Python replace/delete special characters - python

character = (%.,-();'0123456789-—:`’)
character.replace(" ")
character.delete()
I want to delete or replace all the special characters and numbers from my program, I know it can be done in the one string just not sure how to space all the special characters with quotes or anything. Somehow I'm supposed to separate all the special character in the parenthesis just not sure how to break up and keep all the characters stored in the variable.

The translate method is my preferred way of doing this. Create a mapping between the chars you want mapped and then apply that table to your input string.
from string import maketrans
special = r"%.,-();'0123456789-—:`’"
blanks = " " * len(special)
table = maketrans(special, blanks)
input_string.translate(table)

Seems like a good application for filter
>>> s = 'This is a test! It has #1234 and letters?'
>>> filter(lambda i: i.isalpha(), s)
'ThisisatestIthasandletters'

You can have a function with an optional fill value, if not set it will just delete/remove the non alpha characters or you can specify a default replace value:
def delete_replace(s,fill_char = ""):
return "".join([x if x.isalpha() else fill_char for x in s])

Related

How can I implement isalnum() into this Python web scraper to remove special characters? [duplicate]

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.

Print a string without any other characters except letters, and replace the space with an underscore

I need to print a string, using this rules:
The first letter should be capital and make all other letters are lowercase. Only the characters a-z A-Z are allowed in the name, any other letters have to be deleted(spaces and tabs are not allowed and use underscores are used instead) and string could not be longer then 80 characters.
It seems to me that it is possible to do it somehow like this:
name = "hello2 sjsjs- skskskSkD"
string = name[0].upper() + name[1:].lower()
lenght = len(string) - 1
answer = ""
for letter in string:
x = letter.isalpha()
if x == False:
answer = string.replace(letter,"")
........
return answer
I think it's better to use a for loop or isalpha () here, but I can't think of a better way to do it. Can someone tell me how to do this?
For one-to-one and one-to-None mappings of characters, you can use the .translate() method of strings. The string module provides lists (strings) of the various types of characters including one for all letters in upper and lowercase (string.ascii_letters) but you could also use your own constant string such as 'abcdef....xyzABC...XYZ'.
import string
def cleanLetters(S):
nonLetters = S.translate(str.maketrans('','',' '+string.ascii_letters))
return S.translate(str.maketrans(' ','_',nonLetters))
Output:
cleanLetters("hello2 sjsjs- skskskSkD")
'hello_sjsjs_skskskSkD'
One method to accomplish this is to use regular expressions (regex) via the built-in re library. This enables the capturing of only the valid characters, and ignoring the rest.
Then, using basic string tools for the replacement and capitalisation, then a slice at the end.
For example:
import re
name = 'hello2 sjsjs- skskskSkD'
trans = str.maketrans({' ': '_', '\t': '_'})
''.join(re.findall('[a-zA-Z\s\t]', name)).translate(trans).capitalize()[:80]
>>> 'Hello_sjsjs_skskskskd'
Strings are immutable, so every time you do string.replace() it needs to iterate over the entire string to find characters to replace, and a new string is created. Instead of doing this, you could simply iterate over the current string and create a new list of characters that are valid. When you're done iterating over the string, use str.join() to join them all.
answer_l = []
for letter in string:
if letter == " " or letter == "\t":
answer_l.append("_") # Replace spaces or tabs with _
elif letter.isalpha():
answer_l.append(letter) # Use alphabet characters as-is
# else do nothing
answer = "".join(answer_l)
With string = 'hello2 sjsjs- skskskSkD', we have answer = 'hello_sjsjs_skskskSkD';
Now you could also write this using a generator expression instead of creating the entire list and then joining it. First, we define a function that returns the letter or "_" for our first two conditions, and an empty string for the else condition
def translate(letter):
if letter == " " or letter == "\t":
return "_"
elif letter.isalpha():
return letter
else:
return ""
Then,
answer = "".join(
translate(letter) for letter in string
)
To enforce the 80-character limit, just take answer[:80]. Because of the way slices work in python, this won't throw an error even when the length of answer is less than 80.

