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'm very new to Python and am trying to understand how to manipulate strings.
What I want to do is change a string by removing the spaces and alternating the case from upper to lower, IE "This is harder than I thought it would be" to "ThIsIsHaRdErThAnItHoUgHtItWoUlDbE"
I've cobbled together a code to remove the spaces (heavily borrowed from here):
string1 = input("Ask user for something.")
nospace = ""
for a in string1:
if a == " ":
pass
else:
nospace=nospace+a
... but just can't get my head around the caps/lower case part. There are several similar issues on this site and I've tried amending a few of them, with no joy. I realise I need to define a range and iterate through it, but that's where I draw a blank.
for c in nospace[::]:
d = ""
c = nospace[:1].lower()
d = d + c
c = nospace[:1].upper
print d
All I am getting is a column of V's. I'm obviously getting this very wrong. Please can someone advise where? Thanks in advance.
Here is a cutesie way to do this:
>>> s = "This is harder than I thought it would be"
>>> from itertools import cycle
>>> funcs = cycle([str.upper, str.lower])
>>> ''.join(next(funcs)(c) for c in s if c != ' ')
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
>>>
Or, as suggested by Moses in the comments, you can use str.isspace, which will take care of not just a single space ' '
>>> ''.join(next(funcs)(c) for c in s if not c.isspace())
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
This approach only does a single pass on the string. Although, a two-pass method is likely performant enough.
Now, if you were starting with a nospace string, the best way is to convert to some mutable type (e.g. a list) and use slice-assignment notation. It's a little bit inefficient because it builds intermediate data structures, but slicing is fast in Python, so it may be quite performant. You have to ''.join at the end, to bring it back to a string:
>>> nospace
'ThisisharderthanIthoughtitwouldbe'
>>> nospace = list(nospace)
>>> nospace[0::2] = map(str.upper, nospace[0::2])
>>> nospace[1::2] = map(str.lower, nospace[1::2])
>>> ''.join(nospace)
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
>>>
You're trying to do everything at once. Don't. Break your program into steps.
Read the string.
Remove the spaces from the string (as #A.Sherif just demonstrated here)
Go over the string character by character. If the character is in an odd position, convert it to uppercase. Otherwise, convert to lowercase.
So your 2nd loop is where you're breaking it, because the original list isn't being shortened, the c=nospace[:1] grabs the first character of the string and that's the only character that's ever printed. So a solution would be as follows.
string1 = str(input("Ask user for something."))
nospace = ''.join(string1.split(' '))
for i in range(0, len(nospace)):
if i % 2 == 0:
print(nospace[i].upper(), end="")
else:
print(nospace[i].lower(), end="")
Could also replace the if/else statement with a ternary opperator.
for i in range(0, len(nospace)):
print(nospace[i].upper() if (i % 2 == 0) else nospace[i].lower(), end='')
Final way using enumerate as commented about
for i, c in enumerate(nospace):
print(c.upper() if (i % 2 == 0) else c.lower(), end='')
This is how the string splitting works for me right now:
output = string.encode('UTF8').split('}/n}')[0]
output += '}\n}'
But I am wondering if there is a more pythonic way to do it.
The goal is to get everything before this '}/n}' including '}/n}'.
This might be a good use of str.partition.
string = '012za}/n}ddfsdfk'
parts = string.partition('}/n}')
# ('012za', '}/n}', 'ddfsdfk')
''.join(parts[:-1])
# 012za}/n}
Or, you can find it explicitly with str.index.
repl = '}/n}'
string[:string.index(repl) + len(repl)]
# 012za}/n}
This is probably better than using str.find since an exception will be raised if the substring isn't found, rather than producing nonsensical results.
It seems like anything "more elegant" would require regular expressions.
import re
re.search('(.*?}/n})', string).group(0)
# 012za}/n}
It can be done with with re.split() -- the key is putting parens around the split pattern to preserve what you split on:
import re
output = "".join(re.split(r'(}/n})', string.encode('UTF8'))[:2])
However, I doubt that this is either the most efficient nor most Pythonic way to achieve what you want. I.e. I don't think this is naturally a split sort of problem. For example:
tag = '}/n}'
encoded = string.encode('UTF8')
output = encoded[:encoded.index(tag)] + tag
or if you insist on a one-liner:
output = (lambda string, tag: string[:string.index(tag)] + tag)(string.encode('UTF8'), '}/n}')
or returning to regex:
output = re.match(r".*}/n}", string.encode('UTF8')).group(0)
>>> string_to_split = 'first item{\n{second item'
>>> sep = '{\n{'
>>> output = [item + sep for item in string_to_split.split(sep)]
NOTE: output = ['first item{\n{', 'second item{\n{']
then you can use the result:
for item_with_delimiter in output:
...
