CODEWITHSUNDEEP
pythonstringpython-2.7
user3423572
user3423572

Reputation: 559

Python 2.7 edit a string of characters

I'm trying to figure out how to delete everything from a string except the square brackets

If a user enters "Hello, my name is [lol] bob [not[really]] john[son]" or enters "[[[[]]][[][][]"

If the user enters something like the first statement I want to delete everything except the brackets. I do not want to implement input validation. Cheers

Edit: Thanks for the help guys I love you

Upvotes: 2

Views: 137

Answers (3)

Duncan
Duncan

Reputation: 95622

When in doubt about the answer it is worth trying a few and timing them:

PS C:\python33> .\python.exe lib\timeit.py -s @"
s = 'Hello, my name is [lol] bob [not[really]]  john[son]'
"@ @"
for x in frozenset(s)-frozenset('[]'):
    s = s.replace(x, '')
"@

1000000 loops, best of 3: 0.844 usec per loop

PS C:\python33> .\python.exe lib\timeit.py -s @"
s = 'Hello, my name is [lol] bob [not[really]]  john[son]'
"@ "''.join(c for c in s if c in '[]')"

100000 loops, best of 3: 4.79 usec per loop

PS C:\python33> .\python.exe lib\timeit.py -s @"
s = 'Hello, my name is [lol] bob [not[really]]  john[son]'
"@ @"
s.translate(str.maketrans('','',''.join(set(s)-set('[]'))))
"@

100000 loops, best of 3: 8.5 usec per loop

PS C:\python33> .\python.exe lib\timeit.py -s @"
s = 'Hello, my name is [lol] bob [not[really]]  john[son]'
"@ "''.join(filter(lambda x: x in '[]', s))"


100000 loops, best of 3: 9.67 usec per loop

PS C:\python33> .\python.exe lib\timeit.py -s @"
s = 'Hello, my name is [lol] bob [not[really]]  john[son]'
import re
"@ "re.sub('[^][]', '', s)"

100000 loops, best of 3: 15.9 usec per loop

So @guess's now deleted answer using s.replace() over the set of characters to remove seems to be fastest and regular expression substitution the slowest (but that's Python 3.3 and my machine, others may differ).

Also note that if you replace the setup string in all of the above with:

s = 'Hello, my name is [lol] bob [not[really]]  john[son]'*1000

Then the times I get are:

10000 loops, best of 3: 190 usec per loop
100 loops, best of 3: 3.6 msec per loop
100 loops, best of 3: 4.25 msec per loop
100 loops, best of 3: 8.72 msec per loop
100 loops, best of 3: 7.13 msec per loop

So re.sub scales better than the lambda solution, but otherwise the others stay in the same relative positions.

Upvotes: 1

anon582847382
anon582847382

Reputation: 20351

Using lambda and filter- just another way to do it, but I find it more beautiful.

>>> your_input = "Hello, my name is [lol] bob [not[really]] john[son]"
>>> ''.join(filter(lambda x: x in '[]', your_input))
'[][[]][]'

Upvotes: 2

Christian Tapia
Christian Tapia

Reputation: 34146

You can use join():

s = "Hello, my name is [lol] bob [not[really]] john[son]"
s = ''.join(c for c in s if c in '[]')
print s

Output:

[][[]][]

Upvotes: 6

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