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Remove all types of brackets in a string in python (with content) [duplicate]

I have a very long string of text with () and [] in it. I'm trying to remove the characters between the parentheses and brackets but I cannot figure out how.
The list is similar to this:
x = "This is a sentence. (once a day) [twice a day]"
This list isn't what I'm working with but is very similar and a lot shorter.

You can use re.sub function.
>>> import re
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("([\(\[]).*?([\)\]])", "\g<1>\g<2>", x)
'This is a sentence. () []'
If you want to remove the [] and the () you can use this code:
>>> import re
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("[\(\[].*?[\)\]]", "", x)
'This is a sentence. '
Important: This code will not work with nested symbols
Explanation
The first regex groups ( or [ into group 1 (by surrounding it with parentheses) and ) or ] into group 2, matching these groups and all characters that come in between them. After matching, the matched portion is substituted with groups 1 and 2, leaving the final string with nothing inside the brackets. The second regex is self explanatory from this -> match everything and substitute with the empty string.
-- modified from comment by Ajay Thomas

Run this script, it works even with nested brackets.
Uses basic logical tests.
def a(test_str):
ret = ''
skip1c = 0
skip2c = 0
for i in test_str:
if i == '[':
skip1c += 1
elif i == '(':
skip2c += 1
elif i == ']' and skip1c > 0:
skip1c -= 1
elif i == ')'and skip2c > 0:
skip2c -= 1
elif skip1c == 0 and skip2c == 0:
ret += i
return ret
x = "ewq[a [(b] ([c))]] This is a sentence. (once a day) [twice a day]"
x = a(x)
print x
print repr(x)
Just incase you don't run it,
Here's the output:
>>>
ewq This is a sentence.
'ewq This is a sentence. '

Here's a solution similar to #pradyunsg's answer (it works with arbitrary nested brackets):
def remove_text_inside_brackets(text, brackets="()[]"):
count = [0] * (len(brackets) // 2) # count open/close brackets
saved_chars = []
for character in text:
for i, b in enumerate(brackets):
if character == b: # found bracket
kind, is_close = divmod(i, 2)
count[kind] += (-1)**is_close # `+1`: open, `-1`: close
if count[kind] < 0: # unbalanced bracket
count[kind] = 0 # keep it
else: # found bracket to remove
break
else: # character is not a [balanced] bracket
if not any(count): # outside brackets
saved_chars.append(character)
return ''.join(saved_chars)
print(repr(remove_text_inside_brackets(
"This is a sentence. (once a day) [twice a day]")))
# -> 'This is a sentence. '

This should work for parentheses. Regular expressions will "consume" the text it has matched so it won't work for nested parentheses.
import re
regex = re.compile(".*?\((.*?)\)")
result = re.findall(regex, mystring)
or this would find one set of parentheses, simply loop to find more:
start = mystring.find("(")
end = mystring.find(")")
if start != -1 and end != -1:
result = mystring[start+1:end]

You can split, filter, and join the string again. If your brackets are well defined the following code should do.
import re
x = "".join(re.split("\(|\)|\[|\]", x)[::2])

You can try this. Can remove the bracket and the content exist inside it.
import re
x = "This is a sentence. (once a day) [twice a day]"
x = re.sub("\(.*?\)|\[.*?\]","",x)
print(x)
Expected ouput :
This is a sentence.

For anyone who appreciates the simplicity of the accepted answer by jvallver, and is looking for more readability from their code:
>>> import re
>>> x = 'This is a sentence. (once a day) [twice a day]'
>>> opening_braces = '\(\['
>>> closing_braces = '\)\]'
>>> non_greedy_wildcard = '.*?'
>>> re.sub(f'[{opening_braces}]{non_greedy_wildcard}[{closing_braces}]', '', x)
'This is a sentence. '
Most of the explanation for why this regex works is included in the code. Your future self will thank you for the 3 additional lines.
(Replace the f-string with the equivalent string concatenation for Python2 compatibility)

