以相同的顺序同时洗牌两个列表

Question

我正在使用nltk库的movie_reviews语料库，其中包含大量文档。我的任务是在数据预处理和不预处理的情况下获得这些评论的预测性能。但是有一个问题，在列表documents和documents2中，我有相同的文档，我需要将它们洗牌，以便在两个列表中保持相同的顺序。我不能单独洗牌，因为每次洗牌列表时，我都会得到其他结果。这就是为什么我需要立即以相同的顺序洗牌，因为我需要在最后比较它们（这取决于顺序）。我正在使用python 2.7

示例（在real中，字符串是标记化的，但不是相对的）：

documents = [(['plot : two teen couples go to a church party , '], 'neg'),
             (['drink and then drive . '], 'pos'),
             (['they get into an accident . '], 'neg'),
             (['one of the guys dies'], 'neg')]

documents2 = [(['plot two teen couples church party'], 'neg'),
              (['drink then drive . '], 'pos'),
              (['they get accident . '], 'neg'),
              (['one guys dies'], 'neg')]

我需要在洗牌两个列表后得到这个结果：

documents = [(['one of the guys dies'], 'neg'),
             (['they get into an accident . '], 'neg'),
             (['drink and then drive . '], 'pos'),
             (['plot : two teen couples go to a church party , '], 'neg')]

documents2 = [(['one guys dies'], 'neg'),
              (['they get accident . '], 'neg'),
              (['drink then drive . '], 'pos'),
              (['plot two teen couples church party'], 'neg')]

我有以下代码：

def cleanDoc(doc):
    stopset = set(stopwords.words('english'))
    stemmer = nltk.PorterStemmer()
    clean = [token.lower() for token in doc if token.lower() not in stopset and len(token) > 2]
    final = [stemmer.stem(word) for word in clean]
    return final

documents = [(list(movie_reviews.words(fileid)), category)
             for category in movie_reviews.categories()
             for fileid in movie_reviews.fileids(category)]

documents2 = [(list(cleanDoc(movie_reviews.words(fileid))), category)
             for category in movie_reviews.categories()
             for fileid in movie_reviews.fileids(category)]

random.shuffle( and here shuffle documents and documents2 with same order) # or somehow

Answer 1

您可以这样做：

import random

a = ['a', 'b', 'c']
b = [1, 2, 3]

c = list(zip(a, b))

random.shuffle(c)

a, b = zip(*c)

print a
print b

[OUTPUT]
['a', 'c', 'b']
[1, 3, 2]

当然，这是一个简单列表的例子，但是对于您的情况，改编将是相同的

希望能有帮助。祝你好运

Answer 2

我有一个简单的方法来做这件事

import numpy as np
a = np.array([0,1,2,3,4])
b = np.array([5,6,7,8,9])

indices = np.arange(a.shape[0])
np.random.shuffle(indices)

a = a[indices]
b = b[indices]
# a, array([3, 4, 1, 2, 0])
# b, array([8, 9, 6, 7, 5])

Answer 3

from sklearn.utils import shuffle

a = ['a', 'b', 'c','d','e']
b = [1, 2, 3, 4, 5]

a_shuffled, b_shuffled = shuffle(np.array(a), np.array(b))
print(a_shuffled, b_shuffled)

#random output
#['e' 'c' 'b' 'd' 'a'] [5 3 2 4 1]