删除包含 2 个连续元音的单词

Question

What i want is remove the words which have more than two consecutive vowels in it. So input:

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

Output:

[boat,rain,near,mysterious] 

So here is my code. I was just wondering if there is any better way to do this or is this efficient enough.And if you can do this with python dict or lists are ok? :) I'm new to python so yeah. :) comments would be nice.

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

Alternatively, you could always use regular expressions and list comprehension to get the list of words:

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

s.split() splits the sentence into a list of words. The expression [x for x in s.split()] considers each word in this list in turn.

The re.search(r'[aeiou]{2}', x) part of the expression searches each word for two consecutive letters from the group [aeiou] . Only if two consecutive vowels are found is the word put in the new list.

Answer 2

using sets:

First method using set.intersection will only find non identical consecutive pairs so oo would not be a match:

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']

Method 2 uses set.issubset so now identical consecutive pairs will be considered a match.

using set.issubset with a function using the yield from python 3 syntax which might be more appropriate and indeed to catch repeated identical vowels :

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))])

Or again in a single list comp:

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]))))])

Finally make vowels a set and check if it is a set.issuperset of any pair of chars:

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 3

Using pairwise iteration:

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 4

Use regular expressions instead:

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 5

Using product instead:

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']

or if you don't need to use any external library:

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']

This one is even quicker than regex solution!

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"))