I have a string like this
ABC/AAAA DEF/78kkk OBJ/89KKK KLE/67899
and I pass the substring to find and replace after. so If I pass DEF/00012 and the original string
should be replaced as like this
ABC/AAAA DEF/00012 OBJ/89KKK KLE/67899
I have tried with string.replace('DEF', 'DEF/00012')
I would get the output as
ABC/AAAA DEF/00012/78kkk OBJ/89KKK KLE/67899
any suggestions would be highly appreciated.
Thanks
I would do:
txt = 'ABC/AAAA DEF/78kkk OBJ/89KKK KLE/67899'
change = 'DEF'
changeto = 'DEF/00012'
newtxt = ' '.join(changeto if i.startswith(change) else i for i in txt.split(' '))
print(newtxt)
Output:
ABC/AAAA DEF/00012 OBJ/89KKK KLE/67899
I splitted at spaces and changed part beginning with DEF
string.replace('DEF/78kkk', 'DEF/00012')
If you mean by "substring" is that the succeeding characters after "DEF" is not fixed to a specific value, use regular expressions instead.
result = re.sub("DEF/\w+", "DEF/00012", string)
Assuming there really is a blank space after every "substring" you will have to use re:
import re
your_string = re.sub("DEF/*$", "DEF/00012", your_string)
I am trying to write a function to remove all punctuation characters from a string. I've tried several permutations on translate, replace, strip, etc. My latest attempt uses a brute force approach:
def clean_lower(sample):
punct = list(string.punctuation)
for c in punct:
sample.replace(c, ' ')
return sample.split()
That gets rid of almost all of the punctuation but I'm left with // in front of one of the words. I can't seem to find any way to remove it. I've even tried explicitly replacing it with sample.replace('//', ' ').
What do I need to do?
using translate is the fastest way to remove punctuations, this will remove // too:
import string
s = "This is! a string, with. punctuations? //"
def clean_lower(s):
return s.translate(str.maketrans('', '', string.punctuation))
s = clean_lower(s)
print(s)
Use regular expressions
import re
def clean_lower(s):
return(re.sub(r'\W','',s))
Above function erases any symbols except underscore
Perhaps you should approach it from the perspective of what you want to keep:
For example:
import string
toKeep = set(string.ascii_letters + string.digits + " ")
toRemove = set(string.printable) - toKeep
cleanUp = str.maketrans('', '', "".join(toRemove))
usage:
s = "Hello! world of / and dice".translate(cleanUp)
# s will be 'Hello world of and dice'
as suggested by #jasonharper you need to redefine "sample" and it should work:
import string
sample='// Hello?) // World!'
print(sample)
punct=list(string.punctuation)
for c in punct:
sample=sample.replace(c,'')
print(sample.split())
I am massaging strings so that the 1st letter of the string and the first letter following either a dash or a slash needs to be capitalized.
So the following string:
test/string - this is a test string
Should look look like so:
Test/String - This is a test string
So in trying to solve this problem my 1st idea seems like a bad idea - iterate the string and check every character and using indexing etc. determine if a character follows a dash or slash, if it does set it to upper and write out to my new string.
def correct_sentence_case(test_phrase):
corrected_test_phrase = ''
firstLetter = True
for char in test_phrase:
if firstLetter:
corrected_test_phrase += char.upper()
firstLetter = False
#elif char == '/':
else:
corrected_test_phrase += char
This just seems VERY un-pythonic. What is a pythonic way to handle this?
Something along the lines of the following would be awesome but I can't pass in both a dash and a slash to the split:
corrected_test_phrase = ' - '.join(i.capitalize() for i in test_phrase.split(' - '))
Which I got from this SO:
Convert UPPERCASE string to sentence case in Python
Any help will be appreciated :)
I was able to accomplish the desired transformation with a regular expression:
import re
capitalized = re.sub(
'(^|[-/])\s*([A-Za-z])', lambda match: match[0].upper(), phrase)
The expression says "anywhere you match either the start of the string, ^, or a dash or slash followed by maybe some space and a word character, replace the word character with its uppercase."