It might be useful to look up os.linesep if you're not sure what the line ending will be. os.linesep is whatever the line ending is under your current OS, so '\r\n' under Windows or '\n' under Linux or Mac. It depends where input data is from, and how flexible your code needs to be across environments.
Adapted from Slice a string after a certain phrase?, you can combine find and slice to get the first part of the string and retain }/n}.
str = "012za}/n}ddfsdfk"
str[:str.find("}/n}")+4]
Will result in 012za}/n}
Basically, I'm asking the user to input a string of text into the console, but the string is very long and includes many line breaks. How would I take the user's string and delete all line breaks to make it a single line of text. My method for acquiring the string is very simple.
string = raw_input("Please enter string: ")
Is there a different way I should be grabbing the string from the user? I'm running Python 2.7.4 on a Mac.
P.S. Clearly I'm a noob, so even if a solution isn't the most efficient, the one that uses the most simple syntax would be appreciated.
How do you enter line breaks with raw_input? But, once you have a string with some characters in it you want to get rid of, just replace them.
>>> mystr = raw_input('please enter string: ')
please enter string: hello world, how do i enter line breaks?
>>> # pressing enter didn't work...
...
>>> mystr
'hello world, how do i enter line breaks?'
>>> mystr.replace(' ', '')
'helloworld,howdoienterlinebreaks?'
>>>
In the example above, I replaced all spaces. The string '\n' represents newlines. And \r represents carriage returns (if you're on windows, you might be getting these and a second replace will handle them for you!).
basically:
# you probably want to use a space ' ' to replace `\n`
mystring = mystring.replace('\n', ' ').replace('\r', '')
Note also, that it is a bad idea to call your variable string, as this shadows the module string. Another name I'd avoid but would love to use sometimes: file. For the same reason.
You can try using string replace:
string = string.replace('\r', '').replace('\n', '')
You can split the string with no separator arg, which will treat consecutive whitespace as a single separator (including newlines and tabs). Then join using a space:
In : " ".join("\n\nsome text \r\n with multiple whitespace".split())
Out: 'some text with multiple whitespace'
https://docs.python.org/2/library/stdtypes.html#str.split
The canonic answer, in Python, would be :
s = ''.join(s.splitlines())
It splits the string into lines (letting Python doing it according to its own best practices). Then you merge it. Two possibilities here:
replace the newline by a whitespace (' '.join())
or without a whitespace (''.join())
updated based on Xbello comment:
string = my_string.rstrip('\r\n')
read more here
Another option is regex:
>>> import re
>>> re.sub("\n|\r", "", "Foo\n\rbar\n\rbaz\n\r")
'Foobarbaz'
If anybody decides to use replace, you should try r'\n' instead '\n'
mystring = mystring.replace(r'\n', ' ').replace(r'\r', '')
A method taking into consideration
additional white characters at the beginning/end of string
additional white characters at the beginning/end of every line
various end-line characters
it takes such a multi-line string which may be messy e.g.
test_str = '\nhej ho \n aaa\r\n a\n '
and produces nice one-line string
>>> ' '.join([line.strip() for line in test_str.strip().splitlines()])
'hej ho aaa a'
UPDATE:
To fix multiple new-line character producing redundant spaces:
' '.join([line.strip() for line in test_str.strip().splitlines() if line.strip()])
This works for the following too
test_str = '\nhej ho \n aaa\r\n\n\n\n\n a\n '
Regular expressions is the fastest way to do this
s='''some kind of
string with a bunch\r of
extra spaces in it'''
re.sub(r'\s(?=\s)','',re.sub(r'\s',' ',s))
result:
'some kind of string with a bunch of extra spaces in it'
The problem with rstrip() is that it does not work in all cases (as I myself have seen few). Instead you can use
text = text.replace("\n"," ")
This will remove all new line '\n' with a space.
You really don't need to remove ALL the signs: lf cr crlf.
# Pythonic:
r'\n', r'\r', r'\r\n'
Some texts must have breaks, but you probably need to join broken lines to keep particular sentences together.
Therefore it is natural that line breaking happens after priod, semicolon, colon, but not after comma.
My code considers above conditions. Works well with texts copied from pdfs.
Enjoy!:
def unbreak_pdf_text(raw_text):
""" the newline careful sign removal tool
Args:
raw_text (str): string containing unwanted newline signs: \\n or \\r or \\r\\n
e.g. imported from OCR or copied from a pdf document.
Returns:
_type_: _description_
"""
pat = re.compile((r"[, \w]\n|[, \w]\r|[, \w]\r\n"))
breaks = re.finditer(pat, raw_text)
processed_text = raw_text
raw_text = None
for i in breaks:
processed_text = processed_text.replace(i.group(), i.group()[0]+" ")
return processed_text