或者用相应的索引替换字符串中的char

Question

How to replace the alternative characters in the string with the corresponding index without iterating? For example:

'abcdefghijklmnopqrstuvwxyz'

should be returned as:

'a1c3e5g7i9k11m13o15q17s19u21w23y25'

I have the below code to achieve this. But is there a way to skip the loop or, more pythonic way to achieve this :

string = 'abcdefghijklmnopqrstuvwxyz'
new_string = ''
for i, c in enumerate(string):
    if i % 2:
        new_string += str(i)
    else:
        new_string += c

where new_string hold my required value

Answer 1

You could use a list comprehension , and re-join the characters with str.join() ; the latter avoids repeated (slow) string concatenation):

newstring = ''.join([str(i) if i % 2 else l for i, l in enumerate(string)])

Note that you can't evade iteration here. Even if you defined a large list of pre-stringified odd numbers here ( odds = [str(i) for i in range(1, 1000, 2)] ) then re-use that to use slice assignment on a list string_list[1::2] = odds[:len(string) // 2] Python has to iterate under the hood to re-assign the indices. That's the nature of working with an arbitrary-length sequence.