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我想像这样在 python 中创建二维数组：

     n1 n2 n3 n4 n5

w1   1  4  0  1 10

w2   3  0  7  0  3

w3   0  12 9  5  4

w4   9  0  0  9  7

其中 w1 w2... 是不同的词，而 n1 n2 n3 是不同的博客。
我怎样才能做到这一点？

4

5 回答 5

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假设每个博客中的文本都以字符串形式提供，并且您在 中提供了此类字符串的列表blogs，这就是您创建矩阵的方式。

import re
# Sample input for the following code.
blogs = ["This is a blog.","This is another blog.","Cats? Cats are awesome."]
# This is a list that will contain dictionaries counting the wordcounts for each blog
wordcount = []
# This is a list of all unique words in all blogs.
wordlist = []
# Consider each blog sequentially
for blog in blogs:
    # Remove all the non-alphanumeric, non-whitespace characters,
    # and then split the string at all whitespace after converting to lowercase.
    # eg: "That's not mine." -> "Thats not mine" -> ["thats","not","mine"]
    words = re.sub("\s+"," ",re.sub("[^\w\s]","",blog)).lower().split(" ")
    # Add a new dictionary to the list. As it is at the end,
    # it can be referred to by wordcount[-1]
    wordcount.append({})
    # Consider each word in the list generated above.
    for word in words:
        # If that word has been encountered before, increment the count
        if word in wordcount[-1]: wordcount[-1][word]+=1
        # Else, create a new entry in the dictionary
        else: wordcount[-1][word]=1
        # If it is not already in the list of unique words, add it.
        if word not in wordlist: wordlist.append(word)

# We now have wordlist, which has a unique list of all words in all blogs.
# and wordcount, which contains len(blogs) dictionaries, containing word counts.
# Matrix is the table that you need of wordcounts. The number of rows will be
# equal to the number of unique words, and the number of columns = no. of blogs.
matrix = []
# Consider each word in the unique list of words (corresponding to each row)
for word in wordlist:
    # Add as many columns as there are blogs, all initialized to zero.
    matrix.append([0]*len(wordcount))
    # Consider each blog one by one
    for i in range(len(wordcount)):
        # Check if the currently selected word appears in that blog
        if word in wordcount[i]:
            # If yes, increment the counter for that blog/column
            matrix[-1][i]+=wordcount[i][word]

# For printing matrix, first generate the column headings
temp = "\t"
for i in range(len(blogs)):
    temp+="Blog "+str(i+1)+"\t"

print temp
# Then generate each row, with the word at the starting, and tabs between numbers.

for i in range(len(matrix)):
    temp = wordlist[i]+"\t"
    for j in matrix[i]: temp += str(j)+"\t"
    print temp

现在，将包含该单词在 blog 中出现matrix[i][j]的次数。wordlist[i]blogs[j]

于 2012-10-20T14:28:00.923 回答
0

如果列表或字典中的元组不起作用，请考虑使用pandas：

from pandas import *
In [554]: print DataFrame({'n1':[1,3,0,9], 'n2':[4,0,12,0], 'n3':[0,7,9,0], 'n4':[1,0,5,9], 'n5':[10,3,4,7]},index=['w1','w2','w3','w4'])
    n1  n2  n3  n4  n5
w1   1   4   0   1  10
w2   3   0   7   0   3
w3   0  12   9   5   4
w4   9   0   0   9   7
于 2012-10-20T13:45:55.273 回答
0

我根本不会创建任何列表，也不会创建二维数组，而是创建一个由 x 和 y 标头作为键的字典作为元组。如：

data["w1", "n1"] = 1

这可以被认为是一种“稀疏矩阵”表示。根据您要对数据执行的操作，您可能还需要一个 dict 的 dict，其中外部 dict 的键是 xheader 或 yheader，而内部 dict 的键是相反的。

假设元组作为键表示，将您的数据表作为输入：

text = """\
     n1 n2 n3 n4 n5

w1   1  4  0  1 10

w2   3  0  7  0  3

w3   0  12 9  5  4

w4   9  0  0  9  7
"""

data = {}
lines = text.splitlines()
xheaders = lines.pop(0).split()
for line in lines:
    if not line.strip():
        continue
    elems = line.split()
    yheader = elems[0]
    for (xheader, datum) in zip(xheaders, elems[1:]):
        data[xheader, yheader] = int(datum)
print data
print sorted(data.items())

打印产生：

{('n3', 'w4'): 0, ('n4', 'w2'): 0, ('n2', 'w2'): 0, ('n1', 'w4'): 9, ('n3', 'w3'): 9, ('n2', 'w3'): 12, ('n3', 'w2'): 7, ('n2', 'w4'): 0, ('n5', 'w3'): 4, ('n2', 'w1'): 4, ('n4', 'w1'): 1, ('n5', 'w2'): 3, ('n5', 'w1'): 10, ('n4', 'w3'): 5, ('n4', 'w4'): 9, ('n1', 'w3'): 0, ('n1', 'w2'): 3, ('n5', 'w4'): 7, ('n1', 'w1'): 1, ('n3', 'w1'): 0}
[(('n1', 'w1'), 1), (('n1', 'w2'), 3), (('n1', 'w3'), 0), (('n1', 'w4'), 9), (('n2', 'w1'), 4), (('n2', 'w2'), 0), (('n2', 'w3'), 12), (('n2', 'w4'), 0), (('n3', 'w1'), 0), (('n3', 'w2'), 7), (('n3', 'w3'), 9), (('n3', 'w4'), 0), (('n4', 'w1'), 1), (('n4', 'w2'), 0), (('n4', 'w3'), 5), (('n4', 'w4'), 9), (('n5', 'w1'), 10), (('n5', 'w2'), 3), (('n5', 'w3'), 4), (('n5', 'w4'), 7)]
于 2012-10-20T14:05:08.383 回答
0

一种方法是使用numpy：

>>> from numpy import array
>>> array( [ (1,4,0,1,10), (3,0,7,0,3), (0,12,9,5,4), (9,0,0,9,7) ] )
array([[ 1,  4,  0,  1, 10],
   [ 3,  0,  7,  0,  3],
   [ 0, 12,  9,  5,  4],
   [ 9,  0,  0,  9,  7]])
于 2012-10-20T15:06:02.440 回答
0

如果你只是想要没有任何解析的二维数组，你可以这样写：

a = [
    [1, 4, 0, 1, 10],
    [3, 0, 7, 0, 3],
    [0, 12, 9, 5, 4],
    [9, 0, 0, 9, 7]
]
于 2012-10-21T07:16:33.247 回答