The RegEx \(.*?\)|\[.*?\] removes bracket content by finding pairs, first it remove paranthesis and then square brackets. I also works fine for the nested brackets as it acts in sequence. Ofcourse, it would break in case of bad brackets scenario.
_brackets = re.compile("\(.*?\)|\[.*?\]")
_spaces = re.compile("\s+")
_b = _brackets.sub(" ", "microRNAs (miR) play a role in cancer ([1], [2])")
_s = _spaces.sub(" ", _b.strip())
print(_s)
# OUTPUT: microRNAs play a role in cancer

Regex replace in Spyder with case conversion [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
Example:
>>> convert('CamelCase')
'camel_case'

Camel case to snake case
import re
name = 'CamelCaseName'
name = re.sub(r'(?<!^)(?=[A-Z])', '_', name).lower()
print(name) # camel_case_name
If you do this many times and the above is slow, compile the regex beforehand:
pattern = re.compile(r'(?<!^)(?=[A-Z])')
name = pattern.sub('_', name).lower()
To handle more advanced cases specially (this is not reversible anymore):
def camel_to_snake(name):
name = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
return re.sub('([a-z0-9])([A-Z])', r'\1_\2', name).lower()
print(camel_to_snake('camel2_camel2_case')) # camel2_camel2_case
print(camel_to_snake('getHTTPResponseCode')) # get_http_response_code
print(camel_to_snake('HTTPResponseCodeXYZ')) # http_response_code_xyz
To add also cases with two underscores or more:
def to_snake_case(name):
name = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
name = re.sub('__([A-Z])', r'_\1', name)
name = re.sub('([a-z0-9])([A-Z])', r'\1_\2', name)
return name.lower()
Snake case to pascal case
name = 'snake_case_name'
name = ''.join(word.title() for word in name.split('_'))
print(name) # SnakeCaseName

There's an inflection library in the package index that can handle these things for you. In this case, you'd be looking for inflection.underscore():
>>> inflection.underscore('CamelCase')
'camel_case'

I don't know why these are all so complicating.
for most cases, the simple expression ([A-Z]+) will do the trick
>>> re.sub('([A-Z]+)', r'_\1','CamelCase').lower()
'_camel_case'
>>> re.sub('([A-Z]+)', r'_\1','camelCase').lower()
'camel_case'
>>> re.sub('([A-Z]+)', r'_\1','camel2Case2').lower()
'camel2_case2'
>>> re.sub('([A-Z]+)', r'_\1','camelCamelCase').lower()
'camel_camel_case'
>>> re.sub('([A-Z]+)', r'_\1','getHTTPResponseCode').lower()
'get_httpresponse_code'
To ignore the first character simply add look behind (?!^)
>>> re.sub('(?!^)([A-Z]+)', r'_\1','CamelCase').lower()
'camel_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','CamelCamelCase').lower()
'camel_camel_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','Camel2Camel2Case').lower()
'camel2_camel2_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','getHTTPResponseCode').lower()
'get_httpresponse_code'
If you want to separate ALLCaps to all_caps and expect numbers in your string you still don't need to do two separate runs just use | This expression ((?<=[a-z0-9])[A-Z]|(?!^)[A-Z](?=[a-z])) can handle just about every scenario in the book
>>> a = re.compile('((?<=[a-z0-9])[A-Z]|(?!^)[A-Z](?=[a-z]))')
>>> a.sub(r'_\1', 'getHTTPResponseCode').lower()
'get_http_response_code'
>>> a.sub(r'_\1', 'get2HTTPResponseCode').lower()
'get2_http_response_code'
>>> a.sub(r'_\1', 'get2HTTPResponse123Code').lower()
'get2_http_response123_code'
>>> a.sub(r'_\1', 'HTTPResponseCode').lower()
'http_response_code'
>>> a.sub(r'_\1', 'HTTPResponseCodeXYZ').lower()
'http_response_code_xyz'
It all depends on what you want so use the solution that best suits your needs as it should not be overly complicated.
nJoy!

