It seems like there should be a simpler way than:
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)
Is there?
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
Regular expressions are simple enough, if you know them.
import re
s = "string. With. Punctuation?"
s = re.sub(r'[^\w\s]','',s)
For the convenience of usage, I sum up the note of striping punctuation from a string in both Python 2 and Python 3. Please refer to other answers for the detailed description.
Python 2
import string
s = "string. With. Punctuation?"
table = string.maketrans("","")
new_s = s.translate(table, string.punctuation) # Output: string without punctuation
Python 3
import string
s = "string. With. Punctuation?"
table = str.maketrans(dict.fromkeys(string.punctuation)) # OR {key: None for key in string.punctuation}
new_s = s.translate(table) # Output: string without punctuation
myString.translate(None, string.punctuation)
Not necessarily simpler, but a different way, if you are more familiar with the re family.
import re, string
s = "string. With. Punctuation?" # Sample string
out = re.sub('[%s]' % re.escape(string.punctuation), '', s)
string.punctuation is ASCII only! A more correct (but also much slower) way is to use the unicodedata module:
# -*- coding: utf-8 -*-
from unicodedata import category
s = u'String — with - «punctation »...'
s = ''.join(ch for ch in s if category(ch)[0] != 'P')
print 'stripped', s
You can generalize and strip other types of characters as well:
''.join(ch for ch in s if category(ch)[0] not in 'SP')
It will also strip characters like ~*+§$ which may or may not be "punctuation" depending on one's point of view.
I usually use something like this:
>>> s = "string. With. Punctuation?" # Sample string
>>> import string
>>> for c in string.punctuation:
... s= s.replace(c,"")
...
>>> s
'string With Punctuation'
For Python 3 str or Python 2 unicode values, str.translate() only takes a dictionary; codepoints (integers) are looked up in that mapping and anything mapped to None is removed.
To remove (some?) punctuation then, use:
import string
remove_punct_map = dict.fromkeys(map(ord, string.punctuation))
s.translate(remove_punct_map)
The dict.fromkeys() class method makes it trivial to create the mapping, setting all values to None based on the sequence of keys.
To remove all punctuation, not just ASCII punctuation, your table needs to be a little bigger; see J.F. Sebastian's answer (Python 3 version):
import unicodedata
import sys
remove_punct_map = dict.fromkeys(i for i in range(sys.maxunicode)
if unicodedata.category(chr(i)).startswith('P'))
string.punctuation misses loads of punctuation marks that are commonly used in the real world. How about a solution that works for non-ASCII punctuation?
import regex
s = u"string. With. Some・Really Weird、Non？ASCII。 「（Punctuation）」?"
remove = regex.compile(ur'[\p{C}|\p{M}|\p{P}|\p{S}|\p{Z}]+', regex.UNICODE)
remove.sub(u" ", s).strip()
Personally, I believe this is the best way to remove punctuation from a string in Python because:
It removes all Unicode punctuation
It's easily modifiable, e.g. you can remove the \{S} if you want to remove punctuation, but keep symbols like $.
You can get really specific about what you want to keep and what you want to remove, for example \{Pd} will only remove dashes.
This regex also normalizes whitespace. It maps tabs, carriage returns, and other oddities to nice, single spaces.
This uses Unicode character properties, which you can read more about on Wikipedia.
I haven't seen this answer yet. Just use a regex; it removes all characters besides word characters (\w) and number characters (\d), followed by a whitespace character (\s):
import re
s = "string. With. Punctuation?" # Sample string
out = re.sub(ur'[^\w\d\s]+', '', s)
Here's a one-liner for Python 3.5:
import string
"l*ots! o(f. p#u)n[c}t]u[a'ti\"on#$^?/".translate(str.maketrans({a:None for a in string.punctuation}))
This might not be the best solution however this is how I did it.
import string
f = lambda x: ''.join([i for i in x if i not in string.punctuation])
import re
s = "string. With. Punctuation?" # Sample string
out = re.sub(r'[^a-zA-Z0-9\s]', '', s)
Here is a function I wrote. It's not very efficient, but it is simple and you can add or remove any punctuation that you desire:
def stripPunc(wordList):
"""Strips punctuation from list of words"""
puncList = [".",";",":","!","?","/","\\",",","#","#","$","&",")","(","\""]
for punc in puncList:
for word in wordList:
wordList=[word.replace(punc,'') for word in wordList]
return wordList
>>> s = "string. With. Punctuation?"
