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.
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.
Strings in python are immutable, so you cannot treat them as a list and assign to indices.
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.
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.
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)
Turn the string into a list; then you can change the characters individually. Then you can put it back together with
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
You can do the below, to replace any char with a respective char at a given index, if you wish not to use
word = 'python' index = 4 char = 'i' word = word[:index] + char + word[index + 1:] print word o/p: pythin
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
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",'*')
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
name = "babble" front= name [0:1] fromSecondCharacter = name [1:] back=fromSecondCharacter.replace(front,'*') return front+back
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')
To replace a character at a specific index, the function is as follows:
def replace_char(s , n , c): n-=1 s = s[0:n] + s[n:n+1].replace(s[n] , c) + s[n+1:] return s
where s is a string, n is index and c is a character.
I wrote this method to replace characters or replace strings at a specific instance. instances start at 0 (this can easily be changed to 1 if you change the optional inst argument to 1, and test_instance variable to 1.
def replace_instance(some_word, str_to_replace, new_str='', inst=0): return_word = '' char_index, test_instance = 0, 0 while char_index < len(some_word): test_str = some_word[char_index: char_index + len(str_to_replace)] if test_str == str_to_replace: if test_instance == inst: return_word = some_word[:char_index] + new_str + some_word[char_index + len(str_to_replace):] break else: test_instance += 1 char_index += 1 return return_word
to use the .replace() method effectively on string without creating a separate list
for example take a look at the list username containing string with some white space, we want to replace the white space with an underscore in each of the username string.
names = ["Joey Tribbiani", "Monica Geller", "Chandler Bing", "Phoebe Buffay"] usernames = 
to replace the white spaces in each username consider using the range function in python.
for name in names: usernames.append(name.lower().replace(' ', '_')) print(usernames)
Or if you want to use one list:
for user in range(len(names)): names[user] = names[user].lower().replace(' ', '_') print(names)
You can do this:
string = "this; is a; sample; ; python code;!;" #your desire string result = "" for i in range(len(string)): s = string[i] if (s == ";" and i in [4, 18, 20]): #insert your desire list s = ":" result = result + s print(result)
names = ["Joey Tribbiani", "Monica Geller", "Chandler Bing", "Phoebe Buffay"] usernames =  for i in names: if " " in i: i = i.replace(" ", "_") print(i)
My problem was that I had a list of numbers, and I only want to replace a part of that number, soy I do this:
original_list = ['08113', '09106', '19066', '17056', '17063', '17053'] # With this part I achieve my goal cves_mod =  for i in range(0,len(res_list)): cves_mod.append(res_list[i].replace(res_list[i][2:], '999')) cves_mod # Result cves_mod ['08999', '09999', '19999', '17999', '17999', '17999']
Even more simpler:
input = "a:b:c:d" output ='' for c in input: if c==':': output +='/' else: output+=c print(output)
i tried using this instead as a 2 in 1
usernames = ["Joey Tribbiani", "Monica Geller", "Chandler Bing", "Phoebe Buffay"] # write your for loop here for user in range(0,len(usernames)): usernames[user] = usernames[user].lower().replace(' ', '_') print(usernames)
Cleaner way to replace character at a specific index
def replace_char(str , index , c): return str[:index]+c+str[index+1:]