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
Related
So I have, p.e, this string: ' I love python ' and I want to convert all the spaces to '_'. My problem is that I also need to delete the outside spaces so I dont finish with the result: '_I_love_python__' and more like this 'I_love_python'
I searched and found out that I can develop it with a single line of code mystring.strip().replace(" ", "_") which is unfortunaly is sintax that I cant apply in my essay.
So what I landed with was this:
frase= str(input('Introduza: '))
aux=''
for car in frase:
if car==' ':
car='_'
aux+=car
else:
aux+=car
print(aux)
My problem now is on deleting those outside spaces. What I thought about was runing another for i in in the start and another on the final of the string and to stop until they found a non space caracter. But unfortunaly I havent been able to do that...
Apreciate all the help you can suply!
I came up with following solution:
You iterate over the string, but instead of replacing the space with underscore as soon as it appears, you store the amount of spaces encountered. Then, once a non-space-character is reached, you add the amount of spaces found to the string. So if the string ends with lots of spaces, it will never reach a non-space-character and therefore never add the underscores.
For cutting off the spaces at the beginning, I just added a condition to add the underscores being: "Have I encountered a non-space-character before?"
Here is the code:
text = " I love python e "
out = ""
string_started = False
underscores_to_add = 0
for c in text:
if c == " ":
underscores_to_add += 1
else:
if string_started:
out += "_" * underscores_to_add
underscores_to_add = 0
string_started = True
out += c
print(out) # prints "I_love___python____e"
You can use the following trick to remove leading and trailing spaces in your string:
s = ' I love python '
ind1 = min(len(s) if c == ' ' else n for n, c in enumerate(s))
ind2 = max(0 if c == ' ' else n for n, c in enumerate(s))
s = ''.join('_' if c == ' ' else c for c in s[ind1:ind2 + 1])
print('*', s, '*', sep='')
Output:
*I_love_python*
If you are not allowed to use strip() method
def find(text):
for i, s in enumerate(text):
if s != " ":
break
return i
text = " I love python e "
text[find(text):len(text)-find(text[::-1])].replace(" ","_")
texts = [" I love python e ","I love python e"," I love python e","I love python e ", "I love python e"]
for text in texts:
print (text[find(text):len(text)-find(text[::-1])].replace(" ","_"))
output:
I_love___python____e
I_love___python____e
I_love___python____e
I_love___python____e
I_love___python____e
Given a string find will find the first non space character in the string
Use find to find the first nonspace character and the last nonspace character
Get the substring using above found indices
Replace all spaces with _ in the above substring
I need to get a user to enter a sentence for an assignment. Using a for loop, I then need to replace all spaces with %20 in order to prep the string to be used as a URL. I cannot for the life of me figure it out.
sentence = str(input("Please enter sentence:"))
space = (" ")
for space in sentence:
space.replace(sentence, space, "%20", 500)
print(sentence)
This is what I have entered so far but it is completely wrong.
String in Python can't be modified. The replace() function returns a new string with all the replacements done. You need to assign this result somewhere. So you can do:
sentence = sentence.replace(" ", "%20")
If you want to do it with a loop, you need to build the result in another variable.
new_sentence = ""
for char in sentence:
if char == " ":
new_sentence += "%20"
else:
new_sentence += char
There's no point in using replace() in the loop.
Replace everything after the first line with print(sentence.replace(" ","%20"))
or, if you really must use a loop:
s = ''
for x in sentence:
if x == ' ':
s += "%20"
else:
s += x
print(s)
you've got a few problems here:
the variable space that you define before the loop is equal to " " but then you're using the same name for the for loop variable, which is going to iterate through the string the sentence string, in the loop, it's going to use the value defined by the loop.
You're giving the replace method too many arguments, it's a method of the string object so you don't need to pass it the large string to operate on Replace Method
You're using the replace method on the space string and not on the sentence string
Good way to do it in real life:
old = ' '
new = '%20'
sentence = str(input("Please enter sentence:"))
print(sentence.replace(old,new))
The way your teacher is probably after:
sentence = str(input("Please enter sentence:"))
old = ' '
new = '%20'
newSentence = ''
for letter in sentence:
if letter == ' ':
newSentence += new
else:
newSentence += letter
print(newSentence)
I have to enter a string, remove all spaces and print the string without vowels. I also have to print a string of all the removed vowels.
I have gotten very close to this goal, but for some reason when I try to remove all the vowels it will not remove two vowels in a row. Why is this? Please give answers for this specific block of code, as solutions have helped me solve the challenge but not my specific problem
# first define our function
def disemvowel(words):
# separate the sentence into separate letters in a list
no_v = list(words.lower().replace(" ", ""))
print no_v
# create an empty list for all vowels
v = []
# assign the number 0 to a
a = 0
for l in no_v:
# if a letter in the list is a vowel:
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
# add it to the vowel list
v.append(l)
#print v
# delete it from the original list with a
del no_v[a]
print no_v
# increment a by 1, in order to keep a's position in the list moving
else:
a += 1
# print both lists with all spaces removed, joined together
print "".join(no_v)
print "".join(v)
disemvowel(raw_input(""))
Mistakes
So there are a lot of other, and perhaps better approaches to solve this problem. But as you mentioned I just discuss your failures or what you can do better.
