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Check if character in string has space behind

I am trying to write a function that breaks up camel casing by using a space between words. How do I check if char already has space behind?
def solution(s):
space = ' '
for chr in s:
if chr.isupper() == True:
new_str = s.replace(chr, space + chr)
return new_str
Input:
"camelCaseWord" # a word in camelCasing
Output:
"camel Case Word" # separated by spaces where word starts with capital leter
My solution only gives me "camelCase Word"

Your solution does not work, because you always use the base s as "source":
s = "someWordWithCases"
will replace "W" with " W"and store it in new_str ... then throw that result away and replace "C" with " C" - using the original s again - "W" are back in with no space added before them.
Creating strings by adding to them is wasteful. Strings are immutable so the old one is thrown away after you create a new one.
The solution is to split at capital letters into a list and then join the list elements with spaces:
Yours minimally altered:
def solution(s):
r = []
for chr in s:
# if chr is capital and ( r is empty OR the last one is not a space)
if chr.isupper() and (not r or r[-1] != " "):
# add space then the capital letter
r.append(" "+chr)
else:
# only add letter
r.append(chr)
return ''.join(r)
Or version using slicing:
def solution(s):
k = []
start = 0
for i,c in enumerate(s):
if c.isupper() and start != i:
k.append(s[start:i])
start = i
if c == " ":
k.append(s[start:i])
start = i+1
if i > start:
k.append(s[start:])
return ' '.join(k)
# some test cases are "more" then pure camelCaseSpelling
tests = ["startsLowerThenHasWords", "StartsCapitalThenHasWords",
" starts with spaces no capitals", " Has some Capitals",
"has Capitals ABC and Others that areNotCapitals"]
maxL = max(len(t) for t in tests)
for t in tests:
print(f"{t:<{maxL}} ==> {solution(t)}")
to get
startsLowerThenHasWords ==> starts Lower Then Has Words
StartsCapitalThenHasWords ==> Starts Capital Then Has Words
starts with spaces no capitals ==> starts with spaces no capitals
Has some Capitals ==> Has some Capitals
has Capitals ABC and Others that areNotCapitals ==> has Capitals A B C and Others that are Not Capitals

How about this one? I used enumerate to get the index of iteration.
def solution(s):
space = ' '
space_positions = []
for index, chr in enumerate(s):
print(chr)
if chr != space and chr.isupper() == True:
if s[index - 1] != space:
space_positions.insert(0, index)
for index in space_positions:
s = s[:index] + " " + s[index:]
return s

How to insert space by punctuation?

I have strings like so: hey what is up!, "what did you say?", "he said 'well'", etc. and a regex expression like so: [!%&'\(\)$#\"\/\\*+,-.:;<=>?#\[\]^_´{|}~]´. These are my delimiters and into the strings shown a space shall be inserted like so: "hey what is up !", "what did you say ?", "he said ' well '". So if one of the delimiters is in front of another character sequence, add a space, and if its is after, add space as well.
How can I achieve this? I do not want to split by these delimiters.
Here's my solution but I would be curious how to solve it with regex.
space = set("[!%&'()$#\"/\*+,-.:;<=>?#[]^_´`{|}~]")
for sent in self.sentences:
sent = list(sent)
for i, char in enumerate(sent):
# Make sure to respect length of string when indexing
if i != 0:
# insert space in front if char is punctuation
if sent[i] in space and sent[i - 1] != " ":
sent.insert(i, " ")
if i != len(sent)-1:
# insert space after if char is punctuation
if sent[i] in space and sent[i + 1] != " ":
sent.insert(i + 1, " ")

You could expand your pattern to catch optional spaces and then replace by capture group plus spaces before and after (loop only for demo, not neccessary):
import re
strings = ["hey what is up!", "what did you say?", "he said 'well'"]
pattern = r'(\s?[!%&\'\(\)$#\"\/\\*+,-.:;<=>?#\[\]^_´{|}~]\s?)'
for string in strings:
print(re.sub(pattern, r' \1 ', string))
This will give this output:
hey what is up !
what did you say ?
he said ' well '