>>> s = re.sub(r'[^\w\s]','',s)
>>> re.split(r'\s*', s)
['string', 'With', 'Punctuation']
Just as an update, I rewrote the #Brian example in Python 3 and made changes to it to move regex compile step inside of the function. My thought here was to time every single step needed to make the function work. Perhaps you are using distributed computing and can't have regex object shared between your workers and need to have re.compile step at each worker. Also, I was curious to time two different implementations of maketrans for Python 3
table = str.maketrans({key: None for key in string.punctuation})
vs
table = str.maketrans('', '', string.punctuation)
Plus I added another method to use set, where I take advantage of intersection function to reduce number of iterations.
This is the complete code:
import re, string, timeit
s = "string. With. Punctuation"
def test_set(s):
exclude = set(string.punctuation)
return ''.join(ch for ch in s if ch not in exclude)
def test_set2(s):
_punctuation = set(string.punctuation)
for punct in set(s).intersection(_punctuation):
s = s.replace(punct, ' ')
return ' '.join(s.split())
def test_re(s): # From Vinko's solution, with fix.
regex = re.compile('[%s]' % re.escape(string.punctuation))
return regex.sub('', s)
def test_trans(s):
table = str.maketrans({key: None for key in string.punctuation})
return s.translate(table)
def test_trans2(s):
table = str.maketrans('', '', string.punctuation)
return(s.translate(table))
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("sets2 :",timeit.Timer('f(s)', 'from __main__ import s,test_set2 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("translate2 :",timeit.Timer('f(s)', 'from __main__ import s,test_trans2 as f').timeit(1000000))
print("replace :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000))
This is my results:
sets : 3.1830138750374317
sets2 : 2.189873124472797
regex : 7.142953420989215
translate : 4.243278483860195
translate2 : 2.427158243022859
replace : 4.579746678471565
A one-liner might be helpful in not very strict cases:
''.join([c for c in s if c.isalnum() or c.isspace()])
Here's a solution without regex.
import string
input_text = "!where??and!!or$$then:)"
punctuation_replacer = string.maketrans(string.punctuation, ' '*len(string.punctuation))
print ' '.join(input_text.translate(punctuation_replacer).split()).strip()
Output>> where and or then
Replaces the punctuations with spaces
Replace multiple spaces in between words with a single space
Remove the trailing spaces, if any with
strip()
Why none of you use this?
''.join(filter(str.isalnum, s))
Too slow?
I was looking for a really simple solution. here's what I got:
import re
s = "string. With. Punctuation?"
s = re.sub(r'[\W\s]', ' ', s)
print(s)
'string With Punctuation '
Here's one other easy way to do it using RegEx
import re
punct = re.compile(r'(\w+)')
sentence = 'This ! is : a # sample $ sentence.' # Text with punctuation
tokenized = [m.group() for m in punct.finditer(sentence)]
sentence = ' '.join(tokenized)
print(sentence)
'This is a sample sentence'
# FIRST METHOD
# Storing all punctuations in a variable
punctuation='!?,.:;"\')(_-'
newstring ='' # Creating empty string
word = raw_input("Enter string: ")
for i in word:
if(i not in punctuation):
newstring += i
print ("The string without punctuation is", newstring)
# SECOND METHOD
word = raw_input("Enter string: ")
punctuation = '!?,.:;"\')(_-'
newstring = word.translate(None, punctuation)
print ("The string without punctuation is",newstring)
# Output for both methods
Enter string: hello! welcome -to_python(programming.language)??,
The string without punctuation is: hello welcome topythonprogramminglanguage
with open('one.txt','r')as myFile:
str1=myFile.read()
print(str1)
punctuation = ['(', ')', '?', ':', ';', ',', '.', '!', '/', '"', "'"]
for i in punctuation:
str1 = str1.replace(i," ")
myList=[]
myList.extend(str1.split(" "))
print (str1)
for i in myList:
print(i,end='\n')
print ("____________")
Try that one :)
regex.sub(r'\p{P}','', s)
The question does not have a lot of specifics, so the approach I took is to come up with a solution with the simplest interpretation of the problem: just remove the punctuation.
