Question

类似,但我想就外部来源提供的任意名单或文字开展这项任务,根据用户的投入,这些内容可能发生变化。例如,我可以:

input = [ annotate , color , october , cellular , wingding , appeasement , sorta ]

产出应当

output = [ color , october , wingding , sorta ]

感谢任何帮助!

Answer 1

你可以定期表达这一点。

list(filter(lambda x: not  re.search(r (w)1 , x), input))

Answer 2

你们可以把一个词与自己进行比较。这使清单简明扼要:

words = [ annotate , color , october , cellular , wingding , appeasement , sorta ]

[w for w in words if all(a != b for a, b in zip(w, w[1:]))]    
# [ color ,  october ,  wingding ,  sorta ]

zip(w, w[1:])将构成诸如[(a, n ), (n , n ), (n , o .]等信的编号,然后由您进行比较,凡在一行中注明相同字母的。

Answer 3

你可以做这样简单的事情。答案是,你所说的职位是为了达到你的要求。

import re

arr = [ annotate , color , october , cellular , wingding , appeasement , sorta ]
result = [w for w in arr if not re.search(r (w)1+ , w)]

print(result)

It works for word characters, which includes: a-z, A-Z, 0-9, _.
Also: it is a better idea to not keep the array variable name to a keyword (input), as you have in your post.

Answer 4

www.un.org/Depts/DGACM/index_spanish.htm 使用封套检查每字的双字母:

这种解决办法可能比基于“Ex”的解决办法更容易理解。

words = [ annotate , color , october , cellular , wingding , appeasement , sorta ]

for index, word in enumerate(words):
    last = None
    for letter in word:
        if letter == last:
            del words[index]  # Double letters, so delete from the list.
        last = letter
print(words)

产出:

[ color ,  october ,  wingding ,  sorta ]

该守则只是对每字进行两封信的核对,如果它有两封信,则从<条码>中删除。 <>注>如果你指定一个可变的<代码>input,将产生问题。

Answer 5

import re

def filter_words(input_list):
    output_list = []
    pattern = r (.)1   # Regex pattern to match consecutive identical characters
    
    for word in input_list:
        if not re.search(pattern, word):
            output_list.append(word)
    
    return output_list

# Example usage:
input = [ annotate ,  color ,  october ,  cellular ,  wingding ,  appeasement ,  sorta ]
output = filter_words(input)
print(output)

产出:

[ color ,  october ,  wingding ,  sorta ]

友情链接