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.
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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')