Without the aid of the re module you could simply do this:
punctuation = "!%&'()$#\"/\\*+,-.:;<=>?#[]^_´{|}~"
mystring = "Well hello! How are you?"
mylist = list(mystring)
i = 0
for c in mystring:
if c in punctuation:
mylist.insert(i, ' ')
i += 2
else:
i += 1
print(''.join(mylist))

You can make a loop that goes through your strings and when it finds a ponctuation character use the slice function to cut your string in half and concatenate with a space in between.
For example:
for i in yourString:
if yourString[i] == '!':
newString = yourString.slice(0, i) + " " + yourString.slice(i + 1)
It only checks for "!" but you could replace it with a dictionnary of ponctuation characters

Convert the string to a string in which the words are separated by spaces and only the first word starts with an uppercase letter [duplicate]

This question already has answers here:
Split a string at uppercase letters
(22 answers)
Closed 2 years ago.
I am trying to make a script that will accept a string as input in which all of the words are run together, but the first character of each word is uppercase. It should convert the string to a string in which the words are separated by spaces and only the first word starts with an uppercase letter.
For Example (The Input):
"StopWhateverYouAreDoingInterestingIDontCare"
The expected output:
"Stop whatever you are doing interesting I dont care"
Here is the one I wrote so far:
string_input = "StopWhateverYouAreDoingInterestingIDontCare"
def organize_string():
start_sentence = string_input[0]
index_of_i = string_input.index("I")
for i in string_input[1:]:
if i == "I" and string_input[index_of_i + 1].isupper():
start_sentence += ' ' + i
elif i.isupper():
start_sentence += ' ' + i.lower()
else:
start_sentence += i
return start_sentence
While this takes care of some parts, I am struggling with differentiating if the letter "I" is single or a whole word. Here is my output:
"Stop whatever you are doing interesting i dont care"
Single "I" needs to be uppercased, while the "I" in the word "Interesting" should be lowercased "interesting".
I will really appreciate all the help!

A regular expression will do in this example.
import re
s = "StopWhateverYouAreDoingInterestingIDontCare"
t = re.sub(r'(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z])', ' ', s)
Explained:
(?<=[a-z])(?=[A-Z]) - a lookbehind for a lowercase letter followed by a lookahead uppercase letter
| - (signifies or)
(?<=[A-Z])(?=[A-Z]) - a lookbehind for a uppercase letter followed by a lookahead uppercase letter
This regex substitutes a space when there is a lowercase letter followed by an uppercase letter, OR, when there is an uppercase letter followed by an uppercase letter.
UPDATE: This doesn't correctly lowercase the words (with the exception of I and the first_word)
UPDATE2: The fix to this is:
import re
s = "StopWhateverYouAreDoingInterestingIDontCare"
first_word, *rest = re.split(r'(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z])', s)
rest = [word.lower() if word != 'I' else word for word in rest]
print(first_word, ' '.join(rest))
Prints:
Stop whatever you are doing interesting I dont care
Update 3: I looked at why your code failed to correctly form the sentence (which I should have done in the first place instead of posting my own solution :-)).
Here is the corrected code with some remarks about the changes.
string_input = "StopWhateverYouAreDoingInterestingIDontCare"
def organize_string():
start_sentence = string_input[0]
#index_of_i = string_input.index("I")
for i, char in enumerate(string_input[1:], start=1):
if char == "I" and string_input[i + 1].isupper():
start_sentence += ' ' + char
elif char.isupper():
start_sentence += ' ' + char.lower()
else:
start_sentence += char
return start_sentence
print(organize_string())
!. I commented out the line index_of_i = string_input.index("I") as it doesn't do what you need (it finds the index of the first capital I and not an I that should stand alone (it finds the index of the I in Interesting instead of the IDont further in the string_input string). It is not a correct statement.
for i, char in enumerate(string_input[1:], 1) enumerate states the index of the letters in the string starting at 1 (since string_input[1:] starts at index 1 so they are in sync). i is the index of a letter in string_input.
I changed the i's to char to make it clearer that char is the character. Other than these changes, the code stands as you wrote it.
Now the program gives the correct output.

