Example- For Given string ‘Hello World’ returned string is ‘H#l#o W#r#d’.
i tried this code but spaces are also included in this . i want spaces to be maintain in between words
def changer():
ch=[]
for i in 'Hello World':
ch.append(i)
for j in range(1,len(ch),2):
ch[j]= '#'
s=''
for k in ch:
s=s+k
print(s)
changer()
Output - H#l#o#W#r#d
Output i want = H#l#o W#r#d
You can str.split on whitespace to get substrings, then for each substring replace all the odd characters with '#' while preserving the even characters. Then str.join the replaced substrings back together.
>>> ' '.join(''.join('#' if v%2 else j for v,j in enumerate(i)) for i in s.split())
'H#l#o W#r#d'
you can control the increment, by default 2 but, in case of spaces 1 to jump it and continue evaluating the next word
def changer():
ch=[]
increment = 2
for i in 'Hello World':
ch.append(i)
for j in range(1,len(ch),increment):
if not ch[j].isspace():
ch[j]= '#'
increment = 2
else:
increment = 1
s=''
for k in ch:
s=s+k
print(s)
changer()
isspace() +1
increment doesn't do anything.Since you said you don't want spaces to be included in the output, don't include them:
ch=[]
for i in 'Hello World':
ch.append(i)
for j in range(1,len(ch),2):
if ch[j] != " ": # don't 'include' spaces
ch[j]= '#'
s=''
for k in ch:
s=s+k
print(s)
There are a lot of very inconsistent answers here. I think we need a little more info to get you the solution you are expecting. Can you give a string with more words in it to confirm your desired output. You said you want every successive character to be a #, and gave an example of H#l#o W#r#d. Do you want the space to be included in determining what the next character should be? Or should the space be written, but skipped over as a determining factor for the next character? The other option would be 'H#l#o #o#l#' where the space is included in the new text, but is ignored when determining the next character.
Some of the answers give something like this:
string = "Hello World This Is A Test"
'H#l#o W#r#d T#i# I# A T#s#'
'H#l#o W#r#d T#i# #s A T#s#'
'H#l#o W#r#d T#i# I# A T#s# '
This code gives the output: 'H#l#o W#r#d T#i# #s A T#s#'
string = 'Hello World This Is A Test'
solution = ''
c = 0
for letter in string:
if letter == ' ':
solution += ' '
c += 1
elif c % 2:
solution += "#"
c += 1
else:
solution += letter
c += 1
If you actually want the desired outcome if including the whitespace, but not having them be a factor in determing the next character, alls you need to do is remove the counter first check so the spaces do not affect the succession. The solution would be: 'H#l#o #o#l# T#i# I# A #e#t'
You could use accumulate from itertools to build the resulting string progressively
from itertools import accumulate
s = "Hello World"
p = "".join(accumulate(s,lambda r,c:c if {r[-1],c}&set(" #") else "#"))
print(p)
Using your algorithm, you can process each word individually, this way you don't run into issues with spaces. Here's an adaptation of your algorithm where each word is concatenated to a string after being processed:
my_string = 'Hello World'
my_processed_string = ''
for word in my_string.split(' '):
ch = []
for i in word:
ch.append(i)
for j in range(1, len(ch), 2):
ch[j] = '#'
for k in ch:
my_processed_string += k
my_processed_string += ' '
You can maintain a count separate of whitespace and check its lowest bit, replacing the character with hash depending on even or odd.
def changer():
ch=[]
count = 0 # hash even vals (skips 0 because count advanced before compare)
for c in 'Hello World':
if c.isspace():
count = 0 # restart count
else:
count += 1
if not count & 1:
c = '#'
ch.append(c)
s = ''.join(ch)
print(s)
changer()
Result
H#l#o W#r#d
I have not made much changes to your code. so i think this maybe easy for you to understand.
enter code here
def changer():
ch=[]
h='Hello World' #I have put your string in a variable
for i in h:
ch.append(i)
for j in range(1,len(ch),2):
if h[j]!=' ':
ch[j]= '#'
s=''
for k in ch:
s=s+k
print(s)
changer()
list('Hello World')instead of the first loop?