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.
How can I copy data from changing string?
I tried to slice, but length of slice is changing.
For example in one case I should copy number 128 from string '"edge_liked_by":{"count":128}', in another I should copy 15332 from "edge_liked_by":{"count":15332}
You could use a regular expression:
import re
string = '"edge_liked_by":{"count":15332}'
number = re.search(r'{"count":(\d*)}', string).group(1)
Really depends on the situation, however I find regular expressions to be useful.
To grab the numbers from the string without caring about their location, you would do as follows:
import re
def get_string(string):
return re.search(r'\d+', string).group(0)
>>> get_string('"edge_liked_by":{"count":128}')
'128'
To only get numbers from the *end of the string, you can use an anchor to ensure the result is pulled from the far end. The following example will grab any sequence of unbroken numbers that is both preceeded by a colon and ends within 5 characters of the end of the string:
import re
def get_string(string):
rval = None
string_match = re.search(r':(\d+).{0,5}$', string)
if string_match:
rval = string_match.group(1)
return rval
>>> get_string('"edge_liked_by":{"count":128}')
'128'
>>> get_string('"edge_liked_by":{"1321":1}')
'1'
In the above example, adding the colon will ensure that we only pick values and don't match keys such as the "1321" that I added in as a test.
If you just want anything after the last colon, but excluding the bracket, try combining split with slicing:
>>> '"edge_liked_by":{"count":128}'.split(':')[-1][0:-1]
'128'
Finally, considering this looks like a JSON object, you can add curly brackets to the string and treat it as such. Then it becomes a nested dict you can query:
>>> import json
>>> string = '"edge_liked_by":{"count":128}'
>>> string = '{' + string + '}'
>>> string = json.loads(string)
>>> string.get('edge_liked_by').get('count')
128
The first two will return a string and the final one returns a number due to being treated as a JSON object.
It looks like the type of string you are working with is read from JSON, maybe you are getting it as the output of some API you are working with?
If it is JSON, you've probably gone one step too far in atomizing it to a string like this. I'd work with the original output, if possible, if I were you.
If not, to make it more JSON like, I'd convert it to JSON by wrapping it in {}, and then working with the json.loads module.
import json
string = '"edge_liked_by":{"count":15332}'
string = "{"+string+"}"
json_obj = json.loads(string)
count = json_obj['edge_liked_by']['count']
count will have the desired output. I prefer this option to using regular expressions because you can rely on the structure of the data and reuse the code in case you wish to parse out other attributes, in a very intuitive way. With regular expressions, the code you use will change if the data are decimal, or negative, or contain non-numeric characters.
Does this help ?
a='"edge_liked_by":{"count":128}'
import re
b=re.findall(r'\d+', a)[0]
b
Out[16]: '128'
I have been noticing a problem I am having whenever I try to make a function that takes changes a string or a list then returns it.
I will give you an example of this happening with a code I just wrote:
def remove_exclamation(string):
string.split(' ')
for i in string:
i.split()
for char in i:
if char == '!':
del char
''.join(i)
' '.join(string)
return string
For instance, I create this code to take a string as its parameter, remove any exclamation in it, the return it changed. The input and output should look like this:
>>>remove_exclamation('This is an example!')
'This is an example'
But instead I get this:
>>>remove_exclamation('This is an example!')
'This is an example!'
The function is not removing the exclamation in the output, and is not doing what I intended for it to day.
How can I keep avoiding this when I make for loops, nested for loops etc?
You write your code and formulate your question as if it was possible to modify strings in Python. It is not possible.
Strings are immutable. All functions which operate on strings return new strings. They do not modify existing strings.
This returns a list of strings, but you are not using the result:
string.split(' ')
This also:
i.split()
This deletes the variable named char. It does not affect the char itself:
del char
This creates a new string which you do not use:
''.join(i)
This also:
' '.join(string)
All in all, almost every line of the code is wrong.
You probably wanted to do this:
def remove_exclamation(string):
words = string.split(' ')
rtn_words = []
for word in words:
word_without_exclamation = ''.join(ch for ch in word if ch != '!')
rtn_words.append(word_without_exclamation)
return ' '.join(rtn_words)
But in the end, this does the same thing:
def remove_exclamation(string):
return string.replace('!', '')
Without clearly knowing the intentions of your function and what you are attempting to do. I have an alternative to the answer that zvone gave.
This option is to remove any characters that you have not defined in an allowed characters list:
characters = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ "
test_string = "This is an example!"
test_string = ''.join(list(filter(lambda x: x in characters, test_string)))
print(test_string)
This outputs:
This is an example
Note, this is the Python 3 version.
Python 2, you do not need the ''.join(list())
Doing it this way would allow you to define any character that you do not want present in your string, and it will remove them.
You can even do the reverse:
ignore_characters= "!"
test_string = "This is an example!"
test_string = ''.join(list(filter(lambda x: x not in ignore_characters, test_string)))
print(test_string)
Strings are immutable in Python. And you cannot change them. You can however, re-assign there values.
That is where your problem lies. You never reassign the value of your strings, when you call .split() on them.
