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I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
How do I do this properly?
Strings in Python are immutable (can't be changed). Because of this, the effect of line.replace(...) is just to create a new string, rather than changing the old one. You need to rebind (assign) it to line in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use str.translate, (see Python 3 answer below):
line = line.translate(None, '!##$')
or regular expression replacement with re.sub
import re
line = re.sub('[!##$]', '', line)
The characters enclosed in brackets constitute a character class. Any characters in line which are in that class are replaced with the second parameter to sub: an empty string.
Python 3 answer
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a comment on one of the answers, and it's noted in the documentation for str.translate.
When calling the translate method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass None as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling ord on them) to the ordinal values of the characters which should replace them, or—usefully to us—None to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
translation_table = dict.fromkeys(map(ord, '!##$'), None)
unicode_line = unicode_line.translate(translation_table)
Here dict.fromkeys and map are used to succinctly generate a dictionary containing
{ord('!'): None, ord('#'): None, ...}
Even simpler, as another answer puts it, create the translation table in place:
unicode_line = unicode_line.translate({ord(c): None for c in '!##$'})
Or, as brought up by Joseph Lee, create the same translation table with str.maketrans:
unicode_line = unicode_line.translate(str.maketrans('', '', '!##$'))
* for compatibility with earlier Pythons, you can create a "null" translation table to pass in place of None:
import string
line = line.translate(string.maketrans('', ''), '!##$')
Here string.maketrans is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.
Am I missing the point here, or is it just the following:
string = "ab1cd1ef"
string = string.replace("1", "")
print(string)
# result: "abcdef"
Put it in a loop:
a = "a!b#c#d$"
b = "!##$"
for char in b:
a = a.replace(char, "")
print(a)
# result: "abcd"
>>> line = "abc##!?efg12;:?"
>>> ''.join( c for c in line if c not in '?:!/;' )
'abc##efg12'
With re.sub regular expression
Since Python 3.5, substitution using regular expressions re.sub became available:
import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
Example
import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
'QDoIwriteNo'
Explanation
In regular expressions (regex), | is a logical OR and \ escapes spaces and special characters that might be actual regex commands. Whereas sub stands for substitution, in this case with the empty string ''.
The asker almost had it. Like most things in Python, the answer is simpler than you think.
>>> line = "H E?.LL!/;O:: "
>>> for char in ' ?.!/;:':
... line = line.replace(char,'')
...
>>> print line
HELLO
You don't have to do the nested if/for loop thing, but you DO need to check each character individually.
For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator [^ABCabc]. For example, to remove everything except ascii letters, digits, and the hyphen:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
From the python regular expression documentation:
Characters that are not within a range can be matched by complementing
the set. If the first character of the set is '^', all the characters
that are not in the set will be matched. For example, [^5] will match
any character except '5', and [^^] will match any character except
'^'. ^ has no special meaning if it’s not the first character in the
set.
line = line.translate(None, " ?.!/;:")
>>> s = 'a1b2c3'
>>> ''.join(c for c in s if c not in '123')
'abc'
Strings are immutable in Python. The replace method returns a new string after the replacement. Try:
for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
This is identical to your original code, with the addition of an assignment to line inside the loop.
Note that the string replace() method replaces all of the occurrences of the character in the string, so you can do better by using replace() for each character you want to remove, instead of looping over each character in your string.
I was surprised that no one had yet recommended using the builtin filter function.
import operator
import string # only for the example you could use a custom string
s = "1212edjaq"
Say we want to filter out everything that isn't a number. Using the filter builtin method "...is equivalent to the generator expression (item for item in iterable if function(item))" [Python 3 Builtins: Filter]
sList = list(s)
intsList = list(string.digits)
obj = filter(lambda x: operator.contains(intsList, x), sList)))
In Python 3 this returns
>> <filter object # hex>
To get a printed string,
nums = "".join(list(obj))
print(nums)
>> "1212"
I am not sure how filter ranks in terms of efficiency but it is a good thing to know how to use when doing list comprehensions and such.
UPDATE
Logically, since filter works you could also use list comprehension and from what I have read it is supposed to be more efficient because lambdas are the wall street hedge fund managers of the programming function world. Another plus is that it is a one-liner that doesnt require any imports. For example, using the same string 's' defined above,
num = "".join([i for i in s if i.isdigit()])
That's it. The return will be a string of all the characters that are digits in the original string.