Avoiding libraries and regular expressions:
def camel_to_snake(s):
return ''.join(['_'+c.lower() if c.isupper() else c for c in s]).lstrip('_')
>>> camel_to_snake('ThisIsMyString')
'this_is_my_string'

stringcase is my go-to library for this; e.g.:
>>> from stringcase import pascalcase, snakecase
>>> snakecase('FooBarBaz')
'foo_bar_baz'
>>> pascalcase('foo_bar_baz')
'FooBarBaz'

I think this solution is more straightforward than previous answers:
import re
def convert (camel_input):
words = re.findall(r'[A-Z]?[a-z]+|[A-Z]{2,}(?=[A-Z][a-z]|\d|\W|$)|\d+', camel_input)
return '_'.join(map(str.lower, words))
# Let's test it
test_strings = [
'CamelCase',
'camelCamelCase',
'Camel2Camel2Case',
'getHTTPResponseCode',
'get200HTTPResponseCode',
'getHTTP200ResponseCode',
'HTTPResponseCode',
'ResponseHTTP',
'ResponseHTTP2',
'Fun?!awesome',
'Fun?!Awesome',
'10CoolDudes',
'20coolDudes'
]
for test_string in test_strings:
print(convert(test_string))
Which outputs:
camel_case
camel_camel_case
camel_2_camel_2_case
get_http_response_code
get_200_http_response_code
get_http_200_response_code
http_response_code
response_http
response_http_2
fun_awesome
fun_awesome
10_cool_dudes
20_cool_dudes
The regular expression matches three patterns:
[A-Z]?[a-z]+: Consecutive lower-case letters that optionally start with an upper-case letter.
[A-Z]{2,}(?=[A-Z][a-z]|\d|\W|$): Two or more consecutive upper-case letters. It uses a lookahead to exclude the last upper-case letter if it is followed by a lower-case letter.
\d+: Consecutive numbers.
By using re.findall we get a list of individual "words" that can be converted to lower-case and joined with underscores.

Personally I am not sure how anything using regular expressions in python can be described as elegant. Most answers here are just doing "code golf" type RE tricks. Elegant coding is supposed to be easily understood.
def to_snake_case(not_snake_case):
final = ''
for i in xrange(len(not_snake_case)):
item = not_snake_case[i]
if i < len(not_snake_case) - 1:
next_char_will_be_underscored = (
not_snake_case[i+1] == "_" or
not_snake_case[i+1] == " " or
not_snake_case[i+1].isupper()
)
if (item == " " or item == "_") and next_char_will_be_underscored:
continue
elif (item == " " or item == "_"):
final += "_"
elif item.isupper():
final += "_"+item.lower()
else:
final += item
if final[0] == "_":
final = final[1:]
return final
>>> to_snake_case("RegularExpressionsAreFunky")
'regular_expressions_are_funky'
>>> to_snake_case("RegularExpressionsAre Funky")
'regular_expressions_are_funky'
>>> to_snake_case("RegularExpressionsAre_Funky")
'regular_expressions_are_funky'

''.join('_'+c.lower() if c.isupper() else c for c in "DeathToCamelCase").strip('_')
re.sub("(.)([A-Z])", r'\1_\2', 'DeathToCamelCase').lower()

Here's my solution:
def un_camel(text):
""" Converts a CamelCase name into an under_score name.
>>> un_camel('CamelCase')
'camel_case'
>>> un_camel('getHTTPResponseCode')
'get_http_response_code'
"""
result = []
pos = 0
while pos < len(text):
if text[pos].isupper():
if pos-1 > 0 and text[pos-1].islower() or pos-1 > 0 and \
pos+1 < len(text) and text[pos+1].islower():
result.append("_%s" % text[pos].lower())
else:
result.append(text[pos].lower())
else:
result.append(text[pos])
pos += 1
return "".join(result)
It supports those corner cases discussed in the comments. For instance, it'll convert getHTTPResponseCode to get_http_response_code like it should.

