index = [0, 2, 5] s = "I am like stackoverflow-python" for i in index: s = s[i].upper() print(s) IndexError: string index out of range I understand that in the first iteration the string, s, become just the first character, an uppercase "I" in this particular case. But, I have tried to do it without the "s = " , using swapchcase() instead, but it's not working.
Basically, I'm trying to print the s string with the index letters as uppercase using Python 3.X
Strings are immutable in Python, so you need to create a new string object. One way to do it:
indices = set([0, 7, 12, 25]) s = "i like stackoverflow and python" print("".join(c.upper() if i in indices else c for i, c in enumerate(s))) printing
I like StackOverflow and Python Here is my solution. It doesn't iterate over every character, but I'm not sure if converting the string to a list and back to a string is any more efficient.
>>> indexes = set((0, 7, 12, 25)) >>> chars = list('i like stackoverflow and python') >>> for i in indexes: ... chars[i] = chars[i].upper() ... >>> string = ''.join(chars) >>> string 'I like StackOverflow and Python' 
set() in this case. string = list(..) is misleading, you could use chars (less misleading).
MutableString docstring: A faster and better solution is to rewrite your program using lists. 2. indexes = [0, 2, 5] ... OP want to add an index ... indexes.append(1).set's docstring: "Build an unordered collection of unique elements." It doesn't need to be ordered and every index should be unique since it wouldn't make sense to have duplicates.