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How to reverse the words of a string considering the punctuation?

Here is what I have so far:
def reversestring(thestring):
words = thestring.split(' ')
rev = ' '.join(reversed(words))
return rev
stringing = input('enter string: ')
print(reversestring(stringing))
I know I'm missing something because I need the punctuation to also follow the logic.
So let's say the user puts in Do or do not, there is no try.. The result should be coming out as .try no is there , not do or Do, but I only get try. no is there not, do or Do. I use a straightforward implementation which reverse all the characters in the string, then do something where it checks all the words and reverses the characters again but only to the ones with ASCII values of letters.

Try this (explanation in comments of code):
s = "Do or do not, there is no try."
o = []
for w in s.split(" "):
puncts = [".", ",", "!"] # change according to needs
for c in puncts:
# if a punctuation mark is in the word, take the punctuation and add it to the rest of the word, in the beginning
if c in w:
w = c + w[:-1] # w[:-1] gets everthing before the last char
o.append(w)
o = reversed(o) # reversing list to reverse sentence
print(" ".join(o)) # printing it as sentence
#output: .try no is there ,not do or Do

Your code does exactly what it should, splitting on space doesn't separator a dot ro comma from a word.
I'd suggest you use re.findall to get all words, and all punctation that interest you
import re
def reversestring(thestring):
words = re.findall(r"\w+|[.,]", thestring)
rev = ' '.join(reversed(words))
return rev
reversestring("Do or do not, there is no try.") # ". try no is there , not do or Do"

You can use regular expressions to parse the sentence into a list of words and a list of separators, then reverse the word list and combine them together to form the desired string. A solution to your problem would look something like this:
import re
def reverse_it(s):
t = "" # result, empty string
words = re.findall(r'(\w+)', s) # just the words
not_s = re.findall(r'(\W+)', s) # everything else
j = len(words)
k = len(not_s)
words.reverse() # reverse the order of word list
if re.match(r'(\w+)', s): # begins with a word
for i in range(k):
t += words[i] + not_s[i]
if j > k: # and ends with a word
t += words[k]
else: # begins with punctuation
for i in range(j):
t += not_s[i] + words[i]
if k > j: # ends with punctuation
t += not_s[j]
return t #result
def check_reverse(p):
q = reverse_it(p)
print("\"%s\", \"%s\"" % (p, q) )
check_reverse('Do or do not, there is no try.')
Output
"Do or do not, there is no try.", "try no is there, not do or Do."
It is not a very elegant solution but sure does work!

How does this code only print out the initials of a string?

This is the function:
def initials(phrase):
words = phrase.split()
result = ""
for word in words:
result += word[0]
return result.upper()
This is an exercise on my online course. The objective is to return the first initials of a string capitalized. For example, initials ("Universal Serial Bus") should return "USB".

phrase is a str type object.
str objects can have functions applied to them through their methods. split is a function that returns a list containing multiple str objects. This is stored in words
the for word in words takes each element of words and puts it in the variable word for each iteration of the loop.
The += function adds the first letter of word to result by accessing the first character of the str by using the [0] index of word.
Then the upper function is applied to the result.
I hope this clears it up for you.

def initials(phrase):
words = phrase.split()
result = ""
for word in words:
result += word[0]
return result.upper()
This:
Splits the phrase at every space (" "), with phrase.split(). .split() returns a list which is assigned to words
Iterates through the list words and adds the first letter of each word (word[0]) to the result variable.
Returns result converted to uppercase (result.upper())

def initials(phrase):
words = phrase.split()
result = ""
for word in words:
result += word[0].upper()
return result
print(ShortName("Active Teens Taking Initiative To Understand Driving Experiences"))
Should be: ATTITUDE

def initials(phrase):
words =phrase.split()
result=""+""
for word in words:
result += word[0].upper()
return result
print(initials("Universal Serial Bus")) # Should be: USB
print(initials("local area network")) # Should be: LAN
print(initials("Operating system")) # Should be: OS
Here is output:
USB
LAN
OS
This:
Splits the phrase at every space (" "+" ") and concatenate next one first letter,with phrase.split() returns a list which is assigned to words Iterates through the list words and adds the first letter of each word (word[0]) to the result variable.
Returns result converted to uppercase (result.upper())

strong text
def initials(phrase):
words = phrase.split()
result = ""
for word in words:
result += word[0].uppper()
return result

How do I swap a letter with its immediate neighbor in a string?

