Как найти самое короткое слово в питоне

Yes, the examples given are very different.

The first example effectively says:

Take the string s, split it by spaces, and then take each word, x, found and return the minimum value of just the length of x.

The second example effectively says:

Find the minimum value in the list generated by len(x) for x in s.split().

That first example generates an error because the min function expects to compare at least 2 or more elements, and only 1 is provided.

That second example works because the list that is generated by len(x) for x in s.split() converts a string, like say "Python types with ducks?" to a list of word lengths (in my example, it would convert the string to [6, 5, 4, 6]). That list that is generated (this is also why it’s called a generator), is what the min function then uses to find the minimum value inside said list.

Another way to write that first example so that it works like you would expect is like this

def find_short(s):
    min_length = float("inf")

    for x in s.split():
        if len(x) < min_length:
            min_length = len(x)

    return min_length

However, notice how you have to keep track of a variable that you do not have to define using the list generator method in your second example. Although this is not a big deal when you are learning programming for the first time, it becomes a bigger deal when you start making larger, more complex programs.

Sidenote:

Any value that follows the return keyword is what a function «outputs», and thus no more code gets executed.

For example, in your first example (and assuming that the error was not generated), your loop would only ever execute once regardless of the string you give it because it does not check that you actually have found the value you want. What I mean by that is that any time your code encounters a return statement, it means that your function is done.

That is why in my example find_short function, I have an if statement to check that I have the value that I want before committing to the return statement that exits the function entirely.

Можно сделать так:

import re
s = "Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum."
words = re.sub("[^w]", " ",  s).split()
#  если просто найти все минимальные слова
t = [v for v in words if len(v) == len(min(words))]

# если все слова надо разделить по длине
from itertools import groupby
t1 = {k:list(v) for k,v in groupby(sorted(words, key=len), len)}

print(t1[min(t1)], t)

Я работаю над заданием, где мне нужно, чтобы кто-то ввел строку слов, и моя программа должна затем выбрать длину самого короткого и самого длинного слова в этой строке. Я успешно сделал это, чтобы выбрать длину самого длинного слова:

def longWord(string):
    alist = []
    length = 0
    for letter in string:
        if letter != " ":
            length +=1
        else:
            alist.append(length)
            length = 0
        return alist

Поэтому, если ввод «счастливый день», моя функция печати сообщает мне, что самое длинное слово состоит из 5 символов.

Тем не менее, я не могу понять, как написать что-то, что выберет длину самого короткого слова во входной строке. Как бы я это сделал?