I have a string for items separated by comma. Each item is surrounded by quotes ("), but the items can also contain commas (,). So using split(',') creates problems.
How can I split this text properly in Python?
An example of such string
"coffee", "water, hot"
What I want to achieve
["coffee", "water, hot"]
You can split on separators that contain more than one character. '"coffee", "water, hot"'.split('", "') gives ['"coffee','water, hot"']. From there you can remove the initial and terminal quote mark.
import ast
s = '"coffee", "water, hot"'
result = ast.literal_eval(f'[{s}]')
print(result)
You can use re.findall
import re
s = '"coffee", "water, hot"'
re.findall('"(.*?)"', s) # ['coffee', 'water, hot']
Firstly, I defined a function 'del_quote' to remove unnecessary quotes and spaces.
Then I split it taking '",' as a separator. then the result was mapped to remove quotes then converted into list.
def del_quote(s):
return s.replace('"','').strip()
x='"coffee", "water, hot"'
result=list(map(del_quote,x.split('",')))
print(result)
You can almost use csv for this:
import csv
from io import StringIO
sio = StringIO()
sio.write('"coffee", "water, hot"')
sio.seek(0)
reader = csv.reader(sio)
print(next(reader))
# Prints ['coffee', ' "water', ' hot"']
The problem is that there is a space before the opening quote of "water, hot". If you replace '", "' with ",", then csv will work, and you will get ['coffee', 'water, hot'].
How about:
string = '"coffee", "water, hot"'
stringList = string.split("'")
string = stringList[0]
print(string)
Unclear on how to frame the following function correctly:
Creating a function that will take in a string and return the string in camel case without spaces (or pascal case if the first letter was already capital), removing special characters
text = "This-is_my_test_string,to-capitalize"
def to_camel_case(text):
# Return 1st letter of text + all letters after
return text[:1] + text.title()[1:].replace(i" ") if not i.isdigit()
# Output should be "ThisIsMyTestStringToCapitalize"
the "if" statement at the end isn't working out, and I wrote this somewhat experimentally, but with a syntax fix, could the logic work?
Providing the input string does not contain any spaces then you could do this:
from re import sub
def to_camel_case(text, pascal=False):
r = sub(r'[^a-zA-Z0-9]', ' ', text).title().replace(' ', '')
return r if pascal else r[0].lower() + r[1:]
ts = 'This-is_my_test_string,to-capitalize'
print(to_camel_case(ts, pascal=True))
print(to_camel_case(ts))
Output:
ThisIsMyTestStringToCapitalize
thisIsMyTestStringToCapitalize
Here is a short solution using regex. First it uses title() as you did, then the regex finds non-alphanumeric-characters and removes them, and finally we take the first character to handle pascal / camel case.
import re
def to_camel_case(s):
s1 = re.sub('[^a-zA-Z0-9]+', '', s.title())
return s[0] + s1[1:]
text = "this-is2_my_test_string,to-capitalize"
print(to_camel_case(text)) # ThisIsMyTestStringToCapitalize
The below should work for your example.
Splitting apart your example by anything that isn's alphanumeric or a space. Then capitalizing each word. Finally, returning the re-joined string.
import re
def to_camel_case(text):
words = re.split(r'[^a-zA-Z0-9\s]', text)
return "".join([word.capitalize() for word in words])
text_to_camelcase = "This-is_my_test_string,to-capitalize"
print(to_camel_case(text_to_camelcase))
use the split function to split between anything that is not a letter or a whitespace and the function .capitalize() to capitalize single words
import re
text_to_camelcase = "This-is_my_test_string,to-capitalize"
def to_camel_case(text):
split_text = re.split(r'[^a-zA-Z0-9\s]', text)
cap_string = ''
for word in split_text:
cap_word = word.capitalize()
cap_string += cap_word
return cap_string
print(to_camel_case(text_to_camelcase))
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())
How can I know if
word = " a "
has a whitespace around it (which it does here, but word is dynamic) and then, iff it does, then strip() it?
There is no point in checking before stripping. Just use str.strip(); it is safe to do so if there is no whitespace around the text:
word.strip()
If you really need to test, you could use str.startswith() and str.endswith() with tuples:
whitespace = tuple(' \n\r\t')
word.startswith(whitespace) or word.endswith(whitespace)
is true if there is whitespace at the start or end.
Can you not just strip() all words, will you not get the same result?
Compare the length before and after?
word = " a "
stripped = word.strip()
if len(word) != len(stripped):
return stripped
else:
return word
That may sound pretty silly, but as you're talking about only one space, why not do like this:
>>> a=" a "
>>> a[0]==' ' and a[-1]==' '
True
And then
if a and a[0]==' ' and a[-1]==' ':
#do the stuff you want to do
I would look into the startswith and endswith methods:
http://www.tutorialspoint.com/python/string_startswith.htm
http://www.tutorialspoint.com/python/string_endswith.htm
if word.startswith(' ') and word.endswith(' ') then
...
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