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String substitution performance in python


String substitution performance in python

By : Vincent Lopez
Date : November 21 2020, 07:38 AM
To fix the issue you can do A lot of the performance overhead of regular expressions comes from compiling the regular expressions. You should move the compilation of the regular expression out of the loop.
This should give you a considerable improvement:
code :
pattern1 = re.compile('[^0-9a-zA-Z]+')
pattern2 = re.compile('\s+')
for k in range(len(titles)):
    #remove all non-alphanumeric characters 
    s = re.sub(pattern1,' ',titles[k])

    #remove extra white space
    s = re.sub(pattern2,' ', s).strip()
import re
def noncompiled():
    with open("wordlist.txt",'r') as f:
        titles = f.readlines()
    titles = ["".join([title,nonalpha]) for title in titles for nonalpha in "!@#$%"]
    for k in range(len(titles)):
        #remove all non-alphanumeric characters 
        s = re.sub('[^0-9a-zA-Z]+', ' ',titles[k])

        #remove extra white space
        s = re.sub( '\s+', ' ', s).strip()

def compiled():
    with open("wordlist.txt",'r') as f:
        titles = f.readlines()
    titles = ["".join([title,nonalpha]) for title in titles for nonalpha in "!@#$%"]
    pattern1=re.compile('[^0-9a-zA-Z]+')
    pattern2 = re.compile( '\s+')
    for k in range(len(titles)):
        #remove all non-alphanumeric characters 
        s = pattern1.sub('',titles[k])

        #remove extra white space
        s = pattern2.sub('', s)



In [2]: %timeit noncompiled()
1 loops, best of 3: 292 ms per loop

In [3]: %timeit compiled()
10 loops, best of 3: 176 ms per loop
def with_excludes():
    with open("wordlist.txt",'r') as f:
        titles = f.readlines()
    titles = ["".join([title,nonalpha]) for title in titles for nonalpha in "!@#$%"]
    pattern1=re.compile('[^0-9a-zA-Z]+')
    pattern2 = re.compile( '\s+')
    excludes = ["shit","poo","ass","love","boo","ch"]
    excludes_regex = re.compile('|'.join(excludes))
    for k in range(len(titles)):
        #remove all non-alphanumeric characters 
        s = pattern1.sub('',titles[k])

        #remove extra white space
        s = pattern2.sub('', s)
        #remove bad words
        s = pattern2.sub('', s)
In [2]: %timeit with_excludes()
1 loops, best of 3: 251 ms per loop
def master():
    with open("wordlist.txt",'r') as f:
        titles = f.readlines()
    titles = ["".join([title,nonalpha]) for title in titles for nonalpha in "!@#$%"]
    excludes = ["shit","poo","ass","love","boo","ch"]
    nonalpha='[^0-9a-zA-Z]+'
    whitespace='\s+'
    badwords = '|'.join(excludes)
    master_regex=re.compile('|'.join([nonalpha,whitespace,badwords]))

    for k in range(len(titles)):
        #remove all non-alphanumeric characters 
        s = master_regex.sub('',titles[k])
In [2]: %timeit master()
10 loops, best of 3: 148 ms per loop
    result = [master_regex.sub('',item) for item in titles]


In [4]: %timeit list_comp()
10 loops, best of 3: 139 ms per loop
def baseline():
    with open("wordlist.txt",'r') as f:
        titles = f.readlines()
    titles = ["".join([title,nonalpha]) for title in titles for nonalpha in "!@#$%"]

In [2]: %timeit baseline()
10 loops, best of 3: 24.8 ms per loop


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python string substitution

python string substitution


By : shanwxu
Date : March 29 2020, 07:55 AM
will help you Is there a simple way of passing a list as the parameter to a string substitution in python ? Something like: , Just convert the list to a tuple:
code :
w = ['a', 'b', 'c']
s = '%s\t%s\t%s\n' % tuple(w)
How can Python do string substitution without %s?

How can Python do string substitution without %s?


By : Jaco De Beer
Date : March 29 2020, 07:55 AM
it helps some times I bet you'll find later in the code that there is a DB API execute statement that takes update_dict as a parameter. The DB API then does the substitution instead of Python string formatting and thus properly handles binding.
Have a look at this: http://furius.ca/pubcode/pub/antiorm/lib/python/dbapiext.html#escaping-in-the-dbapi-2-0
Python, SQlite and string substitution

Python, SQlite and string substitution


By : greenfern
Date : March 29 2020, 07:55 AM
should help you out First, you can make a dict that maps replacement values to lists of 'original values':
code :
replacements = {
    'Apple':['Iphone','Ipod'],
    'Samsung':['Galaxy','Nexus']
}
for replacement, replacables in replacements.iteritems():
    query = 'update foo set value=? where value in ({})'.format(",".join("?"*len(replacables)))
    c.execute(query, [replacement]+replacables)
dynamic string substitution using string.Template in Python

dynamic string substitution using string.Template in Python


By : rafa
Date : March 29 2020, 07:55 AM
To fix the issue you can do You need to call f by doing f(). f means the function; f() means its return value.
String substitution in Python 3?

String substitution in Python 3?


By : Nightrord
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You have two three options for formatting:
The old python2 style % operator that works with C-printf-style format strings The new python3 style format() method of the string. Because of backporting you can use this option in both python2 and python3. The even newer python 3.6+ f-strings that aren't backported to python2: these allow you to specify expressions inside your string literals The % formatter that accepts C-printf-style format strings:
code :
"Hello %s" % subs
"Hello %s%s" % (subs, '.')
my_dict = {'placeholder_name': 'world', 'period': '.'}
"Hello %(placeholder_name)s%(period)s" % my_dict
"Hello {}".format(subs)
"Hello {placeholder_name}".format(placeholder_name=subs)
"Hello {subs}".format(subs=subs)
my_tuple = ('world', '.')
"Hello {}{}".format(*my_tuple)
my_dict = {'subs': 'world', 'period': '.'}
"Hello {subs}{period}".format(**my_dict)
>>> x=5
>>> y=6
>>> f'woof {x*y} meow'
'woof 30 meow'
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