？ 我從label.get_position（）獲得的值沒有意義。

import numpy as np
import matplotlib.pyplot as plt

def f(t):
    return np.exp(-t) * np.cos(2*np.pi*t)

t1 = np.arange(0.0, 5.0, 0.1)
t2 = np.arange(0.0, 5.0, 0.02)

# fig, ax = plt.figure(1)
fig, ax = plt.subplots()
plt.plot(t1, f(t1), 'bo', t2, f(t2), 'k')

# plt.show()
print(fig)
print(ax.get_position())

# ------------------------------------------------
# positions of the tick labels, incorrect (0,0) returned
# ------------------------------------------------
print([text.get_position() for text in ax.get_xticklabels()])
# correct tick label values
print(ax.get_xticks())

上面代碼的輸出是：

Figure(640x480)
Bbox('array([[ 0.125,  0.1  ],\n       [ 0.9  ,  0.9  ]])')
[(0.0, 0.0), (0.0, 0.0), (0.0, 0.0), (0.0, 0.0), (0.0, 0.0), (0.0, 0.0)] <-- incorrect positions
[ 0.  1.  2.  3.  4.  5.]

我如何獲得xtick主要標簽的位置？ 我從label.get_position（）獲得的值沒有意義。 有一種我不知道的轉變嗎？ 最終我想要（x，y）圖像像素單位中文本框的位置。

Answer 1

如果需要像素坐標，則需要圖形坐標，然后對其進行轉換。

如果您需要有關轉換的更多信息：請查看此matplotlib轉換教程： ref

編輯：為了完整性，我添加了指定dpi的選項，這將影響您的圖形尺寸

import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt

def f(t):
    return np.exp(-t) * np.cos(2*np.pi*t)

t1 = np.arange(0.0, 5.0, 0.1)
t2 = np.arange(0.0, 5.0, 0.02)

# set the dpi you want in your final figure
dpi = 300
mpl.rcParams['figure.dpi'] = dpi

fig, ax = plt.subplots()
plt.plot(t1, f(t1), 'bo', t2, f(t2), 'k')

# saving the figure: don't forget the dpi option!
fig.savefig('./out.png', format='png', dpi=dpi)

xtickslocs = ax.get_xticks()
ymin, _ = ax.get_ylim()
print('xticks pixel coordinates')
print(ax.transData.transform([(xtick, ymin) for xtick in xtickslocs]))
print('label bounding boxes')
print([l.get_window_extent() for l in ax.get_xticklabels()])

xticks pixel coordinates
array([[  60. ,   40. ],
       [ 134.4,   40. ],
       [ 208.8,   40. ],
       [ 283.2,   40. ],
       [ 357.6,   40. ],
       [ 432. ,   40. ]])
label bounding boxes
[Bbox([[56.4375, 25.5555555556], [63.5625, 35.5555555556]]),
 Bbox([[130.8375, 25.5555555556], [137.9625, 35.5555555556]]),
 Bbox([[205.2375, 25.5555555556], [212.3625, 35.5555555556]]),
 Bbox([[279.6375, 25.5555555556], [286.7625, 35.5555555556]]),
 Bbox([[354.0375, 25.5555555556], [361.1625, 35.5555555556]]),
 Bbox([[428.4375, 25.5555555556], [435.5625, 35.5555555556]])]

Answer 2

由於您需要“（x，y）中文本框的位置”，在評論/代碼中根據您的請求調整此答案 ，我們有：

import numpy as np
import matplotlib.pyplot as plt

def f(t):
    return np.exp(-t) * np.cos(2*np.pi*t)

t1 = np.arange(0.0, 5.0, 0.1)
t2 = np.arange(0.0, 5.0, 0.02)

# fig, ax = plt.figure(1)
fig, ax = plt.subplots()
plt.plot(t1, f(t1), 'bo', t2, f(t2), 'k')

print(fig)
print(ax.get_position())


plt.gcf().canvas.draw()
ticks = [tick for tick in plt.gca().get_xticklabels()]

for i, t in enumerate(ticks):
    print("Label ", i, ", data: ", t.get_text(), " ; ", t.get_window_extent())

print(ax.get_xticks())
plt.show()

哪個印刷品：

>>> 
 RESTART: C:/Users/Vinicius/AppData/Local/Programs/Python/Python35-32/stack21.py 
Figure(640x480)
Bbox(x0=0.125, y0=0.10999999999999999, x1=0.9, y1=0.88)
Label  0 , data:    ;  Bbox(x0=102.54545454545455, y0=43.077777777777776, x1=102.54545454545455, y1=43.077777777777776)
Label  1 , data:  0  ;  Bbox(x0=98.17045454545455, y0=29.077777777777776, x1=106.92045454545455, y1=43.077777777777776)
Label  2 , data:  1  ;  Bbox(x0=188.65194870390656, y0=29.077777777777776, x1=197.52694870390656, y1=43.077777777777776)
Label  3 , data:  2  ;  Bbox(x0=279.25844286235855, y0=29.077777777777776, x1=288.00844286235855, y1=43.077777777777776)
Label  4 , data:  3  ;  Bbox(x0=369.80243702081054, y0=29.077777777777776, x1=378.55243702081054, y1=43.077777777777776)
Label  5 , data:  4  ;  Bbox(x0=460.34643117926254, y0=29.077777777777776, x1=469.09643117926254, y1=43.077777777777776)
Label  6 , data:  5  ;  Bbox(x0=550.8904253377145, y0=29.077777777777776, x1=559.6404253377145, y1=43.077777777777776)
Label  7 , data:    ;  Bbox(x0=102.54545454545455, y0=43.077777777777776, x1=102.54545454545455, y1=43.077777777777776)
[-1.  0.  1.  2.  3.  4.  5.  6.]