I am trying to create a function that takes a string as an argument and returns a list of all of the generated words by swapping a letter with its immediate neighbor.
I first take each letter and create a list of strings and each string contains one letter.
Then I iterate through the new list of letters and try to swap them.
Then I join the letters together to form a string.
Then I append the new string to the list that I return.
Here is my code. Please tell me how to fix it. I don't want it to display the passed word in the list. Thank you.
def mixedString(word):
word = word.lower()
letters = []
newArray = []
for n in word:
letter = f"{n}"
letters.append(letter)
newList = []
for i in range(len(letters)):
newWord = ""
newArray = letters[i:] + letters[:i]
newWord = "".join(newArray)
newList.append(newWord)
return newList
myWord = "Dog"
print(mixedString(myWord))

Hint: there are only n - 1 distinct words where one letter of the original word has been swapped. To see why, note that ab only has ba as result.
If a word has the letters are position i and i+1 swapped the letters before i are unchanged and the letters after i + 1 also are unchanged.
def swap(s, i):
return s[:i] + s[i+1] + s[i] + s[i+2:]
def neighbors(s):
return [swap(s, i) for i in range(len(s)-1)]

You can swap letters like this:
def swap(string, place_1, place_2):
string = list(string)
string[place_1], string[place_2] = string[place_2], string[place_1]
return ''.join(string)
a = '1234'
print(swap(a, 1, 2))
>>> 1324

There are many points of improvement, but I will be addressing the one you ask for
I don't want it to display the passed word in the list.
Just skip the first iteration in the relevant loop, e.g. by replacing this:
for i in range(len(letters)):
with this:
for i in range(1, len(letters)):

What would be the easiest way to search through a list?

It's actually a string but I just converted it to a list because the answer is supposed to be returned as a list. I've been looking at this problem for hours now and cannot get it. I'm supposed to take a string, like "Mary had a little lamb" for example and another string such as "ab" for example and search through string1 seeing if any of the letters from string2 occur. So if done correctly with the two example it would return
["a=4","b=1"]
I have this so far:
def problem3(myString, charString):
myList = list(myString)
charList = list(charString)
count = 0
newList = []
newString = ""
for i in range(0,len(myList)):
for j in range(0,len(charList)):
if charList[j] == myList[i]:
count = count + 1
newString = charList[j] + "=" + str(count)
newList.append(newString)
return newList
Which returns [a=5] I know it's something with the newList.append(string) and where it should be placed, anyone have any suggestions?

You can do this very easily with list comprehensions and the count function that strings (and lists!) have:
Split the search string into a list of chars.
For each character in the search string, loop over the input string and determine how much it occurs (via count).
Example:
string = 'Mary had a little lamb'
search_string = 'ab'
search_string_chars = [char for char in search_string]
result = []
for char in search_string_chars:
result.append('%s=%d' % (char, string.count(char)))
Result:
['a=4', 'b=1']
Note that you don't need to split the search_string ('ab') into a list of characters, as strings are already lists of characters - the above was done that way to illustrate the concept. Hence, a reduced version of the above could be (which also yields the same result):
string = 'Mary had a little lamb'
search_string = 'ab'
result = []
for char in search_string:
result.append('%s=%d' % (char, string.count(char)))

