小编ihm*_*all的帖子

我正在使用Python 2.7并尝试让PyBrain工作.

但即使安装了scipy,我也会收到此错误 -

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python2.7/site-packages/PyBrain-0.3.1-
py2.7.egg/pybrain/__init__.py", line 1, in <module>
    from pybrain.structure.__init__ import *
  File "/usr/local/lib/python2.7/site-packages/PyBrain-0.3.1-py2.7.egg/pybrain/structure/__init__.py", line 1, in <module>
    from pybrain.structure.connections.__init__ import *
  File "/usr/local/lib/python2.7/site-packages/PyBrain-0.3.1-py2.7.egg/pybrain/structure/connections/__init__.py", line 1, in <module>
    from pybrain.structure.connections.full import FullConnection
  File "/usr/local/lib/python2.7/site-packages/PyBrain-0.3.1-py2.7.egg/pybrain/structure/connections/full.py", line 3, in <module>
    from scipy import reshape, dot, outer
ImportError: No module named scipy

我用这个命令安装了scipy -

sudo apt-get install python-scipy

我明白了 -

Reading package lists... Done
Building dependency tree       
Reading state …

我有一个目录中的文本文件列表。

我想创建一个矩阵，其中包含每个文件中整个语料库中每个单词的频率。（语料库是目录中每个文件中的每个唯一单词。）

例子：

File 1 - "aaa", "xyz", "cccc", "dddd", "aaa"  
File 2 - "abc", "aaa"
Corpus - "aaa", "abc", "cccc", "dddd", "xyz"  

输出矩阵：

[[2, 0, 1, 1, 1],
 [1, 1, 0, 0, 0]]

我的解决方案是使用collections.Counter每个文件，获取包含每个单词计数的字典，并初始化和大小为n的列表列表 × m（n = 文件数，m = 语料库中唯一单词的数量）。然后，我再次遍历每个文件以查看对象中每个单词的频率，并用它填充每个列表。