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 need to output any repeated character to refer to the previous character.
For example: a(-1)rdv(-4)(-4)k or hel(-1)o
This is my code so far:
text= 'aardvark'
i=0
j=0
for i in range(len(text)-1):
for j in range(i+1, len(text)):
if text[j]==text[i]:
sub= text[j]
val2=text.find(sub, i+1, len(text))
p=val2+1
val=str(i-j)
text= text[:val2] + val + text[p:]
break
print(text)
Output: a-1rdva-4k
The second 'a' is not recognised. And I'm not sure how to include brackets in my print.
By updating the text in-place each time you find a back-reference, you muck up your indices (your text gets longer each time) and you never process the last characters properly. You stop checking when you find the first repeat of the 'current' character, so the 3rd a is never processed. This applies to every 3rd repeat in an input string. In addition, if your input text contains any - characters or digits they'll end up being tested against the -offset references you inserted before them too!
For your specific example of aardvark, a string with 8 characters, what happens is this:
You find the second a and set text to a-1rdvark. The text is now 9 characters long, so the last r will never be checked (you loop to i = 6 at most); this would be a problem if your test string ended in a double letter. You break out of the loop, so the j for loop never comes to the 3rd a, and the second a can't be tested for anymore as it has already been replaced.
Your code finds - (not repeated), 1 (not repeated) and then r (repeated once), so now you replace text with a-1rdva-4k. Now you have a string of 10 characters, so -, and 4 will never be tested. Not a big problem anymore, but what if there was a repeat in just the last 3 positions of the string?
Build a new object for the output (adding both letters you haven't seen before and backreferences). That way you won't cause the text you are looping over to grow, and you will continue to find repeats; for the parentheses you could use more string concatenation. You'll need to scan the part of the string before i, not after, for this to work, and go backwards! Testing i - 1, i - 2, etc, down to 0. Naturally, this means your i loop should then range up to the full length:
output = ''
for i in range(len(text)):
current = text[i]
for j in range(i - 1, -1, -1):
if text[j] == current:
current = '(' + str(j - i) + ')'
break
output = output + current
print(output)
I kept the fix to a minimum here, but ideally I'd also make some more changes:
Add all processed characters and references to a new list instead of a string, then use str.join() to join that list into the output afterwards. This is far more efficient than rebuilding the string each iteration.
Using two loops means you check every character in the string again while looping over the text, so the number of steps the algorithm takes grows exponentially with the length of the input. In Computer Science we talk about the time complexity of algorithms, and yours is a O(N^2) (N squared) exponential algorithm. A text with 1000 letters would take up to 1 million steps to process! Rather than loop an exponential number of times, you can use a dictionary to track indices of letters you have seen. If the current character is in the dictionary you can then trivially calculate the offset. Dictionary lookups take constant time (O(1)), making the whole algorithm take linear time (O(N)), meaning that the time the process takes is directly proportional to the length of the input string.
Use enumerate() to add a counter to the loop so you can just loop over the characters directly, no need to use range().
You can use string formatting to build a "(<offset>)" string; Python 3.6 and newer have formatted string literals, where f'...' strings take {} placeholders that are just expressions. f'({some - calculation + or * other})' will execute the expression and put the result in a string that has(and)characters in it too. For earlier Python versions, you can use the [str.format()method](https://docs.python.org/3/library/stdtypes.html#str.format) to get the same result; the syntax then becomes'({})'.format(some - calculation + or * other)`.