string_input = "StopWhateverYouAreDoingInterestingIDontCare"
counter = 1
def organize_string():
global counter
start_sentence = string_input[0]
for i in string_input[1:]:
if i == "I" and string_input[counter+1].isupper():
start_sentence += ' ' + i
elif i.isupper():
start_sentence += ' ' + i.lower()
else:
start_sentence += i
counter += 1
print(start_sentence)
organize_string()
I made some changes to your program. I used a counter to check the index position. I get your expected output:
Stop whatever you are doing interesting I dont care

s = 'StopWhateverYouAreDoingInterestingIDontCare'
ss = ' '
res = ''.join(ss + x if x.isupper() else x for x in s).strip(ss).split(ss)
sr = ''
for w in res:
sr = sr + w.lower() + ' '
print(sr[0].upper() + sr[1:])
output
Stop whatever you are doing interesting i dont care

I hope this will work fine :-
string_input = "StopWhateverYouAreDoingInterestingIDontCare"
def organize_string():
i=0
while i<len(string_input):
if string_input[i]==string_input[i].upper() and i==0 :
print(' ',end='')
print(string_input[i].upper(),end='')
elif string_input[i]==string_input[i].upper() and string_input[i+1]==string_input[i+1].upper():
print(' ',end='')
print(string_input[i].upper(),end='')
elif string_input[i]==string_input[i].upper() and i!=0:
print(' ',end='')
print(string_input[i].lower(),end='')
if string_input[i]!=string_input[i].upper():
print(string_input[i],end='')
i=i+1
organize_string()

Here is one solution utilising the re package to split the string based on the upper case characters. [Docs]
import re
text = "StopWhateverYouAreDoingInterestingIDontCare"
# Split text by upper character
text_splitted = re.split('([A-Z])', text)
print(text_splitted)
As we see in the output below the separator (The upper case character) and the text before and after is kept. This means that the upper case character is always followed by the rest of the word. The empty first string originates from the first upper case character, which is the first separator.
# Output of print
[
'',
'S', 'top',
'W', 'hatever',
'Y', 'ou',
'A', 're',
'D', 'oing',
'I', 'nteresting',
'I', '',
'D', 'ont',
'C', 'are'
]
As we have seen the first character is always followed by the rest of the word. By combining the two we have the splitted words. This also allows us to easily handle your special case with the I
# Remove first character because it is always empty if first char is always upper
text_splitted = text_splitted[1:]
result = []
for i in range(0, len(text_splitted), 2):
word = text_splitted[i]+text_splitted[i+1]
if (i > 0) and (word != 'I') :
word = word.lower()
result.append(word)
result = ' '.join(result)

split the sentence into individual words. If you find the word "I" in this list, leave it alone. Leave the first word alone. All of the other words, you cast to lower case.

You have to use some string manipulation like this:
output=string_input[0]
for l in string_input[1:]:
if l.islower():
new_s+=l
else:
new_s+=' '+l.lower()
print(output)

Cleaning up a string without split/strip/built-in functions

My requirements
Use Python to create a function cleanstring(S) to "clean up" the spaces in a sentence S.
The sentence may have extra spaces at the front and/or at the end and/or between words.
The subroutine returns a new version of the sentence without the extra spaces.
That is, in the new string, the words should be the same but there should be no spaces at the start, only one space between each word and no spaces at the end.
This program is about you writing code to search through a string to find words and so you are not allowed to use the split function in Python.
You can solve this problem with the basic capabilities of the if and while statements and string operations of len and concatentation.
For example: if the input is: " Hello to the world !" then the output should be: "Hello to the world!"
Question
My program deletes more characters in the program than needed.
Input: " Hello World ! "
Output: "HellWorl"
How do I fix the error in my program?
def cleanupstring (S):
newstring = ["", 0]
j = 1
for i in range(len(S) - 1):
if S[i] != " " and S[i+1] != " ":
newstring[0] = newstring[0] + S[i]
else:
newstring[1] = newstring [1] + 1
return newstring
# main program
sentence = input("Enter a string: ")
outputList = cleanupstring(sentence)
print("A total of", outputList[1], "characters have been removed from your
string.")
print("The new string is:", outputList[0])