But there are also others errors in your program such as:
Your indention
The fact that your just returning the string thats passed into the function
Your use of the del statement
etc.
Instead, create a new string by iterating through the old one and filtering out the character(s) you do not want, via list comprehension and ''.join().
def remove_exclamation(string):
return ''.join([char for char in string if char != '!'])
But as #Moses has already said in the comments, why not just use str.replace()?:
string = string.replace('!', '')
def remove_exclamation(string):
#you think you are splitting string into tokens
#but you are not assigning the split anywhere...
string.split(' ')
#and here you are cycling through individual _chars_ in string which was not affected by the split above ;-)
for i in string:
#and now you are splitting a 1-char string and again not assigning it.
i.split()
And string is still your input param, which I assume is of type str. And immutable.
On top of which, if you were import/using the string module, you would be shadowing it
A big part of your confusion is knowing when the methods mutate the objects and when they return a new object. In the case of strings, they never mutate, you need to assign the results to a new variable.
On a list however, and the join() somewhere makes me think you want to use a list, then methods generally change the object in place.
Anyway, on to your question:
def remove_exclamation(inputstring, to_remove="!"):
return "".join([c for c in inputstring if c != to_remove])
print (remove_exclamation('This is an example!'))
output:
This is an example
I have a string opening with { and closing with }. This brackets are always at first and at last and must appear, they can not appear in the middle. as following:
{-4,10746,.....,205}
{-3,105756}
what is the most efficient way to remove the brackets to receive:
-4,10746,.....,205
-3,105756
s[1:-1] # skip the first and last character
You can also use replace method.
In [1]: a = 'hello world'
In [3]: a.replace('l','')
Out[3]: 'heo word'
Since you were not clear there are two possibilities it may be a string or a set
If it is a set this might work:
a= {-4, 205, 10746}
",".join([str(s) for s in a])
output='10746,-4,205'
If it is a string this will work:
a= '{-4, 205, 10746}'
a.replace("{","").replace("}","")
output= '-4, 205, 10746'
Since there is no order in set the output is that way
Here's a rather roundabout way of doing exactly what you need:
l = {-3,105756}
new_l = []
for ch in l:
if ch!='{' and ch!= '}':
new_l.append(ch)
for i,val in enumerate(new_l):
length = len(new_l)
if(i==length-1):
print str(val)
else:
print str(val)+',',
I'm sure there are numerous single line codes to give you what you want, but this is kind of what goes on in the background, and will also remove the braces irrespective of their positions in the input string.
Just a side note, answer by #dlask is good to solve your issue.
But if what you really want is to convert that string (that looks like a set) to a set object (or some other data structure) , you can also use ast.literal_eval() function -
>>> import ast
>>> s = '{-3,105756}'
>>> sset = ast.literal_eval(s)
>>> sset
{105756, -3}
>>> type(sset)
<class 'set'>
From documentation -
ast.literal_eval(node_or_string)
Safely evaluate an expression node or a Unicode or Latin-1 encoded string containing a Python literal or container display. The string or node provided may only consist of the following Python literal structures: strings, numbers, tuples, lists, dicts, booleans, and None.
The safest way would be to strip:
'{-4, 205, 10746}'.strip("{}")
Is it possible to replace a single character inside a string that occurs many times?
Input:
Sentence=("This is an Example. Thxs code is not what I'm having problems with.") #Example input
^
Sentence=("This is an Example. This code is not what I'm having problems with.") #Desired output
Replace the 'x' in "Thxs" with an i, without replacing the x in "Example".
You can do it by including some context:
s = s.replace("Thxs", "This")
Alternatively you can keep a list of words that you don't wish to replace:
whitelist = ['example', 'explanation']
def replace_except_whitelist(m):
s = m.group()
if s in whitelist: return s
else: return s.replace('x', 'i')
s = 'Thxs example'
result = re.sub("\w+", replace_except_whitelist, s)
print(result)
Output:
This example
Sure, but you essentially have to build up a new string out of the parts you want:
>>> s = "This is an Example. Thxs code is not what I'm having problems with."
>>> s[22]
'x'
>>> s[:22] + "i" + s[23:]
"This is an Example. This code is not what I'm having problems with."
For information about the notation used here, see good primer for python slice notation.
If you know whether you want to replace the first occurrence of x, or the second, or the third, or the last, you can combine str.find (or str.rfind if you wish to start from the end of the string) with slicing and str.replace, feeding the character you wish to replace to the first method, as many times as it is needed to get a position just before the character you want to replace (for the specific sentence you suggest, just one), then slice the string in two and replace only one occurrence in the second slice.
An example is worth a thousands words, or so they say. In the following, I assume you want to substitute the (n+1)th occurrence of the character.
>>> s = "This is an Example. Thxs code is not what I'm having problems with."
>>> n = 1
>>> pos = 0
>>> for i in range(n):
>>> pos = s.find('x', pos) + 1
...
>>> s[:pos] + s[pos:].replace('x', 'i', 1)
"This is an Example. This code is not what I'm having problems with."
Note that you need to add an offset to pos, otherwise you will replace the occurrence of x you have just found.