Here's a possible solution using Counter as mentioned by coder,
from collections import Counter
s = "Mary had a little lambzzz"
cntr = Counter(s)
test_str = "abxyzzz"
results = []
for letter in test_str:
if letter in s:
occurrances = letter + "=" + str(cntr.get(letter))
else:
occurrances = letter + "=" + "0"
if occurrances not in results:
results.append(occurrances)
print(results)
output
['a=4', 'b=1', 'x=0', 'y=1', 'z=3']

import collections
def count_chars(s, chars):
counter = collections.Counter(s)
return ['{}={}'.format(char, counter[char]) for char in set(chars)]
That's all. Let Counter do the work of actually counting the characters in the string. Then create a list comprehension of format strings using the characters in chars. (chars should be a set and not a list so that if there are duplicate characters in chars, the output will only show one.)

Python: Find the longest word in a string

I'm preparing for an exam but I'm having difficulties with one past-paper question. Given a string containing a sentence, I want to find the longest word in that sentence and return that word and its length. Edit: I only needed to return the length but I appreciate your answers for the original question! It helps me learn more. Thank you.
For example: string = "Hello I like cookies". My program should then return "Cookies" and the length 7.
Now the thing is that I am not allowed to use any function from the class String for a full score, and for a full score I can only go through the string once. I am not allowed to use string.split() (otherwise there wouldn't be any problem) and the solution shouldn't have too many for and while statements. The strings contains only letters and blanks and words are separated by one single blank.
Any suggestions? I'm lost i.e. I don't have any code.
Thanks.
EDIT: I'm sorry, I misread the exam question. You only have to return the length of the longest word it seems, not the length + the word.
EDIT2: Okay, with your help I think I'm onto something...
def longestword(x):
alist = []
length = 0
for letter in x:
if letter != " ":
length += 1
else:
alist.append(length)
length = 0
return alist
But it returns [5, 1, 4] for "Hello I like cookies" so it misses "cookies". Why? EDIT: Ok, I got it. It's because there's no more " " after the last letter in the sentence and therefore it doesn't append the length. I fixed it so now it returns [5, 1, 4, 7] and then I just take the maximum value.
I suppose using lists but not .split() is okay? It just said that functions from "String" weren't allowed or are lists part of strings?

You can try to use regular expressions:
import re
string = "Hello I like cookies"
word_pattern = "\w+"
regex = re.compile(word_pattern)
words_found = regex.findall(string)
if words_found:
longest_word = max(words_found, key=lambda word: len(word))
print(longest_word)

Finding a max in one pass is easy:
current_max = 0
for v in values:
if v>current_max:
current_max = v
But in your case, you need to find the words. Remember this quote (attribute to J. Zawinski):
Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.
Besides using regular expressions, you can simply check that the word has letters. A first approach is to go through the list and detect start or end of words:
current_word = ''
current_longest = ''
for c in mystring:
if c in string.ascii_letters:
current_word += c
else:
if len(current_word)>len(current_longest):
current_longest = current_word
current_word = ''
else:
if len(current_word)>len(current_longest):
current_longest = current_word
A final way is to split words in a generator and find the max of what it yields (here I used the max function):
def split_words(mystring):
current = []
for c in mystring:
if c in string.ascii_letters:
current.append(c)
else:
if current:
yield ''.join(current)
max(split_words(mystring), key=len)

Just search for groups of non-whitespace characters, then find the maximum by length:
longest = len(max(re.findall(r'\S+',string), key = len))

For python 3. If both the words in the sentence is of the same length, then it will return the word that appears first.
def findMaximum(word):
li=word.split()
li=list(li)
op=[]
for i in li:
op.append(len(i))
l=op.index(max(op))
print (li[l])
findMaximum(input("Enter your word:"))