demo
If you don't want to go with a messy splitting-joining logic, go with a regex:
import re
string = 'test/string - this is a test string'
print(re.sub(r'(^([a-z])|(?<=[-/])\s?([a-z]))',
lambda match: match.group(1).upper(), string))
# Test/String - This is a test string
Using double split
import re
' - '.join([i.strip().capitalize() for i in re.split(' - ','/'.join([i.capitalize() for i in re.split('/',test_phrase)]))])
I'm using that:
import string
last = 'pierre-GARCIA'
if last not in [None, '']:
last = last.strip()
if '-' in last:
last = string.capwords(last, sep='-')
else:
last = string.capwords(last, sep=None)
I have characters in the middle of a string that I want to get rid of. These characters are =, p,, and H. Since they are not the leftmost and the rightmost characters in the string, I cannot use strip(). Is there a function that gets rid of a certain character in any location in a string?
The usual tool for this job is str.translate
https://docs.python.org/2/library/stdtypes.html#str.translate
>>> 'hello=potato'.translate(None, '=p')
'hellootato'
Check the .replace() function:
> 'aaba'.replace('a','').replace('b','')
< ''
My usual tool for this is the regular expression.
>>> import re
>>> invalidCharacters = r'[=p H]'
>>> mystring = re.sub(invalidCharacters, '', ' poH==hHoPPp p')
'ohoPP'
If you need to constrain the number (i.e., the count) of characters you remove, see the count argument.
I need to remove all special characters, punctuation and spaces from a string so that I only have letters and numbers.
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)
Shorter way :
import re
cleanString = re.sub('\W+','', string )
If you want spaces between words and numbers substitute '' with ' '
TLDR
I timed the provided answers.
import re
re.sub('\W+','', string)
is typically 3x faster than the next fastest provided top answer.
Caution should be taken when using this option. Some special characters (e.g. ø) may not be striped using this method.
After seeing this, I was interested in expanding on the provided answers by finding out which executes in the least amount of time, so I went through and checked some of the proposed answers with timeit against two of the example strings:
string1 = 'Special $#! characters spaces 888323'
string2 = 'how much for the maple syrup? $20.99? That s ridiculous!!!'
Example 1
'.join(e for e in string if e.isalnum())
string1 - Result: 10.7061979771
string2 - Result: 7.78372597694
Example 2
import re
re.sub('[^A-Za-z0-9]+', '', string)
string1 - Result: 7.10785102844
string2 - Result: 4.12814903259
Example 3
import re
re.sub('\W+','', string)
string1 - Result: 3.11899876595
string2 - Result: 2.78014397621
The above results are a product of the lowest returned result from an average of: repeat(3, 2000000)
Example 3 can be 3x faster than Example 1.
Python 2.*
I think just filter(str.isalnum, string) works
In [20]: filter(str.isalnum, 'string with special chars like !,#$% etcs.')
Out[20]: 'stringwithspecialcharslikeetcs'
Python 3.*
In Python3, filter( ) function would return an itertable object (instead of string unlike in above). One has to join back to get a string from itertable:
''.join(filter(str.isalnum, string))
or to pass list in join use (not sure but can be fast a bit)
''.join([*filter(str.isalnum, string)])
note: unpacking in [*args] valid from Python >= 3.5
#!/usr/bin/python
import re
strs = "how much for the maple syrup? $20.99? That's ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!]',r'',strs)
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)
print nestr
you can add more special character and that will be replaced by '' means nothing i.e they will be removed.
Differently than everyone else did using regex, I would try to exclude every character that is not what I want, instead of enumerating explicitly what I don't want.
For example, if I want only characters from 'a to z' (upper and lower case) and numbers, I would exclude everything else:
import re
s = re.sub(r"[^a-zA-Z0-9]","",s)
This means "substitute every character that is not a number, or a character in the range 'a to z' or 'A to Z' with an empty string".
In fact, if you insert the special character ^ at the first place of your regex, you will get the negation.
