删除其中有 2 个连续元音的单词

Question

我想要的是删除其中包含两个以上连续元音的单词。所以输入：

s = " There was a boat in the rain near the shore, by some mysterious lake"

输出：

[boat,rain,near,mysterious]

这是我的代码。我只是想知道是否有更好的方法来做到这一点，或者如果您可以使用 python 字典或列表来做到这一点，这种方法是否有效 enough.And 是否可以？ :) 我是 python 的新手，是的。 :) 评论会很好。

def change(s):
vowel = ["a","e","i","o","u"]
words = []
a = s[:].replace(",","").split()
for i in vowel:
    s = s.replace(i, "*").replace(",","")
for i,j in enumerate(s.split()):
    if "**" in j:
        words.append(a[i])
return words

Answer 1

或者，您始终可以使用正则表达式和列表理解来获取单词列表：

>>> import re
>>> [x for x in s.split() if re.search(r'[aeiou]{2}', x)]
['boat', 'rain', 'near', 'mysterious']

s.split() 将句子拆分为单词列表。表达式 [x for x in s.split()] 依次考虑此列表中的每个单词。

表达式的 re.search(r'[aeiou]{2}', x) 部分在每个单词中搜索组 [aeiou] 中的两个连续字母。只有找到两个连续的元音才会将单词放入新列表中。

Answer 2

改用正则表达式：

import re

s = 'There was a boat in the rain near the shore, by some mysterious lake'
l = [i for i in s.split(' ') if re.search('[aeiou]{2,}', i)]

print ' '.join(l) # back to string

Answer 3

使用集合：

使用 set.intersection 的第一种方法只会找到不相同的连续对，因此 oo 不会匹配：

s = " There was a boat in the rain near the shore, by some mysterious lake"
vowels = "aeiouAEIOU"
print([x for x in s.split() if any(len(set(x[i:i+2]).intersection(vowels))==  2 for i in range(len(x))) ])
['boat', 'rain', 'near', 'mysterious']

方法 2 使用 set.issubset 因此现在相同的连续对将被视为匹配项。

将 set.issubset 与使用 yield from python 3 语法的函数一起使用，这可能更合适并且确实可以捕获重复的相同元音：

vowels = "aeiouAEIOU"
def get(x, step):
    yield from (x[i:i+step] for i in range(len(x[:-1])))

print([x for x in s.split() if any(set(pr).issubset(vowels) for pr in get(x, 2))])

或者在单个列表中再次组合：

print([x for x in s.split() if any(set(pr).issubset(vowels) for pr in (x[i:i+2] for i in range(len(x[:-1]))))])

最后将元音设为一个集合并检查它是否是任何一对字符的 set.issuperset:

vowels = {'a', 'u', 'U', 'o', 'e', 'i', 'A', 'I', 'E', 'O'}


def get(x, step):
    yield from (x[i:i+step] for i in range(len(x[:-1])))

print([x for x in s.split() if any(vowels.issuperset(pr) for pr in get(x, 2))])

Answer 4

使用成对迭代：

from itertools import tee

def pairwise(iterable):
    a, b = tee(iter(iterable))
    next(b)
    return zip(a,b)

vowels = 'aeiouAEIOU'
[word for word in s.split() if any(
        this in vowels and next in vowels for this,next in pairwise(word))]

Answer 5

改用产品：

from itertools import product

vowels = 'aiueo'
comb = list(product(vowels, repeat=2))
s = " There was a boat in the rain near the shore, by some mysterious lake"


def is2consecutive_vowels(word):
    for i in range(len(word)-1):
        if (word[i], word[i+1]) in comb:
            return True
    return False

print [word for word in s.split() if is2consecutive_vowels(word)]
# ['boat', 'rain', 'near', 'mysterious']

或者如果您不需要使用任何外部库：

vowels = 'aeiou'

def is2consecutive_vowels2(word):
    for i in range(len(word)-1):
        if word[i] in vowels and word[i+1] in vowels:
            return True
    return False

print [word for word in s.split() if is2consecutive_vowels2(word)]
# ['boat', 'rain', 'near', 'mysterious']

这个比正则表达式解决方案还要快！

Answer 6

a=[]
def count(s):
    c=0
    t=s.split()
    for i in t:
        for j in range(len(i)-1):
            w=i[j]
            u=i[j+1]
            if u in "aeiou" and w in "aeiou":
                c+=1
        if(c>=1):
            a.append(i)
        c=0
    return(a)
print(count("There was a boat in the rain near the shore, by some mysterious lake"))

删除其中有 2 个连续元音的单词

Delete words which have 2 consecutive vowels in it

python

algorithm

performance