It's quite simple:
def long_word(s):
n = max(s.split())
return(n)
IN [48]: long_word('a bb ccc dddd')
Out[48]: 'dddd'

found an error in a previous provided solution, he's the correction:
def longestWord(text):
current_word = ''
current_longest = ''
for c in text:
if c in string.ascii_letters:
current_word += c
else:
if len(current_word)>len(current_longest):
current_longest = current_word
current_word = ''
if len(current_word)>len(current_longest):
current_longest = current_word
return current_longest

I can see imagine some different alternatives. Regular expressions can probably do much of the splitting words you need to do. This could be a simple option if you understand regexes.
An alternative is to treat the string as a list, iterate over it keeping track of your index, and looking at each character to see if you're ending a word. Then you just need to keep the longest word (longest index difference) and you should find your answer.

Regular Expressions seems to be your best bet. First use re to split the sentence:
>>> import re
>>> string = "Hello I like cookies"
>>> string = re.findall(r'\S+',string)
\S+ looks for all the non-whitespace characters and puts them in a list:
>>> string
['Hello', 'I', 'like', 'cookies']
Now you can find the length of the list element containing the longest word and then use list comprehension to retrieve the element itself:
>>> maxlen = max(len(word) for word in string)
>>> maxlen
7
>>> [word for word in string if len(word) == maxlen]
['cookies']

This method uses only one for loop, doesn't use any methods in the String class, strictly accesses each character only once. You may have to modify it depending on what characters count as part of a word.
s = "Hello I like cookies"
word = ''
maxLen = 0
maxWord = ''
for c in s+' ':
if c == ' ':
if len(word) > maxLen:
maxWord = word
word = ''
else:
word += c
print "Longest word:", maxWord
print "Length:", len(maxWord)

Given you are not allowed to use string.split() I guess using a regexp to do the exact same thing should be ruled out as well.
I do not want to solve your exercise for you, but here are a few pointers:
Suppose you have a list of numbers and you want to return the highest value. How would you do that? What information do you need to track?
Now, given your string, how would you build a list of all word lengths? What do you need to keep track of?
Now, you only have to intertwine both logics so computed word lengths are compared as you go through the string.

My proposal ...
import re
def longer_word(sentence):
word_list = re.findall("\w+", sentence)
word_list.sort(cmp=lambda a,b: cmp(len(b),len(a)))
longer_word = word_list[0]
print "The longer word is '"+longer_word+"' with a size of", len(longer_word), "characters."
longer_word("Hello I like cookies")

import re
def longest_word(sen):
res = re.findall(r"\w+",sen)
n = max(res,key = lambda x : len(x))
return n
print(longest_word("Hey!! there, How is it going????"))
Output : there
Here I have used regex for the problem. Variable "res" finds all the words in the string and itself stores them in the list after splitting them.
It uses split() to store all the characters in a list and then regex does the work.
findall keyword is used to find all the desired instances in a string. Here \w+ is defined which tells the compiler to look for all the words without any spaces.
Variable "n" finds the longest word from the given string which is now free of any undesired characters.
Variable "n" uses lambda expressions to define the key len() here.
Variable "n" finds the longest word from "res" which has removed all the non-string charcters like %,&,! etc.
>>>#import regular expressions for the problem.**
>>>import re
>>>#initialize a sentence
>>>sen = "fun&!! time zone"
>>>res = re.findall(r"\w+",sen)
>>>#res variable finds all the words and then stores them in a list.
>>>res
Out: ['fun','time','zone']
>>>n = max(res)
Out: zone
>>>#Here we get "zone" instead of "time" because here the compiler
>>>#sees "zone" with the higher value than "time".
>>>#The max() function returns the item with the highest value, or the item with the highest value in an iterable.
>>>n = max(res,key = lambda x:len(x))
>>>n
Out: time
Here we get "time" because lambda expression discards "zone" as it sees the key is for len() in a max() function.

list1 = ['Happy', 'Independence', 'Day', 'Zeal']
listLen = []
for i in list1:
listLen.append(len(i))
print list1[listLen.index(max(listLen))]
Output - Independence