如何通过 lambda 函数从列表中生成 n 个连续元素？

Question

例如，我有： list = ['你好，你今天过得怎么样'，'你觉得怎么样'] 我们有一个数字 n（连续元素的数量）？

例如n=2，我想得到：['hello how', 'how is', 'is your', 'your day', 'what do', 'do you', 'you think', '想想'，'关于它']

我想使用 lambda 函数：list(map(lambda x: ..., list))
我知道 in ... 必须是 x.split() 你能帮忙吗？

Answer 1

听起来你想要一个滑动窗口。我会使用 itertools 并使用窗口函数对此感到懒惰。 recipe from old python version

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

sentences = ['hello how is your day', 'what do you think about it']

res = [" ".join(words) for words in window((word for sentence in sentences for word in sentence.split()))]

['hello how',
 'how is',
 'is your',
 'your day',
 'day what',
 'what do',
 'do you',
 'you think',
 'think about',
 'about it']

Answer 2

list = ['hello how is your day', 'what do you think about it']
list_consecutive =[]
n=3

#working on each element of list one by one
for i in list:
  sp = i.split()
  count = 0
  
  while count < (len(sp)-n+1):
    #variable to store the word pair eg. 'hello how is'
    word_pair = ''

    #making pair according to value of 'n'
    for j in range(0,n):
      word_pair += sp[count+j]+' '


    #remove extra space from end
    word_pair = word_pair[0:-1]
    
    #appending the created word pair to list
    list_consecutive.append(word_pair)
    count+=1
print(list_consecutive)

输出

['hello how is', 'how is your', 'is your day', 'what do you', 'do you think', 'you think about', 'think about it']