如何在python中从逗号分隔的字符串中查找最长字符串

Question

我有一个用逗号分隔的字符串,。我想从给定的字符串中找出最长的字符串

words = 'run,barn,abcdefghi,yellow,barracuda,shark,fish,swim'

到目前为止我做了什么

print(max(words.split(','), key=len))

我得到的输出是abcdefghi，但是你可以看到abcdefghi和barracuda的长度相同。那么，为什么我只得到一个而不是两个或全部呢?

还

words = 'fishes,sam,gollum,sauron,frodo,balrog'

在上面的string

中，有许多字的长度相同。我要返回每一个

Answer 1

您可以将word的ziplen改为word，然后从len创建dict，并返回最大的len，如下所示:

>>> from collections import defaultdict
>>> words = 'run,barn,abcdefghi,yellow,barracuda,shark,fish,swim'
>>> dct = defaultdict(list)
>>> lstWrdSplt = words.split(',')
>>> for word, length in (zip(lstWrdSplt,(map(len,lstWrdSplt)))):
...    dct[length].append(word)

>>> dct[max(dct)]
['abcdefghi', 'barracuda']
# for more explanation
>>> dct 
defaultdict(list,
{3: ['run'],
4: ['barn', 'fish', 'swim'],
9: ['abcdefghi', 'barracuda'],
6: ['yellow'],
5: ['shark']})

你可以用它作为function，用regex来查找words，如下所示:

from collections import defaultdict
import re
def mxLenWord(words):
dct = defaultdict(list)
lstWrdSplt = re.findall('w+', words)
for word, length in (zip(lstWrdSplt,(map(len,lstWrdSplt)))):
dct[length].append(word.strip())
return dct[max(dct)]

words = 'rUnNiNg ,swimming, eating,biking, climbing'
mxLenWord(words)

输出:

['swimming', 'climbing']

Answer 2

试试下面的

from collections import defaultdict
data = defaultdict(list)
words = 'run,barn,abcdefghi,yellow,barracuda,shark,fish,swim'
for w in words.split(','):
data[len(w)].append(w)
word_len = sorted(data.keys(),reverse=True)
for wlen in word_len:
print(f'{wlen} -> {data[wlen]}')

输出

9 -> ['abcdefghi', 'barracuda']
6 -> ['yellow']
5 -> ['shark']
4 -> ['barn', 'fish', 'swim']
3 -> ['run']

Answer 3

对于这样一个简单的任务，我发现有很多方法太复杂了。您可以使用sorted()和groupby()的组合来解决它:

from itertools import groupby
words = 'run,barn,abcdefghi,yellow,barracuda,shark,fish,swim'
_, (*longest,) = next(groupby(sorted(words.split(","), key=len, reverse=True), len))
print(longest)

要查找所有相同长度的单词，可以使用下一行:

from itertools import groupby
words = 'fishes,sam,gollum,sauron,frodo,balrog'
words_len = {l: list(w) for l, w in groupby(sorted(words.split(","), key=len), len)}
print(words_len)