Split Strings into words with multiple word boundary delimiters

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
    print("This prints once a minute.")
    time.sleep(60) # Delay for 1 minute (60 seconds).

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

• os.remove() removes a file.

• os.rmdir() removes an empty directory.

• shutil.rmtree() deletes a directory and all its contents.

Path objects from the Python 3.4+ pathlib module also expose these instance methods:

• pathlib.Path.unlink() removes a file or symbolic link.

• pathlib.Path.rmdir() removes an empty directory.

From an efficiency perspective, you"re not going to beat

s.translate(None, string.punctuation)

For higher versions of Python use the following code:

s.translate(str.maketrans("", "", string.punctuation))

It"s performing raw string operations in C with a lookup table - there"s not much that will beat that but writing your own C code.

If speed isn"t a worry, another option though is:

exclude = set(string.punctuation)
s = "".join(ch for ch in s if ch not in exclude)

This is faster than s.replace with each char, but won"t perform as well as non-pure python approaches such as regexes or string.translate, as you can see from the below timings. For this type of problem, doing it at as low a level as possible pays off.

Timing code:

import re, string, timeit

s = "string. With. Punctuation"
exclude = set(string.punctuation)
table = string.maketrans("";"")
regex = re.compile("[%s]" % re.escape(string.punctuation))

def test_set(s):
    return "".join(ch for ch in s if ch not in exclude)

def test_re(s):  # From Vinko"s solution, with fix.
    return regex.sub("", s)

def test_trans(s):
    return s.translate(table, string.punctuation)

def test_repl(s):  # From S.Lott"s solution
    for c in string.punctuation:
        s=s.replace(c,"")
    return s

print "sets      :",timeit.Timer("f(s)", "from __main__ import s,test_set as f").timeit(1000000)
print "regex     :",timeit.Timer("f(s)", "from __main__ import s,test_re as f").timeit(1000000)
print "translate :",timeit.Timer("f(s)", "from __main__ import s,test_trans as f").timeit(1000000)
print "replace   :",timeit.Timer("f(s)", "from __main__ import s,test_repl as f").timeit(1000000)

This gives the following results:

sets      : 19.8566138744
regex     : 6.86155414581
translate : 2.12455511093
replace   : 28.4436721802

This can be done without regex:

>>> string = "Special $#! characters   spaces 888323"
>>> "".join(e for e in string if e.isalnum())
"Specialcharactersspaces888323"

You can use str.isalnum:

S.isalnum() -> bool

Return True if all characters in S are alphanumeric
and there is at least one character in S, False otherwise.

If you insist on using regex, other solutions will do fine. However note that if it can be done without using a regular expression, that"s the best way to go about it.

Here is a regex to match a string of characters that are not a letters or numbers:

[^A-Za-z0-9]+

Here is the Python command to do a regex substitution:

re.sub("[^A-Za-z0-9]+", "", mystring)