Welcome to Stackoverflow. When I started reading I though this was going to be a "please answer my homework" question, but you've actually made a pretty fair effort at solving the problem, so I'm happy to try and help (only you can say whether I actually do).
It's sometimes difficult when you are learning a new language to drop techniques that are much more appropriate in other languages. Doing it character by character you normally just use for c in s rather than incrementing index values like you would in C (though either approach works, index incrementation where not necessary is sometimes regarded as "unpythonic"). Your basic idea seems to be to detect a space followed by another space, otherwise copying characters from the input to the output.
The logic can be simplified by retaining the last character you sent to the output. If it's a space, don't send any more spaces. A loop at the front gets rid of any leading spaces, and since there can be at most one space at the end it can be eliminated easily if present.
I'm not sure why you use a list to keep your results in, as it makes the code much more difficult to understand. If you need to return multiple pieces of information it's much easier to compute them in individual variables and then construct the result in the return statement.
So one desirable modification would be to replace newstring[0] with, say, out_s and newstring[1] with, say count. That will make it a bit clearer what's going on. Then at the end return [out_s, count] if you really need a list. A tuple using return out_s, count would be more usual.
def cleanupstring (s):
out_s = ''
count = 0
last_out = ' '
for c in s:
if c != ' ' or last_out != ' ':
last_out = c
out_s += c
else:
count += 1
if last_out == ' ':
count -= 1
out_s = out_s[:-1]
return out_s, count
# main program
sentence = input("Enter a string: ")
outputList = cleanupstring(sentence)
print("A total of", outputList[1], "characters have been removed from your string.")
print("The new string is:", outputList[0])
Sometimes you just don't have certain pieces of information that would help you to answer the question extremely succinctly. You most likely haven't yet been taught about the strip and replace methods, and so I imagine the following (untested) code
def cleanupstring(s):
out_s = s
while ' ' in out_s:
out_s = out_s.strip().replace(' ', ' ')
return out_s, len(s)-len(out_s)
would be right out.
Also, you can use an "unpacking assignment" to bind the different elements of the function's output directly to names by writing
s, c = cleanupstring(...)
I'm sure you will agree that
print("A total of", c, "characters have been removed from your string.")
print("The new string is:", s)
is rather easier to read. Python values readability so highly because with readable code it's easier to understand the intent of the author. If your code is hard to understand there's a good chance you still have some refactoring to do!

If the "space" it's literally spaces rather than whitespace then the following would work:
import re
def clean_string(value):
return re.sub('[ ]{2,}', ' ', value.strip())
If the stripped values contains consecutive spaces then replace with one space.

My approach would be to keep the last character available and make the decision whether it is a space or not:
def cleanupstring (S):
newstring = ["", 0]
last_character = ' ' # catch initial spaces
for i in range(len(S)-1):
char = S[i]
if char is ' ' and last_character is ' ':
continue # ignore
else:
last_character = char
newstring [0] = newstring[0] + char
return newstring

Count the number of spaces between words in a string

I am doing this problem on Hackerrank,and I came up with the idea, which includes splitting the input and join it afterwards (see my implementation below). However, one of the test cases contains the input (hello< multiple spaces> world), which crashed my code because the input string has more than 1 space between each words. So, I am just wondering if anyone could please help me out fix my code, and I am just wondering how to count how many spaces(esp multiple spaces) in a string in Python. I found how to count spaces in Java, but not in Python. For testcase, I attached the pic.
Thanks in advance.
My implementation:
input_string = input()
splitter = input_string.split()
final = []
for i in range(0,len(splitter)):
for j in range(0,len(splitter[i])):
if(j==0):
final.append(splitter[i][j].upper())
else:
final.append(splitter[i][j])
# Assumed that there is one space btw each words
final.append(' ')
print(''.join(final))
For Test case pic,

You can fix it by splitting with pattern ' ' (whitespace)
splitter = input_string.split(' ')
You can also use .capitalize() method instead of splitting the token again
s = "hello world 4lol"
a = s.split(' ')
new_string = ''
for i in range(0, len(a)) :
new_string = a[i].capitalize() if len(new_string)==0 else new_string +' '+ a[i].capitalize()
print(new_string)
Output:
Hello World 4lol
For counting number of spaces between two words, you can use python's regular expressions module.
import re
s = "hello world loL"
tokens = re.findall('\s+', s)
for i in range(0, len(tokens)) :
print(len(tokens[i]))
Output :
7
2