I don't get idea why using both .sub() calls? :) I'm not regex guru, but I simplified function to this one, which is suitable for my certain needs, I just needed a solution to convert camelCasedVars from POST request to vars_with_underscore:
def myFunc(...):
return re.sub('(.)([A-Z]{1})', r'\1_\2', "iTriedToWriteNicely").lower()
It does not work with such names like getHTTPResponse, cause I heard it is bad naming convention (should be like getHttpResponse, it's obviously, that it's much easier memorize this form).

For the fun of it:
>>> def un_camel(input):
... output = [input[0].lower()]
... for c in input[1:]:
... if c in ('ABCDEFGHIJKLMNOPQRSTUVWXYZ'):
... output.append('_')
... output.append(c.lower())
... else:
... output.append(c)
... return str.join('', output)
...
>>> un_camel("camel_case")
'camel_case'
>>> un_camel("CamelCase")
'camel_case'
Or, more for the fun of it:
>>> un_camel = lambda i: i[0].lower() + str.join('', ("_" + c.lower() if c in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" else c for c in i[1:]))
>>> un_camel("camel_case")
'camel_case'
>>> un_camel("CamelCase")
'camel_case'

Using regexes may be the shortest, but this solution is way more readable:
def to_snake_case(s):
snake = "".join(["_"+c.lower() if c.isupper() else c for c in s])
return snake[1:] if snake.startswith("_") else snake

This is not a elegant method, is a very 'low level' implementation of a simple state machine (bitfield state machine), possibly the most anti pythonic mode to resolve this, however re module also implements a too complex state machine to resolve this simple task, so i think this is a good solution.
def splitSymbol(s):
si, ci, state = 0, 0, 0 # start_index, current_index
'''
state bits:
0: no yields
1: lower yields
2: lower yields - 1
4: upper yields
8: digit yields
16: other yields
32 : upper sequence mark
'''
for c in s:
if c.islower():
if state & 1:
yield s[si:ci]
si = ci
elif state & 2:
yield s[si:ci - 1]
si = ci - 1
state = 4 | 8 | 16
ci += 1
elif c.isupper():
if state & 4:
yield s[si:ci]
si = ci
if state & 32:
state = 2 | 8 | 16 | 32
else:
state = 8 | 16 | 32
ci += 1
elif c.isdigit():
if state & 8:
yield s[si:ci]
si = ci
state = 1 | 4 | 16
ci += 1
else:
if state & 16:
yield s[si:ci]
state = 0
ci += 1 # eat ci
si = ci
print(' : ', c, bin(state))
if state:
yield s[si:ci]
def camelcaseToUnderscore(s):
return '_'.join(splitSymbol(s))
splitsymbol can parses all case types: UpperSEQUENCEInterleaved, under_score, BIG_SYMBOLS and cammelCasedMethods
I hope it is useful

Take a look at the excellent Schematics lib
https://github.com/schematics/schematics
It allows you to created typed data structures that can serialize/deserialize from python to Javascript flavour, eg:
class MapPrice(Model):
price_before_vat = DecimalType(serialized_name='priceBeforeVat')
vat_rate = DecimalType(serialized_name='vatRate')
vat = DecimalType()
total_price = DecimalType(serialized_name='totalPrice')

So many complicated methods...
Just find all "Titled" group and join its lower cased variant with underscore.
>>> import re
>>> def camel_to_snake(string):
... groups = re.findall('([A-z0-9][a-z]*)', string)
... return '_'.join([i.lower() for i in groups])
...
>>> camel_to_snake('ABCPingPongByTheWay2KWhereIsOurBorderlands3???')
'a_b_c_ping_pong_by_the_way_2_k_where_is_our_borderlands_3'
If you don't want make numbers like first character of group or separate group - you can use ([A-z][a-z0-9]*) mask.