Note that solutions presented don't account for contracted words (e.g., you're) or hyphenated words (e.g., anal-retentive)...which is debated as to whether they should or shouldn't be treated as punctuations...nor to account for non-English character set or anything like that...because those specifics were not mentioned in the question. Someone argued that space is punctuation, which is technically correct...but to me it makes zero sense in the context of the question at hand.
# using lambda
''.join(filter(lambda c: c not in string.punctuation, s))
# using list comprehension
''.join('' if c in string.punctuation else c for c in s)
Apparently I can't supply edits to the selected answer, so here's an update which works for Python 3. The translate approach is still the most efficient option when doing non-trivial transformations.
Credit for the original heavy lifting to #Brian above. And thanks to #ddejohn for his excellent suggestion for improvement to the original test.
#!/usr/bin/env python3
"""Determination of most efficient way to remove punctuation in Python 3.
Results in Python 3.8.10 on my system using the default arguments:
set : 51.897
regex : 17.901
translate : 2.059
replace : 13.209
"""
import argparse
import re
import string
import timeit
parser = argparse.ArgumentParser()
parser.add_argument("--filename", "-f", default=argparse.__file__)
parser.add_argument("--iterations", "-i", type=int, default=10000)
opts = parser.parse_args()
with open(opts.filename) as fp:
s = fp.read()
exclude = set(string.punctuation)
table = str.maketrans("", "", string.punctuation)
regex = re.compile(f"[{re.escape(string.punctuation)}]")
def test_set(s):
return "".join(ch for ch in s if ch not in exclude)
def test_regex(s): # From Vinko's solution, with fix.
return regex.sub("", s)
def test_translate(s):
return s.translate(table)
def test_replace(s): # From S.Lott's solution
for c in string.punctuation:
s = s.replace(c, "")
return s
opts = dict(globals=globals(), number=opts.iterations)
solutions = "set", "regex", "translate", "replace"
for solution in solutions:
elapsed = timeit.timeit(f"test_{solution}(s)", **opts)
print(f"{solution:<10}: {elapsed:6.3f}")
For serious natural language processing (NLP), you should let a library like SpaCy handle punctuation through tokenization, which you can then manually tweak to your needs.
For example, how do you want to handle hyphens in words? Exceptional cases like abbreviations? Begin and end quotes? URLs? IN NLP it's often useful to separate out a contraction like "let's" into "let" and "'s" for further processing.
Considering unicode. Code checked in python3.
from unicodedata import category
text = 'hi, how are you?'
text_without_punc = ''.join(ch for ch in text if not category(ch).startswith('P'))
You can also do this:
import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())
When you deal with the Unicode strings, I suggest using PyPi regex module because it supports both Unicode property classes (like \p{X} / \P{X}) and POSIX character classes (like [:name:]).
Just install the package by typing pip install regex (or pip3 install regex) in your terminal and hit ENTER.
In case you need to remove punctuation and symbols of any kind (that is, anything other than letters, digits and whitespace) you can use
regex.sub(r'[\p{P}\p{S}]', '', text) # to remove one by one
regex.sub(r'[\p{P}\p{S}]+', '', text) # to remove all consecutive punctuation/symbols with one go
regex.sub(r'[[:punct:]]+', '', text) # Same with a POSIX character class
See a Python demo online:
import regex
text = 'भारत India <><>^$.,,! 002'
new_text = regex.sub(r'[\p{P}\p{S}\s]+', ' ', text).lower().strip()
# OR
# new_text = regex.sub(r'[[:punct:]\s]+', ' ', text).lower().strip()
print(new_text)
# => भारत india 002
Here, I added a whitespace \s pattern to the character class
I need to replace all non-ASCII (\x00-\x7F) characters with a space. I'm surprised that this is not dead-easy in Python, unless I'm missing something. The following function simply removes all non-ASCII characters:
def remove_non_ascii_1(text):
return ''.join(i for i in text if ord(i)<128)
And this one replaces non-ASCII characters with the amount of spaces as per the amount of bytes in the character code point (i.e. the – character is replaced with 3 spaces):
def remove_non_ascii_2(text):
return re.sub(r'[^\x00-\x7F]',' ', text)
How can I replace all non-ASCII characters with a single space?
Of the myriad of similar SO questions, none address character replacement as opposed to stripping, and additionally address all non-ascii characters not a specific character.