What I suggest doing for the tutorial question is a quick simple solution.
s = input()
print(s.title())
str.title() will capitalise the starting letter of every word in a string.
Now to answer the question for counting spaces you can use str.count()) which will take a string and return the number of occurrences it finds.
s = 'Hello World'
s.count(' ')
There are various other methods as well, such as:
s = input()
print(len(s) - len(''.join(s.split())))
s2 = input()
print(len(s2) - len(s2.replace(' ', '')))
However count is easiest to implement and follow.
Now, count will return the total number, if you're after the number of spaces between each world.
Then something like this should suffice
s = input()
spaces = []
counter = 0
for char in s:
if char== ' ':
counter += 1
elif counter != 0:
spaces.append(counter)
counter = 0
print(spaces)

import re
line = "Hello World LoL"
total = 0
for spl in re.findall('\s+', line):
print len(spl)
total += len(spl) # 4, 2
print total # 6
>>> 4
>>> 2
>>> 6

For you problem with spaces
my_string = "hello world"
spaces = 0
for elem in my_string:
if elem == " ":
#space between quotes
spaces += 1
print(spaces)

you can use count() function to count repeat of a special character
string_name.count('character')
for count space you should :
input_string = input()
splitter = input_string.split()
final = []
for i in range(0, len(splitter)):
for j in range(0, len(splitter[i])):
if(j==0):
final.append(splitter[i][j].upper())
else:
final.append(splitter[i][j])
final.append(' ')
count = input_string.count(' ')
print(''.join(final))
print (count)
good luck

I solved that problem a time ago, just add " " (white space) to the split function and then print each element separated by a white space. Thats all.
for i in input().split(" "):
print(i.capitalize(), end=" ")
The result of the split function with "hello world lol" is
>>> "hello world lol".split(" ")
>>>['hello', '', '', '', 'world', '', '', '', 'lol']
Then print each element + a white space.

Forget the spaces they are not your problem.
You can reduce the string to just the words without the extra spaces using split(None) which will give you a word count and your string i.e.
>>> a = " hello world lol"
>>> b = a.split(None)
>>> len(b)
3
>>> print(" ".join(b))
hello world lol
Edit: After following your link to read the actual question, next time include the relevant details in your question, it makes it easier all round,
your issue still isn't counting the number of spaces, before, between or after the words. The answer that solves the specific task has already been provided, in the form of:
>>> a= " hello world 42 lol"
>>> a.title()
' Hello World 42 Lol'
>>>
See the answer provided by #Steven Summers

Approach
Given a string, the task is to count the number of spaces between words in a string.
Example:
Input: "my name is geeks for geeks"
Output: Spaces b/w "my" and "name": 1
Spaces b/w "name" and "is": 2
Spaces b/w "is" and "geeks": 1
Spaces b/w "geeks" and "for": 1
Spaces b/w "for" and "geeks": 1
Input: "heyall"
Output: No spaces
Steps to be performed
Input string from the user’s and strip the string for the removing unused spaces.
Initialize an empty list
Run a for loop from 0 till the length of the string
Inside for loop, store all the words without spaces
Again Inside for loop, for storing the actual Indexes of the words.
Outside for loop, print the number of spaces b/w words.
Below is the implementation of the above approach:
# Function to find spaces b/w each words
def Spaces(Test_string):
Test_list = [] # Empty list
# Remove all the spaces and append them in a list
for i in range(len(Test_string)):
if Test_string[i] != "":
Test_list.append(Test_string[i])
Test_list1=Test_list[:]
# Append the exact position of the words in a Test_String
for j in range(len(Test_list)):
Test_list[j] = Test_string.index(Test_list[j])
Test_string[j] = None
# Finally loop for printing the spaces b/w each words.
for i in range(len(Test_list)):
if i+1 < len(Test_list):
print(
f"Spaces b/w \"{Test_list1[i]}\" and \"{Test_list1[i+1]}\": {Test_list[i+1]-Test_list[i]}")
# Driver function
if __name__ == "__main__":
Test_string = input("Enter a String: ").strip() # Taking string as input
Test_string = Test_string.split(" ") # Create string into list
if len(Test_string)==1:
print("No Spaces")
else:
Spaces(Test_string) # Call function