matplotlib.scatter（）不适用于Python 3.6上的Numpy

我很难理解为什么matplotlib.scatter（）在使用Python 3.6.3作为解释器时一直抛出以下异常，但在使用MacBook内置的2.7时效果很好：

Traceback (most recent call last):
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/colors.py", line 132, in to_rgba
    rgba = _colors_full_map.cache[c, alpha]
TypeError: unhashable type: 'numpy.ndarray'

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/axes/_axes.py", line 4050, in scatter
    colors = mcolors.to_rgba_array(c)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/colors.py", line 233, in to_rgba_array
    result[i] = to_rgba(cc, alpha)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/colors.py", line 134, in to_rgba
    rgba = _to_rgba_no_colorcycle(c, alpha)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/colors.py", line 189, in _to_rgba_no_colorcycle
    raise ValueError("RGBA sequence should have length 3 or 4")
ValueError: RGBA sequence should have length 3 or 4

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "/Users/thomastiotto/Documents/USI/1 semester/Machine Learning/Assignments/Assignment 1/skeleton.py", line 458, in <module>
    main()
  File "/Users/thomastiotto/Documents/USI/1 semester/Machine Learning/Assignments/Assignment 1/skeleton.py", line 455, in main
    run_part1()
  File "/Users/thomastiotto/Documents/USI/1 semester/Machine Learning/Assignments/Assignment 1/skeleton.py", line 156, in run_part1
    plot_boundary(p, X, T)
  File "/Users/thomastiotto/Documents/USI/1 semester/Machine Learning/Assignments/Assignment 1/skeleton.py", line 142, in plot_boundary
    plot_data(X, targets)
  File "/Users/thomastiotto/Documents/USI/1 semester/Machine Learning/Assignments/Assignment 1/skeleton.py", line 129, in plot_data
    plt.scatter(X[:, 0], X[:, 1], s=40, c=T, cmap=plt.cm.Spectral)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/pyplot.py", line 3357, in scatter
    edgecolors=edgecolors, data=data, **kwargs)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/__init__.py", line 1710, in inner
    return func(ax, *args, **kwargs)
  File "/Users/thomastiotto/python_envs/MachineLearning/lib/python3.6/site-packages/matplotlib/axes/_axes.py", line 4055, in scatter
    raise ValueError(msg.format(c.shape, x.size, y.size))
ValueError: c of shape (11, 1) not acceptable as a color sequence for x with size 11, y with size 11

其中X是一个[11x2]Numpy数组

如果我将解释器设置为MacOS上的内置Python 2.7，代码运行良好，将其设置为Python 3.6.2或3.6.3会导致上述错误。Matplotlib版本在前一种情况下为1.3.1，在后一种情况下为2.1

---

有什么想法吗？

c需要一个一维数组

拉威尔（T.ravel）应该能做到这一点

plt.scatter(X[:, 0], X[:, 1], s=40, c=T, cmap=plt.cm.Spectral)

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在这个函数中，c需要一维数组，如上面的回答中所述，使用T.ravel或T.reformate（400，）

使用np。reformate：

import numpy as np

t1 = np.array([[1,2,3,4,5,6] , [7,8,9,10,11,12]])
t1_single = np.reshape(t1, -1)
print(t1.shape)
print(t1_single.shape)

输出：
（2,6）
（12，）

---

您还可以使用

c=np.squence（T）

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我认为这里的问题实际上是一个更大的python/numpy问题的一部分，即无法推断1D数组的正确用法。这浪费了大量的编码和调试时间。

11x1矩阵和11向量之间存在差异。使用

c=T[：，0]

。完美！不知道为什么它在较旧版本的软件包中工作，而不是在最新版本中。我在Python 2.6.2和Matplotlib 2.1中遇到了完全相同的问题

plt.scatter(X[:, 0], X[:, 1], s=40, c=T, cmap=plt.cm.Spectral)

import numpy as np

t1 = np.array([[1,2,3,4,5,6] , [7,8,9,10,11,12]])
t1_single = np.reshape(t1, -1)
print(t1.shape)
print(t1_single.shape)