A horrendous example using regular expressions (you could easily clean this up :) ):
def f(s):
return s.group(1).lower() + "_" + s.group(2).lower()
p = re.compile("([A-Z]+[a-z]+)([A-Z]?)")
print p.sub(f, "CamelCase")
print p.sub(f, "getHTTPResponseCode")
Works for getHTTPResponseCode though!
Alternatively, using lambda:
p = re.compile("([A-Z]+[a-z]+)([A-Z]?)")
print p.sub(lambda x: x.group(1).lower() + "_" + x.group(2).lower(), "CamelCase")
print p.sub(lambda x: x.group(1).lower() + "_" + x.group(2).lower(), "getHTTPResponseCode")
EDIT: It should also be pretty easy to see that there's room for improvement for cases like "Test", because the underscore is unconditionally inserted.

Lightely adapted from https://stackoverflow.com/users/267781/matth
who use generators.
def uncamelize(s):
buff, l = '', []
for ltr in s:
if ltr.isupper():
if buff:
l.append(buff)
buff = ''
buff += ltr
l.append(buff)
return '_'.join(l).lower()

This simple method should do the job:
import re
def convert(name):
return re.sub(r'([A-Z]*)([A-Z][a-z]+)', lambda x: (x.group(1) + '_' if x.group(1) else '') + x.group(2) + '_', name).rstrip('_').lower()
We look for capital letters that are precedeed by any number of (or zero) capital letters, and followed by any number of lowercase characters.
An underscore is placed just before the occurence of the last capital letter found in the group, and one can be placed before that capital letter in case it is preceded by other capital letters.
If there are trailing underscores, remove those.
Finally, the whole result string is changed to lower case.
(taken from here, see working example online)

Here's something I did to change the headers on a tab-delimited file. I'm omitting the part where I only edited the first line of the file. You could adapt it to Python pretty easily with the re library. This also includes separating out numbers (but keeps the digits together). I did it in two steps because that was easier than telling it not to put an underscore at the start of a line or tab.
Step One...find uppercase letters or integers preceded by lowercase letters, and precede them with an underscore:
Search:
([a-z]+)([A-Z]|[0-9]+)
Replacement:
\1_\l\2/
Step Two...take the above and run it again to convert all caps to lowercase:
Search:
([A-Z])
Replacement (that's backslash, lowercase L, backslash, one):
\l\1

I was looking for a solution to the same problem, except that I needed a chain; e.g.
"CamelCamelCamelCase" -> "Camel-camel-camel-case"
Starting from the nice two-word solutions here, I came up with the following:
"-".join(x.group(1).lower() if x.group(2) is None else x.group(1) \
for x in re.finditer("((^.[^A-Z]+)|([A-Z][^A-Z]+))", "stringToSplit"))
Most of the complicated logic is to avoid lowercasing the first word. Here's a simpler version if you don't mind altering the first word:
"-".join(x.group(1).lower() for x in re.finditer("(^[^A-Z]+|[A-Z][^A-Z]+)", "stringToSplit"))
Of course, you can pre-compile the regular expressions or join with underscore instead of hyphen, as discussed in the other solutions.

Concise without regular expressions, but HTTPResponseCode=> httpresponse_code:
def from_camel(name):
"""
ThisIsCamelCase ==> this_is_camel_case
"""
name = name.replace("_", "")
_cas = lambda _x : [_i.isupper() for _i in _x]
seq = zip(_cas(name[1:-1]), _cas(name[2:]))
ss = [_x + 1 for _x, (_i, _j) in enumerate(seq) if (_i, _j) == (False, True)]
return "".join([ch + "_" if _x in ss else ch for _x, ch in numerate(name.lower())])

Without any library :
def camelify(out):
return (''.join(["_"+x.lower() if i<len(out)-1 and x.isupper() and out[i+1].islower()
else x.lower()+"_" if i<len(out)-1 and x.islower() and out[i+1].isupper()
else x.lower() for i,x in enumerate(list(out))])).lstrip('_').replace('__','_')
A bit heavy, but
CamelCamelCamelCase -> camel_camel_camel_case
HTTPRequest -> http_request
GetHTTPRequest -> get_http_request
getHTTPRequest -> get_http_request