If you have a specific list of acceptable/unacceptable characters you need only adjust the 'if' part of the list comprehension.
target_chars = "".join([i for i in s if i in some_list])
or alternatively,
target_chars = "".join([i for i in s if i not in some_list])
Using filter, you'd just need one line
line = filter(lambda char: char not in " ?.!/;:", line)
This treats the string as an iterable and checks every character if the lambda returns True:
>>> help(filter)
Help on built-in function filter in module __builtin__:
filter(...)
filter(function or None, sequence) -> list, tuple, or string
Return those items of sequence for which function(item) is true. If
function is None, return the items that are true. If sequence is a tuple
or string, return the same type, else return a list.
Try this one:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
This method works well in Python 3
Here's some possible ways to achieve this task:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)
PS2: If you're interested on performance you could measure these attempts with a simple code like:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
In my box you'd get:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
So it seems attempt4 is the fastest one for this particular input.
Here's my Python 2/3 compatible version. Since the translate api has changed.
def remove(str_, chars):
"""Removes each char in `chars` from `str_`.
Args:
str_: String to remove characters from
chars: String of to-be removed characters
Returns:
A copy of str_ with `chars` removed
Example:
remove("What?!?: darn;", " ?.!:;") => 'Whatdarn'
"""
try:
# Python2.x
return str_.translate(None, chars)
except TypeError:
# Python 3.x
table = {ord(char): None for char in chars}
return str_.translate(table)
#!/usr/bin/python
import re
strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr
You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.
*NB: works with Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!##$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.
The output:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:
words = "things"
removed = "%s%s" % (words[:3], words[-1:])
This will result in 'removed' holding the word 'this'.
Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d.
Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.
So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle.
If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.
Examples:
words = "control"
removed = "%s%s" % (words[:2], words[-2:])
removed equals 'cool'.
words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
removed equals 'macs'.
In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position).
Remember, Python starts counting at 0, so you will need to as well.
In Python 3.5
e.g.,
os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
To remove all the number from the string
How about this:
def text_cleanup(text):
new = ""
for i in text:
if i not in " ?.!/;:":
new += i
return new
Below one.. with out using regular expression concept..
ipstring ="text with symbols!##$^&*( ends here"
opstring=''
for i in ipstring:
if i.isalnum()==1 or i==' ':
opstring+=i
pass
print opstring
Recursive split:
s=string ; chars=chars to remove
def strip(s,chars):
if len(s)==1:
return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
example:
print(strip("Hello!","lo")) #He!
You could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.
import re
text = "This is absurd!"
text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
print(text)
Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.
# for each file on a directory, rename filename
file_list = os.listdir (r"D:\Dev\Python")
for file_name in file_list:
os.rename(file_name, re.sub(r'\d+','',file_name))
Even the below approach works
line = "a,b,c,d,e"
alpha = list(line)
while ',' in alpha:
alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
output: abcde
The string method replace does not modify the original string. It leaves the original alone and returns a modified copy.
What you want is something like: line = line.replace(char,'')
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
for char in s:
if ord(char) < 96 or ord(char) > 123:
s = s.replace(char, "")
It will remove all the characters beyond a....z even upper cases.
I want to remove all the text before and including */ in a string.
For example, consider:
string = ''' something
other things
etc. */ extra text.
'''
Here I want extra text. as the output.
I tried:
string = re.sub("^(.*)(?=*/)", "", string)
I also tried:
string = re.sub(re.compile(r"^.\*/", re.DOTALL), "", string)
But when I print string, it did not perform the operation I wanted and the whole string is printing.
I suppose you're fine without regular expressions:
string[string.index("*/ ")+3:]
And if you want to strip that newline:
string[string.index("*/ ")+3:].rstrip()
The problem with your first regex is that . does not match newlines as you noticed. With your second one, you were closer but forgot the * that time. This would work:
string = re.sub(re.compile(r"^.*\*/", re.DOTALL), "", string)
You can also just get the part of the string that comes after your "*/":
string = re.search(r"(\*/)(.*)", string, re.DOTALL).group(2)
Update: After doing some research, I found that the pattern (\n|.) to match everything including newlines is inefficient. I've updated the answer to use [\s\S] instead as shown on the answer I linked.