Your ''.join() expression is filtering, removing anything non-ASCII; you could use a conditional expression instead:
return ''.join([i if ord(i) < 128 else ' ' for i in text])
This handles characters one by one and would still use one space per character replaced.
Your regular expression should just replace consecutive non-ASCII characters with a space:
re.sub(r'[^\x00-\x7F]+',' ', text)
Note the + there.
For you the get the most alike representation of your original string I recommend the unidecode module:
# python 2.x:
from unidecode import unidecode
def remove_non_ascii(text):
return unidecode(unicode(text, encoding = "utf-8"))
Then you can use it in a string:
remove_non_ascii("Ceñía")
Cenia
For character processing, use Unicode strings:
PythonWin 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:57:17) [MSC v.1600 64 bit (AMD64)] on win32.
>>> s='ABC马克def'
>>> import re
>>> re.sub(r'[^\x00-\x7f]',r' ',s) # Each char is a Unicode codepoint.
'ABC def'
>>> b = s.encode('utf8')
>>> re.sub(rb'[^\x00-\x7f]',rb' ',b) # Each char is a 3-byte UTF-8 sequence.
b'ABC def'
But note you will still have a problem if your string contains decomposed Unicode characters (separate character and combining accent marks, for example):
>>> s = 'mañana'
>>> len(s)
6
>>> import unicodedata as ud
>>> n=ud.normalize('NFD',s)
>>> n
'mañana'
>>> len(n)
7
>>> re.sub(r'[^\x00-\x7f]',r' ',s) # single codepoint
'ma ana'
>>> re.sub(r'[^\x00-\x7f]',r' ',n) # only combining mark replaced
'man ana'
If the replacement character can be '?' instead of a space, then I'd suggest result = text.encode('ascii', 'replace').decode():
"""Test the performance of different non-ASCII replacement methods."""
import re
from timeit import timeit
# 10_000 is typical in the project that I'm working on and most of the text
# is going to be non-ASCII.
text = 'Æ' * 10_000
print(timeit(
"""
result = ''.join([c if ord(c) < 128 else '?' for c in text])
""",
number=1000,
globals=globals(),
))
print(timeit(
"""
result = text.encode('ascii', 'replace').decode()
""",
number=1000,
globals=globals(),
))
Results:
0.7208260721400134
0.009975979187503592
What about this one?
def replace_trash(unicode_string):
for i in range(0, len(unicode_string)):
try:
unicode_string[i].encode("ascii")
except:
#means it's non-ASCII
unicode_string=unicode_string[i].replace(" ") #replacing it with a single space
return unicode_string
As a native and efficient approach, you don't need to use ord or any loop over the characters. Just encode with ascii and ignore the errors.
The following will just remove the non-ascii characters:
new_string = old_string.encode('ascii',errors='ignore')
Now if you want to replace the deleted characters just do the following:
final_string = new_string + b' ' * (len(old_string) - len(new_string))
When we use the ascii() it escapes the non-ascii characters and it doesn't change ascii characters correctly. So my main thought is, it doesn't change the ASCII characters, so I am iterating through the string and checking if the character is changed. If it changed then replacing it with the replacer, what you give.
For example: ' '(a single space) or '?' (with a question mark).
def remove(x, replacer):
for i in x:
if f"'{i}'" == ascii(i):
pass
else:
x=x.replace(i,replacer)
return x
remove('hái',' ')
Result: "h i" (with single space between).
Syntax : remove(str,non_ascii_replacer)
str = Here you will give the string you want to work with.
non_ascii_replacer = Here you will give the replacer which you want to replace all the non ASCII characters with.
Pre-processing using Raku (formerly known as Perl_6)
~$ raku -pe 's:g/ <:!ASCII>+ / /;' file
Sample Input:
Peace be upon you
السلام عليكم
שלום עליכם
Paz sobre vosotros
Sample Output:
Peace be upon you
Paz sobre vosotros
Note, you can get extensive information on the matches using the following code:
~$ raku -ne 'say s:g/ <:!ASCII>+ / /.raku;' file
$( )
$(Match.new(:orig("السلام عليكم"), :from(0), :pos(6)), Match.new(:orig("السلام عليكم"), :from(7), :pos(12)))
$(Match.new(:orig("שלום עליכם"), :from(0), :pos(4)), Match.new(:orig("שלום עליכם"), :from(5), :pos(10)))
$( )
$( )
Or more simply, you can just visualize the replacement blank spaces:
~$ raku -ne 'say S:g/ <:!ASCII>+ / /.raku;' file
"Peace be upon you"
" "
" "
"Paz sobre vosotros"
""
https://docs.raku.org/language/regexes#Unicode_properties
https://www.codesections.com/blog/raku-unicode/
https://raku.org
def filterSpecialChars(strInput):
result = []
for character in strInput:
ordVal = ord(character)
if ordVal < 0 or ordVal > 127:
result.append(' ')
else:
result.append(character)
return ''.join(result)
And call it like this:
result = filterSpecialChars('Ceñía mañana')
print(result)
My problem was that my string contained things like Belgià for België and € for the € sign. And I didn't want to replace them with spaces. But wth the right symbol itself.
