Solving problem is about exposing yourself to as many situations as possible like Replacing instances of a character in a string and practice these strategies over and over. With time, it becomes second nature and a natural way you approach any problems in general. Big or small, always start with a plan, use other strategies mentioned here till you are confident and ready to code the solution.
In this post, my aim is to share an overview the topic about Replacing instances of a character in a string, which can be followed any time. Take easy to follow this discuss.
This simple code that simply tries to replace semicolons (at i-specified postions) by colons does not work:
for i in range(0,len(line)):
if (line[i]==";" and i in rightindexarray):
line[i]=":"
It gives the error
line[i]=":"
TypeError: 'str' object does not support item assignment
How can I work around this to replace the semicolons by colons? Using replace does not work as that function takes no index- there might be some semicolons I do not want to replace.
Example
In the string I might have any number of semicolons, eg “Hei der! ; Hello there ;!;”
I know which ones I want to replace (I have their index in the string). Using replace does not work as I’m not able to use an index with it.
Answer #1:
Strings in python are immutable, so you cannot treat them as a list and assign to indices.
Use .replace() instead:
line = line.replace(';', ':')
If you need to replace only certain semicolons, you’ll need to be more specific. You could use slicing to isolate the section of the string to replace in:
line = line[:10].replace(';', ':') + line[10:]
That’ll replace all semi-colons in the first 10 characters of the string.
Answer #2:
You can do the below, to replace any char with a respective char at a given index, if you wish not to use .replace()
word = 'python'
index = 4
char = 'i'
word = word[:index] + char + word[index + 1:]
print word
o/p: pythin
Answer #3:
Turn the string into a list; then you can change the characters individually. Then you can put it back together with .join:
s = 'a;b;c;d'
slist = list(s)
for i, c in enumerate(slist):
if slist[i] == ';' and 0 <= i <= 3: # only replaces semicolons in the first part of the text
slist[i] = ':'
s = ''.join(slist)
print s # prints a:b:c;d
Answer #4:
If you want to replace a single semicolon:
for i in range(0,len(line)):
if (line[i]==";"):
line = line[:i] + ":" + line[i+1:]
Havent tested it though.
Answer #5:
This should cover a slightly more general case, but you should be able to customize it for your purpose
def selectiveReplace(myStr):
answer = []
for index,char in enumerate(myStr):
if char == ';':
if index%2 == 1: # replace ';' in even indices with ":"
answer.append(":")
else:
answer.append("!") # replace ';' in odd indices with "!"
else:
answer.append(char)
return ''.join(answer)
Hope this helps
Answer #6:
You cannot simply assign value to a character in the string.
Use this method to replace value of a particular character:
name = "India"
result=name .replace("d",'*')
Output: In*ia
Also, if you want to replace say * for all the occurrences of the first character except the first character,
eg. string = babble output = ba**le
Code:
name = "babble"
front= name [0:1]
fromSecondCharacter = name [1:]
back=fromSecondCharacter.replace(front,'*')
return front+back
Answer #7:
If you are replacing by an index value specified in variable ‘n’, then try the below:
def missing_char(str, n):
str=str.replace(str[n],":")
return str
Answer #8:
How about this:
sentence = 'After 1500 years of that thinking surpressed'
sentence = sentence.lower()
def removeLetter(text,char):
result = ''
for c in text:
if c != char:
result += c
return text.replace(char,'*')
text = removeLetter(sentence,'a')