The problem is that . in python regex matches everything except newlines. For a regex solution, you can do the following:
import re
strng = ''' something
other things
etc. */ extra text.
'''
print(re.sub("[\s\S]+\*/", "", strng))
# extra text.
Add in a .strip() if you want to remove that remaining leading whitespace.
to keep text until that symbol you can do:
split_str = string.split(' ')
boundary = split_str.index('*/')
new = ' '.join(split_str[0:boundary])
print(new)
which gives you:
something
other things
etc.
string_list = string.split('*/')[1:]
string = '*/'.join(string_list)
print(string)
gives output as
' extra text. \n'
I try to replace commas with semicolons enclosed in curly braces.
Sample string:
text = "a,b,{'c','d','e','f'},g,h"
I am aware that it comes down to lookbehinds and lookaheads, but somehow it won't work like I want it to:
substr = re.sub(r"(?<=\{)(.+?)(,)(?=.+\})",r"\1;", text)
It returns:
a,b,{'c';'d','e','f'},g,h
However, I am aiming for the following:
a,b,{'c';'d';'e';'f'},g,h
Any idea how I can achieve this?
Any help much appreciated :)
You can match the whole block {...} (with {[^{}]+}) and replace commas inside it only with a lambda:
import re
text = "a,b,{'c','d','e','f'},g,h"
print(re.sub(r"{[^{}]+}", lambda x: x.group(0).replace(",", ";"), text))
See IDEONE demo
Output: a,b,{'c';'d';'e';'f'},g,h
By declaring lambda x we can get access to each match object, and get the whole match value using x.group(0). Then, all we need is replace a comma with a semi-colon.
This regex does not support recursive patterns. To use a recursive pattern, you need PyPi regex module. Something like m = regex.sub(r"\{(?:[^{}]|(?R))*}", lambda x: x.group(0).replace(",", ";"), text) should work.
Below I have posted a solution that does not rely on an regular expression. It uses a stack (list) to determine if a character is inside a curly bracket {. Regular expression are more elegant, however, they can be harder to modify when requirements change. Please note that the example below also works for nested brackets.
text = "a,b,{'c','d','e','f'},g,h"
output=''
stack = []
for char in text:
if char == '{':
stack.append(char)
elif char == '}':
stack.pop()
#Check if we are inside a curly bracket
if len(stack)>0 and char==',':
output += ';'
else:
output += char
print output
This gives:
'a,b,{'c';'d';'e';'f'},g,h
You can also rewrite this as a map function if you use a the global variable for stack:
stack = []
def replace_comma_in_curly_brackets(char):
if char == '{':
stack.append(char)
elif char == '}':
stack.pop()
#Check if we are inside a curly bracket
if len(stack)>0 and char==',':
return ';'
return char
text = "a,b,{'c','d','e','f'},g,h"
print ''.join(map(str, map(replace_comma_in_curly_brackets,text)))
Regarding performance, when running the above two methods and the regular expression solution proposed by #stribizhev on the test string at the end of this post, I get the following timings:
Regular expression (#stribizshev): 0.38 seconds
Map function: 26.3 seconds
For loop: 251 seconds
This is the test string that is 55,300,00 characters long:
text = "a,able,about,across,after,all,almost,{also,am,among,an,and,any,are,as,at,be,because},been,but,by,can,cannot,could,dear,did,do,does,either,else,ever,every,for,from,get,got,had,has,have,he,her,hers,him,his,how,however,i,if,in,into,is,it,its,just,least,let,like,likely,may,me,might,most,must,my,neither,no,nor,not,of,off,often,on,only,or,other,our,own,rather,said,say,says,she,should,since,so,some,than,that,the,their,them,then,there,these,they,this,tis,to,too,twas,us,wants,was,we,were,what,when,where,which,while,who,whom,why,will,with,would,yet,you,your" * 100000
If you don't have nested braces, it might be enough to just look ahead at each , if there is a closing } ahead without any opening { in between. Search for
,(?=[^{]*})
and replace with ;
, match a comma literally
(?=...) the lookahead to check
if there's ahead [^{]* any amount of characters, that are not {
followed by a closing curly brace }
See demo at regex101
By definition the JSON string is wrapped with double quote.
In fact:
json.loads('{"v":1}') #works
json.loads("{'v':1}") #doesn't work
But how to deal with the second statements?
