如何乘以向量之类的单词列表？

Question

我有这样的单词列表（这里只列出了 2 个）：

list_1 = ['average', 'reasonable'] 
list_2 = ['fiddle', 'frolic']
list_n = ['etc', 'etc']

我希望将这两个列表相乘以获得这个答案：

obj[l1] * obj[l2] = ['average fiddle', 'average frolic', 'reasonable fiddle', 'reasnable frolic']
obj[l1] * obj[l2] *...* obj[n]

我写了这段代码：

import numpy as np
obj = {}
obj['l1'] = np.array(list_1)
obj['l2'] = np.array(list_2)
print(obj['l1']*obj['l2'])

但这只会给我一个错误：

TypeError: ufunc 'multiply' did not contain a loop with signature matching types dtype('<U10') dtype('<U10') dtype('<U10')

我该怎么办？

编辑： 尝试使用以下用户建议的 itertools：

word_list = ['fair play']
output = {'fair': ['average', 'reasonable'], 'play': ['fiddle', 'frolic']}
result = []
for words in word_list: 
    for word in word_tokenize(words): 
        list_1 = output_set[word]
        result = [(x, y) for x, y in product(list_1, result)]
        result = list(map(' '.join, result))
print(result)

但这只会返回一个空集。有没有办法遍历“无限”列表？

Answer 1

使用 itertools.product 我们可以将这些作为tuples 获取，然后使用' '.join 创建str 的

from itertools import product

list_1 = ['average', 'reasonable'] 
list_2 = ['fiddle', 'frolic']
list_n = ['etc', 'vash']

a = [(x, y, z) for x, y, z in product(list_1, list_2, list_n)]
a = list(map(' '.join, a))
# ['average fiddle etc', 'average fiddle vash', 'average frolic etc', 'average frolic vash', 'reasonable fiddle etc', 'reasonable fiddle vash', 'reasonable frolic etc', 'reasonable frolic vash']

Answer 2

如果你必须用列表来做：

In [86]: list_1 = ['average', 'reasonable'] 
    ...: list_2 = ['fiddle', 'frolic']
In [87]: arr1 = np.array(list_1, object)
In [88]: arr2 = np.array(list_2, object)
In [89]: np.add.outer(arr1, arr2)
Out[89]: 
array([['averagefiddle', 'averagefrolic'],
       ['reasonablefiddle', 'reasonablefrolic']], dtype=object)

通过创建对象数组，而不是字符串 dtype，我强制 add ufunc 使用 Python 字符串的 + 方法。正如@Sandeep 的回答所示，字符串添加是一个连接。字符串乘法是一种复制。

还有第三个数组：

In [90]: arr3 = np.array(['etc', 'etc'], object)
In [91]: np.add.outer(np.add.outer(arr1, arr2),arr3)
Out[91]: 
array([[['averagefiddleetc', 'averagefiddleetc'],
        ['averagefrolicetc', 'averagefrolicetc']],

       [['reasonablefiddleetc', 'reasonablefiddleetc'],
        ['reasonablefrolicetc', 'reasonablefrolicetc']]], dtype=object)

我猜你所说的链接操作是什么意思。

我个人更喜欢@vash 的 itertools 解决方案； numpy 并没有对 Python 的字符串处理增加太多。

In [105]: [' '.join(x) for x in itertools.product(arr1,arr2,arr3)]
Out[105]: 
['average fiddle etc',
 'average fiddle etc',
 'average frolic etc',
 'average frolic etc',
 'reasonable fiddle etc',
 'reasonable fiddle etc',
 'reasonable frolic etc',
 'reasonable frolic etc']