绘制Pandas中的二进制矩阵

Question

我在pandas中有一个数据帧（数据），它有一个datetimeindex（大约25.000天的数据）和527列ID。

                  work_id_10  work_id_100  work_id_1007  work_id_1009
concert_date
1917-01-27             0            0             0             0
1917-01-28             0            0             0             0
1917-01-29             0            0             0             0
1917-01-30             0            0             0             0
1917-01-31             0            0             0             0

每个列ID表示存在或不存在具有0（不存在）或1（存在）的特定ID。所以，基本上我所拥有的是二进制值矩阵。

我现在想要创建一个在x轴上包含所有日期的图表，并为每个列ID创建作为点的存在。我正在使用ipython。

%matplotlib
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
ax.set_yticklabels(data.index)
ax.set_xticklabels(data.columns)
plt.imshow/data, cmap='Greys', interpolation='none')

这给了我一个MemoryError：

Traceback (most recent call last):
  File "C:\Python27\Lib\lib-tk\Tkinter.py", line 1486, in __call__
    return self.func(*args)
  File "C:\Python27\Lib\lib-tk\Tkinter.py", line 533, in callit
    func(*args)
  File "C:\Python27\lib\site-packages\matplotlib\backends\backend_tkagg.py", lin
e 365, in idle_draw
    self.draw()
  File "C:\Python27\lib\site-packages\matplotlib\backends\backend_tkagg.py", lin
e 349, in draw
    FigureCanvasAgg.draw(self)
  File "C:\Python27\lib\site-packages\matplotlib\backends\backend_agg.py", line
469, in draw
    self.figure.draw(self.renderer)
  File "C:\Python27\lib\site-packages\matplotlib\artist.py", line 59, in draw_wr
apper
    draw(artist, renderer, *args, **kwargs)
  File "C:\Python27\lib\site-packages\matplotlib\figure.py", line 1079, in draw
    func(*args)
  File "C:\Python27\lib\site-packages\matplotlib\artist.py", line 59, in draw_wr
apper
    draw(artist, renderer, *args, **kwargs)
  File "C:\Python27\lib\site-packages\matplotlib\axes\_base.py", line 2092, in d
raw
    a.draw(renderer)
  File "C:\Python27\lib\site-packages\matplotlib\artist.py", line 59, in draw_wr
apper
    draw(artist, renderer, *args, **kwargs)
  File "C:\Python27\lib\site-packages\matplotlib\image.py", line 367, in draw
    self._draw_unsampled_image(renderer, gc)
  File "C:\Python27\lib\site-packages\matplotlib\image.py", line 321, in _draw_u
nsampled_image
    self._get_unsampled_image(self._A, extent_in_ic, viewLim_in_ic)
  File "C:\Python27\lib\site-packages\matplotlib\image.py", line 219, in _get_un
sampled_image
    x = (x * 255).astype(np.uint8)
MemoryError

这是正确的方法，为什么我会得到MemoryError？

谢谢！

Answer 1

正如我在评论中提到的，您可能希望将您的数据分解为更易读的块。以下是尺寸为527 x 2500的随机矩阵（1s为蓝色，0s为白色）的示例：

您的数据很可能具有更多结构，但可能仍难以解释。您描述的矩阵将是527 x 25000.您可以按年份（527 x 365）或十年（527 x 3650ish）显示，或者玩游戏并查看最佳效果。

以下是我将如何显示您的数据矩阵（这是针对更小的集合）：

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import datetime

data = pd.read_csv('concertdata.csv')
print data

这会打印我的假数据：

  concert_date  work_id_10  work_id_100  work_id_1007  work_id_1009  \
0   1917-01-27           1            1             0             0   
1   1917-01-28           0            0             1             0   
2   1917-01-29           0            1             1             0   
3   1917-01-30           1            0             0             0   
4   1917-01-31           0            0             0             0   
5   1917-02-01           0            0             1             1   

   work_id_1011  
0             0  
1             0  
2             1  
3             1  
4             1  
5             0  

然后获取标题和值：

id_labels = data.columns[1:]
# take the transpose since you want to see id on y-axis
id_matrix = np.array(data[id_labels].values, dtype=float).T
concert_dates = pd.to_datetime(data['concert_date'])
concert_dates = [d.date() for d in concert_dates]

现在使用imshow（）绘制这个：

fig, ax = plt.subplots()
mat = ax.imshow(id_matrix, cmap='GnBu', interpolation='nearest')
plt.yticks(range(id_matrix.shape[0]), id_labels)
plt.xticks(range(id_matrix.shape[1]), concert_dates)
plt.xticks(rotation=30)
plt.xlabel('Concert Dates')

# this places 0 or 1 centered in the individual squares
for x in xrange(id_matrix.shape[0]):
    for y in xrange(id_matrix.shape[1]):
        ax.annotate(str(id_matrix[x, y])[0], xy=(y, x), 
                    horizontalalignment='center', verticalalignment='center')
plt.show()

你可以玩得更漂亮，但这是一般的想法。