I want to replace a special character with a space

Here is the code i have until now :
dex = tree.xpath('//div[#class="cd-timeline-topic"]/text()')
names = filter(lambda n: n.strip(), dex)
table = str.maketrans(dict.fromkeys('?:,'))
for index, name in enumerate(dex, start = 0):
print('{}.{}'.format(index, name.strip().translate(table)))
The problem is that the output will print also strings with one special character "My name is/Richard". So what i need it's to replace that special character with a space and in the end the printing output will be "My name is Richard". Can anyone help me ?
Thanks!
Your call to dict.fromkeys() does not include the character / in its argument.
If you want to map all the special characters to None, just passing your list of special chars to dict.fromkeys() should be enough. If you want to replace them with a space, you could then iterate over the dict and set the value to for each key.
For example:
special_chars = "?:/"
special_char_dict = dict.fromkeys(special_chars)
for k in special_char_dict:
special_char_dict[k] = " "
You can do this by extending your translation table:
dex = ["My Name is/Richard????::,"]
table = str.maketrans({'?':None,':':None,',':None,'/':' '})
for index, name in enumerate(dex, start = 0):
print('{}.{}'.format(index, name.strip().translate(table)))
OUTPUT
0.My Name is Richard
You want to replace most special characters with None BUT forward slash with a space. You could use a different method to replace forward slashes as the other answers here do, or you could extend your translation table as above, mapping all the other special characters to None and forward slash to space. With this you could have a whole bunch of different replacements happen for different characters.
Alternatively you could use re.sub function following way:
import re
s = 'Te/st st?ri:ng,'
out = re.sub(r'\?|:|,|/',lambda x:' ' if x.group(0)=='/' else '',s)
print(out) #Te st string
Arguments meaning of re.sub is as follows: first one is pattern - it informs re.sub which substring to replace, ? needs to be escaped as otherwise it has special meaning there, | means: or, so re.sub will look for ? or : or , or /. Second argument is function which return character to be used in place of original substring: space for / and empty str for anything else. Third argument is string to be changed.
>>> a = "My name is/Richard"
>>> a.replace('/', ' ')
'My name is Richard'
To replace any character or sequence of characters from the string, you need to use `.replace()' method. So the solution to your answer is:
name.replace("/", " ")
here you can find details

Remove the first character of a string

I would like to remove the first character of a string.
For example, my string starts with a : and I want to remove that only. There are several occurrences of : in the string that shouldn't be removed.
I am writing my code in Python.
python 2.x
s = ":dfa:sif:e"
print s[1:]
python 3.x
s = ":dfa:sif:e"
print(s[1:])
both prints
dfa:sif:e
Your problem seems unclear. You say you want to remove "a character from a certain position" then go on to say you want to remove a particular character.
If you only need to remove the first character you would do:
s = ":dfa:sif:e"
fixed = s[1:]
If you want to remove a character at a particular position, you would do:
s = ":dfa:sif:e"
fixed = s[0:pos]+s[pos+1:]
If you need to remove a particular character, say ':', the first time it is encountered in a string then you would do:
s = ":dfa:sif:e"
fixed = ''.join(s.split(':', 1))
Depending on the structure of the string, you can use lstrip:
str = str.lstrip(':')
But this would remove all colons at the beginning, i.e. if you have ::foo, the result would be foo. But this function is helpful if you also have strings that do not start with a colon and you don't want to remove the first character then.
Just do this:
r = "hello"
r = r[1:]
print(r) # ello
deleting a char:
def del_char(string, indexes):
'deletes all the indexes from the string and returns the new one'
return ''.join((char for idx, char in enumerate(string) if idx not in indexes))
it deletes all the chars that are in indexes; you can use it in your case with del_char(your_string, [0])

Python Chosing To Ignore String Value

Using Python v2, is there a way to ignore a value in a string if it is there?
For instance: I want someone to enter a value of $100.00, or they could enter a value of 100.00 without the leading $ symbol. What I want to do is ignore the '$' value if it is typed in.
Any push in the right direction would be appreciated.
Maybe
s = " $100.00 "
f = float(s.strip().lstrip("$"))
The .strip() strips whitespace from the beginning and the end of the string, and the .lstrip("$") strips a dollar sign from the beginning, if present.
If you only want to remove a '$' then s.replace('$', '') will do want you want.
If you want to replace more than one character then you need to chain replace calls together, which gets very ugly very quickly and in that case one of the other solutions is probably better.
Just filter out unwanted characters from the string. There are multiple ways of doing this, for clarity you could use:
def clean(s, wanted = "0123456789."):
"""Returns version of s without undesired characters in it."""
out = ""
for c in s:
if c in wanted:
out += c
return out
To avoid the dynamic string-building, which is costly, you can build a list and then turn the list into a string:
def clean2(s, wanted = "0123456789."):
outlist = [c for c in s if c in wanted]
return "".join(outlist)
You could simply use a regular expression to extract the number from the string.
Or your could be lazy if you just want to remove a leading $:
if s.startswith('$'):
s = s[1:]
If you want to remove multiple $ signs, replace if with while or use s = s.lstrip('$')
PS: You might want to remove trailing $ signs, too. rstrip() or endswith() and s[:-1] are your friends in this case.
Just lstrip $ from the string before you process it.
value = ...
value = value.lstrip( ' $' ) # Strip blank and $
a = "$100.00"
b = ''.join((c for c in a if c != "$"))
of course this is reasonable if you don't know the position of the character you want to remove

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