Very nice RegEx proposed on this site:
(?<!^)(?=[A-Z])
If python have a String Split method, it should work...
In Java:
String s = "loremIpsum";
words = s.split("(?<!^)(?=[A-Z])");

Just in case someone needs to transform a complete source file, here is a script that will do it.
# Copy and paste your camel case code in the string below
camelCaseCode ="""
cv2.Matx33d ComputeZoomMatrix(const cv2.Point2d & zoomCenter, double zoomRatio)
{
auto mat = cv2.Matx33d::eye();
mat(0, 0) = zoomRatio;
mat(1, 1) = zoomRatio;
mat(0, 2) = zoomCenter.x * (1. - zoomRatio);
mat(1, 2) = zoomCenter.y * (1. - zoomRatio);
return mat;
}
"""
import re
def snake_case(name):
s1 = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
return re.sub('([a-z0-9])([A-Z])', r'\1_\2', s1).lower()
def lines(str):
return str.split("\n")
def unlines(lst):
return "\n".join(lst)
def words(str):
return str.split(" ")
def unwords(lst):
return " ".join(lst)
def map_partial(function):
return lambda values : [ function(v) for v in values]
import functools
def compose(*functions):
return functools.reduce(lambda f, g: lambda x: f(g(x)), functions, lambda x: x)
snake_case_code = compose(
unlines ,
map_partial(unwords),
map_partial(map_partial(snake_case)),
map_partial(words),
lines
)
print(snake_case_code(camelCaseCode))

Wow I just stole this from django snippets. ref http://djangosnippets.org/snippets/585/
Pretty elegant
camelcase_to_underscore = lambda str: re.sub(r'(?<=[a-z])[A-Z]|[A-Z](?=[^A-Z])', r'_\g<0>', str).lower().strip('_')
Example:
camelcase_to_underscore('ThisUser')
Returns:
'this_user'
REGEX DEMO

def convert(name):
return reduce(
lambda x, y: x + ('_' if y.isupper() else '') + y,
name
).lower()
And if we need to cover a case with already-un-cameled input:
def convert(name):
return reduce(
lambda x, y: x + ('_' if y.isupper() and not x.endswith('_') else '') + y,
name
).lower()

Not in the standard library, but I found this module that appears to contain the functionality you need.

If you use Google's (nearly) deterministic Camel case algorithm, then one does not need to handle things like HTMLDocument since it should be HtmlDocument, then this regex based approach is simple. It replace all capitals or numbers with an underscore. Note does not handle multi digit numbers.
import re
def to_snake_case(camel_str):
return re.sub('([A-Z0-9])', r'_\1', camel_str).lower().lstrip('_')

def convert(camel_str):
temp_list = []
for letter in camel_str:
if letter.islower():
temp_list.append(letter)
else:
temp_list.append('_')
temp_list.append(letter)
result = "".join(temp_list)
return result.lower()

Use: str.capitalize() to convert first letter of the string (contained in variable str) to a capital letter and returns the entire string.
Example:
Command: "hello".capitalize()
Output: Hello

How do I reverse words in a string with Python

I am trying to reverse words of a string, but having difficulty, any assistance will be appreciated:
S = " what is my name"
def reversStr(S):
for x in range(len(S)):
return S[::-1]
break
What I get now is: eman ym si tahw
However, I am trying to get: tahw is ym eman (individual words reversed)

def reverseStr(s):
return ' '.join([x[::-1] for x in s.split(' ')])

orig = "what is my name"
reverse = ""
for word in orig.split():
reverse = "{} {}".format(reverse, word[::-1])
print(reverse)