my solution was string.encode('Latin1').decode('utf-8')
Potentially for a different question, but I'm providing my version of #Alvero's answer (using unidecode). I want to do a "regular" strip on my strings, i.e. the beginning and end of my string for whitespace characters, and then replace only other whitespace characters with a "regular" space, i.e.
"Ceñíaㅤmañanaㅤㅤㅤㅤ"
to
"Ceñía mañana"
,
def safely_stripped(s: str):
return ' '.join(
stripped for stripped in
(bit.strip() for bit in
''.join((c if unidecode(c) else ' ') for c in s).strip().split())
if stripped)
We first replace all non-unicode spaces with a regular space (and join it back again),
''.join((c if unidecode(c) else ' ') for c in s)
And then we split that again, with python's normal split, and strip each "bit",
(bit.strip() for bit in s.split())
And lastly join those back again, but only if the string passes an if test,
' '.join(stripped for stripped in s if stripped)
And with that, safely_stripped('ㅤㅤㅤㅤCeñíaㅤmañanaㅤㅤㅤㅤ') correctly returns 'Ceñía mañana'.
To replace all non-ASCII (\x00-\x7F) characters with a space:
''.join(map(lambda x: x if ord(x) in range(0, 128) else ' ', text))
To replace all visible characters, try this:
import string
''.join(map(lambda x: x if x in string.printable and x not in string.whitespace else ' ', text))
This will give the same result:
''.join(map(lambda x: x if ord(x) in range(32, 128) else ' ', text))
I would like to replace (and not remove) all punctuation characters by " " in a string in Python.
Is there something efficient of the following flavour?
text = text.translate(string.maketrans("",""), string.punctuation)
This answer is for Python 2 and will only work for ASCII strings:
The string module contains two things that will help you: a list of punctuation characters and the "maketrans" function. Here is how you can use them:
import string
replace_punctuation = string.maketrans(string.punctuation, ' '*len(string.punctuation))
text = text.translate(replace_punctuation)
Modified solution from Best way to strip punctuation from a string in Python
import string
import re
regex = re.compile('[%s]' % re.escape(string.punctuation))
out = regex.sub(' ', "This is, fortunately. A Test! string")
# out = 'This is fortunately A Test string'
This workaround works in python 3:
import string
ex_str = 'SFDF-OIU .df !hello.dfasf sad - - d-f - sd'
#because len(string.punctuation) = 32
table = str.maketrans(string.punctuation,' '*32)
res = ex_str.translate(table)
# res = 'SFDF OIU df hello dfasf sad d f sd'
There is a more robust solution which relies on a regex exclusion rather than inclusion through an extensive list of punctuation characters.
import re
print(re.sub('[^\w\s]', '', 'This is, fortunately. A Test! string'))
#Output - 'This is fortunately A Test string'
The regex catches anything which is not an alpha-numeric or whitespace character
Replace by ''?.
What's the difference between translating all ; into '' and remove all ;?
Here is to remove all ;:
s = 'dsda;;dsd;sad'
table = string.maketrans('','')
string.translate(s, table, ';')
And you can do your replacement with translate.
In my specific way, I removed "+" and "&" from the punctuation list:
all_punctuations = string.punctuation
selected_punctuations = re.sub(r'(\&|\+)', "", all_punctuations)
print selected_punctuations
str = "he+llo* ithis& place% if you * here ##"
punctuation_regex = re.compile('[%s]' % re.escape(selected_punctuations))
punc_free = punctuation_regex.sub("", str)
print punc_free
Result: he+llo ithis& place if you here