在字符串中的每个字符前后放置一个符号

Question

I would like to add brackets to each character in a string. So

"HelloWorld"

should become:

"[H][e][l][l][o][W][o][r][l][d]"

I have used this code:

word = "HelloWorld"
newWord = ""
for letter in word:
    newWord += "[%s]" % letter

which is the most straightforward way to do it but the string concatenations are pretty slow. Any suggestions on speeding up this code.

Answer 1

>>> s = "HelloWorld"
>>> ''.join('[{}]'.format(x) for x in s)
'[H][e][l][l][o][W][o][r][l][d]'

If string is huge then using str.join with a list comprehension will be faster and memory efficient than using a generator expression( https://stackoverflow.com/a/9061024/846892 ):

>>> ''.join(['[{}]'.format(x) for x in s])
'[H][e][l][l][o][W][o][r][l][d]'

---

From Python performance tips :

Avoid this:

s = ""
for substring in list:
    s += substring

Use s = "".join(list) instead. The former is a very common and catastrophic mistake when building large strings.

Answer 2

The most pythonic way would probably be with a generator comprehension:

>>> s = "HelloWorld"
>>> "".join("[%s]" % c for c in s)
'[H][e][l][l][o][W][o][r][l][d]'

Ashwini Chaudhary's answer is very similar, but uses the modern (Python3) string format function. The old string interpolation with % still works fine and is a bit simpler.

A bit more creatively, inserting ][ between each character, and surrounding it all with [] . I guess this might be a bit faster, since it doesn't do as many string interpolations, but speed shouldn't be an issue here.

>>> "[" + "][".join(s) + "]"
'[H][e][l][l][o][W][o][r][l][d]'

Answer 3

If you are concerned about speed and need a fast implementation, try to determine an implementation which offloads the iteration to the underline native module. This is true for at least in CPython.

Suggested Implementation

"[{}]".format(']['.join(s))

Output

'[H][e][l][l][o][W][o][r][l][d]'

Comparing with a competing solution

In [12]: s = "a" * 10000

In [13]: %timeit "[{}]".format(']['.join(s))
1000 loops, best of 3: 215 us per loop

In [14]: %timeit ''.join(['[{}]'.format(x) for x in s])
100 loops, best of 3: 3.06 ms per loop

In [15]: %timeit ''.join('[{}]'.format(x) for x in s)
100 loops, best of 3: 3.26 ms per loop