Extra tip: if you also need to lowercase the result, you can make the regex even faster and easier, as long as you won't find any uppercase now.
import re
s = re.sub(r"[^a-z0-9]","",s.lower())
string.punctuation contains following characters:
'!"#$%&\'()*+,-./:;<=>?#[\]^_`{|}~'
You can use translate and maketrans functions to map punctuations to empty values (replace)
import string
'This, is. A test!'.translate(str.maketrans('', '', string.punctuation))
Output:
'This is A test'
s = re.sub(r"[-()\"#/#;:<>{}`+=~|.!?,]", "", s)
Assuming you want to use a regex and you want/need Unicode-cognisant 2.x code that is 2to3-ready:
>>> import re
>>> rx = re.compile(u'[\W_]+', re.UNICODE)
>>> data = u''.join(unichr(i) for i in range(256))
>>> rx.sub(u'', data)
u'0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz\xaa\xb2 [snip] \xfe\xff'
>>>
The most generic approach is using the 'categories' of the unicodedata table which classifies every single character. E.g. the following code filters only printable characters based on their category:
import unicodedata
# strip of crap characters (based on the Unicode database
# categorization:
# http://www.sql-und-xml.de/unicode-database/#kategorien
PRINTABLE = set(('Lu', 'Ll', 'Nd', 'Zs'))
def filter_non_printable(s):
result = []
ws_last = False
for c in s:
c = unicodedata.category(c) in PRINTABLE and c or u'#'
result.append(c)
return u''.join(result).replace(u'#', u' ')
Look at the given URL above for all related categories. You also can of course filter
by the punctuation categories.
For other languages like German, Spanish, Danish, French etc that contain special characters (like German "Umlaute" as ü, ä, ö) simply add these to the regex search string:
Example for German:
re.sub('[^A-ZÜÖÄa-z0-9]+', '', mystring)
This will remove all special characters, punctuation, and spaces from a string and only have numbers and letters.
import re
sample_str = "Hel&&lo %% Wo$#rl#d"
# using isalnum()
print("".join(k for k in sample_str if k.isalnum()))
# using regex
op2 = re.sub("[^A-Za-z]", "", sample_str)
print(f"op2 = ", op2)
special_char_list = ["$", "#", "#", "&", "%"]
# using list comprehension
op1 = "".join([k for k in sample_str if k not in special_char_list])
print(f"op1 = ", op1)
# using lambda function
op3 = "".join(filter(lambda x: x not in special_char_list, sample_str))
print(f"op3 = ", op3)
Use translate:
import string
def clean(instr):
return instr.translate(None, string.punctuation + ' ')
Caveat: Only works on ascii strings.
This will remove all non-alphanumeric characters except spaces.
string = "Special $#! characters spaces 888323"
''.join(e for e in string if (e.isalnum() or e.isspace()))
Special characters spaces 888323
import re
my_string = """Strings are amongst the most popular data types in Python. We can create the strings by enclosing characters in quotes. Python treats single quotes the
same as double quotes."""
# if we need to count the word python that ends with or without ',' or '.' at end
count = 0
for i in text:
if i.endswith("."):
text[count] = re.sub("^([a-z]+)(.)?$", r"\1", i)
count += 1
print("The count of Python : ", text.count("python"))
After 10 Years, below I wrote there is the best solution.
You can remove/clean all special characters, punctuation, ASCII characters and spaces from the string.
from clean_text import clean
string = 'Special $#! characters spaces 888323'
new = clean(string,lower=False,no_currency_symbols=True, no_punct = True,replace_with_currency_symbol='')
print(new)
Output ==> 'Special characters spaces 888323'
you can replace space if you want.
update = new.replace(' ','')
print(update)
Output ==> 'Specialcharactersspaces888323'
function regexFuntion(st) {
const regx = /[^\w\s]/gi; // allow : [a-zA-Z0-9, space]
st = st.replace(regx, ''); // remove all data without [a-zA-Z0-9, space]
st = st.replace(/\s\s+/g, ' '); // remove multiple space
return st;
}
console.log(regexFuntion('$Hello; # -world--78asdf+-===asdflkj******lkjasdfj67;'));
// Output: Hello world78asdfasdflkjlkjasdfj67
import re
abc = "askhnl#$%askdjalsdk"
ddd = abc.replace("#$%","")
print (ddd)
and you shall see your result as
'askhnlaskdjalsdk