How does .append work in Python 2? - python

lword = (bword.lower())
word = str(lword)
spaces = []
for spaces in word:
if spaces == ' ':
spaces.append (' ')
else:
spaces.append('_')
print (spaces)
How would this look if it was in Python 2.7?

You made a mistake in your code:
lword = (bword.lower())
word = str(lword)
spaces = []
for character in word: # character is the temporally variable inside the for-loop
if character == ' '
spaces.append(' ')
else:
spaces.append('_')
print (spaces)
This code will give the result what you're expecting.

Related

How to write the code to get the required output

I need to write the code so that I get the required output (see image). When I run the code it doesn't show anything on the screen.
word = 'Python'
print("The original string is = ",end = '')
for ch in word : # ch will take each character from name
print(ch , end = ' ')
print("\n")
length = len(word) # len is function to give no. of characters
print("The reversed string is = ",end = '')
for ch in range(-1, (-length - 1), -1) :
print(word[ch], end = ' ')![enter image description here](https://i.stack.imgur.com/kSYlP.jpg)![enter image description here](https://i.stack.imgur.com/ehkHl.jpg)
I think this is what you expect:
word = 'Python'
print("The original string is = ",end = '')
for ch in word : # ch will take each character from name
print(ch , end = ' ')
print("\n")
length = len(word) # len is function to give no. of characters
print("The reversed string is = ",end = '')
for ch in range(-1, (-length - 1), -1) :
print(word[ch], end = ' ')
Please read this to understand how to fix the indentation.
https://www.python.org/dev/peps/pep-0008/#indentation
word = "Python"
stringlength=len(word)
slicedString=word[stringlength::-1]
print ("The original string is =",word)
print ("The reversed string is =",slicedString)

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)

Python iterations mischaracterizes string value

For this problem, I am given strings ThatAreLikeThis where there are no spaces between words and the 1st letter of each word is capitalized. My task is to lowercase each capital letter and add spaces between words. The following is my code. What I'm doing there is using a while loop nested inside a for-loop. I've turned the string into a list and check if the capital letter is the 1st letter or not. If so, all I do is make the letter lowercase and if it isn't the first letter, I do the same thing but insert a space before it.
def amendTheSentence(s):
s_list = list(s)
for i in range(len(s_list)):
while(s_list[i].isupper()):
if (i == 0):
s_list[i].lower()
else:
s_list.insert(i-1, " ")
s_list[i].lower()
return ''.join(s_list)
However, for the test case, this is the behavior:
Input: s: "CodesignalIsAwesome"
Output: undefined
Expected Output: "codesignal is awesome"
Console Output: Empty
You can use re.sub for this:
re.sub(r'(?<!\b)([A-Z])', ' \\1', s)
Code:
import re
def amendTheSentence(s):
return re.sub(r'(?<!\b)([A-Z])', ' \\1', s).lower()
On run:
>>> amendTheSentence('GoForPhone')
go for phone
Try this:
def amendTheSentence(s):
start = 0
string = ""
for i in range(1, len(s)):
if s[i].isupper():
string += (s[start:i] + " ")
start = i
string += s[start:]
return string.lower()
print(amendTheSentence("CodesignalIsAwesome"))
print(amendTheSentence("ThatAreLikeThis"))
Output:
codesignal is awesome
that are like this
def amendTheSentence(s):
new_sentence=''
for char in s:
if char.isupper():
new_sentence=new_sentence + ' ' + char.lower()
else:
new_sentence=new_sentence + char
return new_sentence
new_sentence=amendTheSentence("CodesignalIsAwesome")
print (new_sentence)
result is codesignal is awesome

Python add a space between special characters and words

I am working on a data science project and I have an issue. I have an array full of string like the following string and I want to add a space between the words and between the special characters
sentence[i] = 'This is a⓵⓶⓷string'
and I expect something like that:
sentence[i] = 'This is a ⓵ ⓶ ⓷ string'
My last try:
l=[]
for i in lines:
for j in i:
if j.isalpha() == False:
l.append(i.split())
else:
l.append(i)
print(l)
for i in l:
s = ' '.join(i)
You could simply scan the complete line and selectively add space for each character that is neither alphabet nor a space.
s = 'This is a⓵⓶⓷string'
t = ''
for x in s :
if not str.isalpha(x) and x != ' ' :
if t[-1] != ' ':
t+= ' '
t += x
t += ' '
else: t += x
this works for example you have given.
Use regular expressions to accomplish that task, specifically re.sub together with a backreference in order to surround the matched characters with spaces.

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

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