你能解释为什么情节停在J（指数10）

Question

我正在运行此程序以查找特定文本中的字符分布。

# this is a paragraph from python documentation :)
mytext = 'When a letter is first k encountered, it is missing from the mapping, so the default_factory function calls int() to supply a default count of zero. The increment operation then builds up the count for each letter.The function int() which always returns zero is just a special case of constant functions. A faster and more flexible way to create constant functions is to use a lambda function which can supply any constant value (not just zero):'

d = dict()
ignorelist = ('(',')',' ', ',', '.', ':', '_')

for n in mytext:
    if(n not in ignorelist):
        n = n.lower()
        if n in d.keys():
            d[n] = d[n] + 1
        else:
            d[n] = 1
xx = list(d.keys())
yy = list(d.values())

import matplotlib.pyplot as plt
plt.scatter(xx,yy, marker = '*')
plt.show()

该列表都有25个元素。由于一些奇怪的原因，情节就像这样。它以x轴的'J'结尾。

如果我缩放它，右侧可见，但没有绘制点。

Answer 1

请注意，这将从matplotlib版本2.2

开始修复

您似乎在matplotlib 2.1的新分类功能中发现了一个错误。对于单字母类别，它显然会将其功能限制为10个项目。如果类别包含更多字母，则不会发生这种情况。

在任何情况下，解决方案都是绘制数值（就像在matplotlib 2.1之前需要做的那样）。然后将ticklabels设置为类别。

# this is a paragraph from python documentation :)
mytext = 'When a letter is first k encountered, it is missing from the mapping, so the default_factory function calls int() to supply a default count of zero. The increment operation then builds up the count for each letter.The function int() which always returns zero is just a special case of constant functions. A faster and more flexible way to create constant functions is to use a lambda function which can supply any constant value (not just zero):'

d = dict()
ignorelist = ('(',')',' ', ',', '.', ':', '_')

for n in mytext:
    if(n not in ignorelist):
        n = n.lower()
        if n in d.keys():
            d[n] = d[n] + 1
        else:
            d[n] = 1

xx,yy = zip(*d.items())

import numpy as np
import matplotlib.pyplot as plt

xx_sorted, order = np.unique(xx, return_inverse=True)

plt.scatter(order,yy, marker="o")
plt.xticks(range(len(xx)), xx_sorted)
plt.show()