Put together, that becomes:
def add_backrefs(text):
output = []
seen = {}
for i, character in enumerate(text):
if character in seen:
# add a back-reference, we have seen this already
output.append(f'({seen[character] - i})')
else:
# add the literal character instead
output.append(character)
# record the position of this character for later reference
seen[character] = i
return ''.join(output)
Demo:
>>> add_backrefs('aardvark')
'a(-1)rdv(-4)(-4)k'
>>> add_backrefs('hello')
'hel(-1)o'
text= 'aardvark'
d={} # create a dictionary to keep track of index of element last seen at
new_text='' # new text to be generated
for i in range(len(text)): # iterate in text from index 0 up to length of text
c = text[i] # storing a character in temporary element as used frequently
if c not in d: # check if character which is explored is visited before or not
d[c] = i # if character visited first time then just add index value of it in dictionary
new_text += c # concatenate character to result text
else: # visiting alreaady visited character
new_text += '({0})'.format(d[c]-i) # used string formatting which will print value of difference of last seen repeated character with current index instead of {0}
d[c] = i # change last seen character index
print(new_text)
Output:
a(-1)rdv(-4)(-4)k
I'm working on PythonChallenge #3. I've got a huge block of text that I have to sort through. I am trying to find a sequence in which the first and last three letters are caps, and the middle one is lowercase.
My function loops through the text. The variable block stores the seven letters that are currently being looped through. There's a variable, toPrint, which gets turned on and off based on whether the letters in block correspond to my pattern (AAAaAAA). Based on the last block printed according to my function, my loop stops early in my text. I have no idea why this is happening and if you could help me figure this out, that would be great.
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
words = []
for i in text:
toPrint = True
block = text[text.index(i):text.index(i)+7]
for b in block[:3]:
if b.isupper() == False:
toPrint = False
for b in block[3]:
if b.islower() == False:
toPrint = False
for b in block[4:]:
if b.isupper() == False:
toPrint = False
if toPrint == True and block not in words:
words.append(block)
print (block)
print (words)
With Regex:
This is a really good time to use regex, it's super fast, more clear, and doesn't require a bunch of nested if statements.
import re
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
print(re.search(r"[A-Z]{3}[a-z][A-Z]{3}", text).group(0))
Explanation of regex:
[A-Z]{3] ---> matches any 3 uppercase letters
[a-z] -------> matches a single lowercase letter
[A-Z]{3] ---> matches 3 more uppercase letters
Without Regex:
If you really don't want to use regex this is how you could do it:
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
for i, _ in enumerate(text[:-6]): #loop through index of each char (not including last 6)
sevenCharacters = text[i:i+7] #create chunk of seven characters
shouldBeCapital = sevenCharacters[0:3] + sevenCharacters[4:7] #combine all the chars that should be cap into list
if (all(char.isupper() for char in shouldBeCapital)): #make sure all those characters are indeeed capital
if(sevenCharacters[3].islower()): #make sure middle character is lowercase
print(sevenCharacters)
I think your first problem is that you are using str.index(). Like find(), the .index() method of a string returns the index of the first match that is found.
Thus, in your example, whenever you search for 'x' you will get the index of the first 'x' found, etc. You cannot successfully work with any character that is not unique in the string, or that is not the first occurrence of a repeated character.
In order to keep the same structure (which isn't necessary- there is an answer posted using enumerate that I prefer myself) I implemented a queuing approach with your block variable. Each iteration, a character is dropped from the front of block, while the new character is appended to the end.
I also cleaned up some of your needless comparisons with False. You will find that this is not only inefficient, it is frequently wrong, because many of the "boolean" activities you perform will not be on actual boolean values. Get out of the habit of spelling out True/False. Just use if c or if not c.
Here's the result:
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
words = []
block = '.' + text[0:6]
for i in text[6:]:
block = block[1:] + i # Drop 1st char, append 'i'
toPrint = True
for b in block[:3]:
if not b.isupper():
toPrint = False
if not block[3].islower():
toPrint = False
for b in block[4:]:
if not b.isupper():
toPrint = False
if toPrint and block not in words:
words.append(block)
print (words)
If I understood your question, then according to my opinion there is no need of loop. My this simple code can find required sequence.
# Use this code
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
import re
print(re.findall("[A-Z]{3}[a-z][A-Z]{3}", text))