1. Make a list of input word
There are a lot of thins you could do better
no_v = list(words.lower().replace(" ", ""))
You don't replaces all spaces cause of " " -> " " so just use this instead
no_v = list(words.lower().translate( None, string.whitespace))
2. Replace for loop with while loop
Because if you delete an element of the list the for l in no_v: will go to the next position. But because of the deletion you need the same position, to remove all the vowels in no_v and put them in v.
while a < len(no_v):
l = no_v[a]
3. Return the values
Cause it's a function don't print the values just return them. In this case replace the print no_v print v and just return and print them.
return (no_v,v) # returning both lists as tuple
4. Not a mistake but be prepared for python 3.x
Just try to use always print("Have a nice day") instead of print "Have a nice day"
Your Algorithm without the mistakes
Your algorithm now looks like this
import string
def disemvowel(words):
no_v = list(words.lower().translate( None, string.whitespace))
v = []
a = 0
while a < len(no_v):
l = no_v[a]
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
v.append(l)
del no_v[a]
else:
a += 1
return ("".join(no_v),"".join(v))
print(disemvowel("Stackoverflow is cool !"))
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')
How I would do this in python
Not asked but I give you a solution I would probably use. Cause it has something to do with string replacement, or matching I would just use regex.
def myDisemvowel(words):
words = words.lower().translate( None, string.whitespace)
nv = re.sub("[aeiou]*","", words)
v = re.sub("[^a^e^i^o^u]*","", words)
return (nv, v)
print(myDisemvowel("Stackoverflow is cool !\n"))
I use just a regular expression and for the nv string I just replace all voewls with and empty string. For the vowel string I just replace the group of all non vowels with an empty string. If you write this compact, you could solve this with 2 lines of code (Just returning the replacement)
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')
You are modifying no_v while iterating through it. It'd be a lot simpler just to make two new lists, one with vowels and one without.
Another option is to convert it to a while loop:
while a < len(no_v):
l = no_v[a]
This way you have just a single variable tracking your place in no_v instead of the two you currently have.
For educational purposes, this all can be made significantly less cumbersome.
def devowel(input_str, vowels="aeiou"):
filtered_chars = [char for char in input_str
if char.lower() not in vowels and not char.isspace()]
return ''.join(filtered_chars)
assert devowel('big BOOM') == 'bgBM'
To help you learn, do the following:
Define a function that returns True if a particular character has to be removed.
Using that function, loop through the characters of the input string and only leave eligible characters.
In the above, avoid using indexes and len(), instead iterate over characters, as in for char in input_str:.
Learn about list comprehensions.
(Bonus points:) Read about the filter function.
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
I'm using Grok Learning and the task it give you is 'to select every third letter out of a sentence (starting from the first letter), and print out those letters with spaces in between them.'
This is my code:
text = input("Message? ")
length = len(text)
for i in range (0, length, 3):
decoded = text[i]
print(decoded, end=" ")
Although I it says it isn't correct, it say this is the desired out-put:
Message? cxohawalkldflghemwnsegfaeap
c h a l l e n g e
And my output is the same expect, in my output, I have a space after the last 'e' in challenge. Can anyone think of a way to fix this?
To have spaces only between the characters, you could use a slice to create the string "challenge" then use str.join to add the spaces:
" ".join(text[::3])
Here's Grok's explanation to your question: "So, this question is asking you to loop over a string, and print out every third letter. The easiest way to do this is to use for and range, letting range do all the heavy lifting and hard work! We know that range creates a list of numbers, - we can use these numbers as indexes for the message!"
So if you are going to include functions like print, len, end, range, input, for and in functions, your code should look somewhat similar to this:
line = input('Message? ')
result = line[0]
for i in range(3, len(line), 3):
result += ' ' + line[i]
print(result)
Or this:
line = input('Message? ')
print(line[0], end='')
for i in range(3, len(line), 3):
print(' ' + line[i], end='')
print()
Or maybe this:
code = input ('Message? ') [0::3]
msg = ""
for i in code: msg += " " + i
print (msg [1:])
All of these should work, and I hope this answers your question.
I think Grok is just really picky about the details. (It's also case sensitive)
Maybe try this for an alternative because this one worked for me:
message = input('Message? ')
last_index = len(message) -1
decoded = ''
for i in range(0, last_index, 3):
decoded += message[i] + ' '
print(decoded.rstrip())
You should take another look at the notes on this page about building up a string, and then printing it out all at once, in this case perhaps using rstrip() or output[:-1] to leave off the space on the far right.
Here's an example printing out the numbers 0 to 9 in the same fashion, using both rstrip and slicing.
output = ""
for i in range(10):
output = output + str(i) + ' '
print(output[:-1])
print(output.rstrip())
If you look through the Grok course, there is one page called ‘Step by step, side by side’ (link here at https://groklearning.com/learn/intro-python-1/repeating-things/8/) where it introduces the rstrip function. If you write print(output.rstrip()) it will get rid of whitespace to the right of the string.