I'm looking for a solution different from eval or replace.
Thanks.
If you get a mailformed json why don't you just replace the double quotes with single quotes before
json.load
If you cannot fix the other side you will have to convert invalid JSON into valid JSON. I think the following treats escaped characters properly:
def fixEscapes(value):
# Replace \' by '
value = re.sub(r"[^\\]|\\.", lambda match: "'" if match.group(0) == "\\'" else match.group(0), value)
# Replace " by \"
value = re.sub(r"[^\\]|\\.", lambda match: '\\"' if match.group(0) == '"' else match.group(0), value)
return value
input = "{'vt\"e\\'st':1}"
input = re.sub(r"'(([^\\']|\\.)+)'", lambda match: '"%s"' % fixEscapes(match.group(1)), input)
print json.loads(input)
Not sure if I got your requirements right, but are you looking for something like this?
def fix_json(string_):
if string_[0] == string_[-1] == "'":
return '"' + string_[1:-1] +'"'
return string_
Example usage:
>>> fix_json("'{'key':'val\"'...cd'}'")
"{'key':'val"'...cd'}"
EDIT: it seems that the humour I tried to have in making the example above is not self-explanatory. So, here's another example:
>>> fix_json("'This string has - I'm sure - single quotes delimiters.'")
"This string has - I'm sure - single quotes delimiters."
This examples show how the "replacement" only happens at the extremities of the string, not within it.
you could also achieve the same with a regular expression, of course, but if you are just checking the starting and finishing char of a string, I find using regular string indexes more readable....
unfortunately you have to do this:
f = open('filename.json', 'rb')
json = eval(f.read())
done!
this works, but apparently people don't like the eval function. Let me know if you find a better approach. I used this on some twitter data...
I need a way to remove all whitespace from a string, except when that whitespace is between quotes.
result = re.sub('".*?"', "", content)
This will match anything between quotes, but now it needs to ignore that match and add matches for whitespace..
I don't think you're going to be able to do that with a single regex. One way to do it is to split the string on quotes, apply the whitespace-stripping regex to every other item of the resulting list, and then re-join the list.
import re
def stripwhite(text):
lst = text.split('"')
for i, item in enumerate(lst):
if not i % 2:
lst[i] = re.sub("\s+", "", item)
return '"'.join(lst)
print stripwhite('This is a string with some "text in quotes."')
Here is a one-liner version, based on #kindall's idea - yet it does not use regex at all! First split on ", then split() every other item and re-join them, that takes care of whitespaces:
stripWS = lambda txt:'"'.join( it if i%2 else ''.join(it.split())
for i,it in enumerate(txt.split('"')) )
Usage example:
>>> stripWS('This is a string with some "text in quotes."')
'Thisisastringwithsome"text in quotes."'
You can use shlex.split for a quotation-aware split, and join the result using " ".join. E.g.
print " ".join(shlex.split('Hello "world this is" a test'))
Oli, resurrecting this question because it had a simple regex solution that wasn't mentioned. (Found your question while doing some research for a regex bounty quest.)
Here's the small regex:
"[^"]*"|(\s+)
The left side of the alternation matches complete "quoted strings". We will ignore these matches. The right side matches and captures spaces to Group 1, and we know they are the right spaces because they were not matched by the expression on the left.
Here is working code (and an online demo):
import re
subject = 'Remove Spaces Here "But Not Here" Thank You'
regex = re.compile(r'"[^"]*"|(\s+)')
def myreplacement(m):
if m.group(1):
return ""
else:
return m.group(0)
replaced = regex.sub(myreplacement, subject)
print(replaced)
Reference
How to match pattern except in situations s1, s2, s3
How to match a pattern unless...
Here little longish version with check for quote without pair. Only deals with one style of start and end string (adaptable for example for example start,end='()')
start, end = '"', '"'
for test in ('Hello "world this is" atest',
'This is a string with some " text inside in quotes."',
'This is without quote.',
'This is sentence with bad "quote'):
result = ''
while start in test :
clean, _, test = test.partition(start)
clean = clean.replace(' ','') + start
inside, tag, test = test.partition(end)
if not tag:
raise SyntaxError, 'Missing end quote %s' % end
else:
clean += inside + tag # inside not removing of white space
result += clean
result += test.replace(' ','')
print result