Since everyone else's covered the case where the punctuation moves, I'll cover the one where you don't want the punctuation to move.
import re
def reverse_words(sentence):
return re.sub(r'[a-zA-Z]+', lambda x : x.group()[::-1], sentence)
Breaking this down.
re is python's regex module, and re.sub is the function in that module that handles substitutions. It has three required parameters.
The first is the regex you're matching by. In this case, I'm using r'\w+'. The r denotes a raw string, [a-zA-Z] matches all letters, and + means "at least one".
The second is either a string to substitute in, or a function that takes in a re.MatchObject and outputs a string. I'm using a lambda (or nameless) function that simply outputs the matched string, reversed.
The third is the string you want to do a find in a replace in.
So "What is my name?" -> "tahW si ym eman?"
Addendum:
I considered a regex of r'\w+' initially, because better unicode support (if the right flags are given), but \w also includes numbers and underscores. Matching - might also be desired behavior: the regexes would be r'[a-zA-Z-]+' (note trailing hyphen) and r'[\w-]+' but then you'd probably want to not match double-dashes (ie --) so more regex modifications might be needed.
The built-in reversed outputs a reversed object, which you have to cast back to string, so I generally prefer the [::-1] option.

inplace refers to modifying the object without creating a copy. Yes, like many of us has already pointed out that python strings are immutable. So technically we cannot reverse a python string datatype object inplace. However, if you use a mutable datatype, say bytearray for storing the string characters, you can actually reverse it inplace
#slicing creates copy; implies not-inplace reversing
def rev(x):
return x[-1::-1]
# inplace reversing, if input is bytearray datatype
def rev_inplace(x: bytearray):
i = 0; j = len(x)-1
while i<j:
t = x[i]
x[i] = x[j]
x[j] = t
i += 1; j -= 1
return x
Input:
x = bytearray(b'some string to reverse')
rev_inplace(x)
Output:
bytearray(b'esrever ot gnirts emose')

Try splitting each word in the string into a list (see: https://docs.python.org/2/library/stdtypes.html#str.split).
Example:
>>string = "This will be split up"
>>string_list = string.split(" ")
>>string_list
>>['This', 'will', 'be', 'split', 'up']
Then iterate through the list and reverse each constituent list item (i.e. word) which you have working already.

def reverse_in_place(phrase):
res = []
phrase = phrase.split(" ")
for word in phrase:
word = word[::-1]
res.append(word)
res = " ".join(res)
return res

[thread has been closed, but IMO, not well answered]
the python string.lib doesn't include an in place str.reverse() method.
So use the built in reversed() function call to accomplish the same thing.
>>> S = " what is my name"
>>> ("").join(reversed(S))
'eman ym si tahw'

There is no obvious way of reversing a string "truly" in-place with Python. However, you can do something like:
def reverse_string_inplace(string):
w = len(string)-1
p = w
while True:
q = string[p]
string = ' ' + string + q
w -= 1
if w < 0:
break
return string[(p+1)*2:]
Hope this makes sense.

In Python, strings are immutable. This means you cannot change the string once you have created it. So in-place reverse is not possible.
There are many ways to reverse the string in python, but memory allocation is required for that reversed string.
print(' '.join(word[::-1] for word in string))

s1 = input("Enter a string with multiple words:")
print(f'Original:{s1}')
print(f'Reverse is:{s1[::-1]}')
each_word_new_list = []
s1_split = s1.split()
for i in range(0,len(s1_split)):
each_word_new_list.append(s1_split[i][::-1])
print(f'New Reverse as List:{each_word_new_list}')
each_word_new_string=' '.join(each_word_new_list)
print(f'New Reverse as String:{each_word_new_string}')

If the sentence contains multiple spaces then usage of split() function will cause trouble because you won't know then how many spaces you need to rejoin after you reverse each word in the sentence. Below snippet might help:
# Sentence having multiple spaces
given_str = "I know this country runs by mafia "
tmp = ""
tmp_list = []
for i in given_str:
if i != ' ':
tmp = tmp + i
else:
if tmp == "":
tmp_list.append(i)
else:
tmp_list.append(tmp)
tmp_list.append(i)
tmp = ""
print(tmp_list)
rev_list = []
for x in tmp_list:
rev = x[::-1]
rev_list.append(rev)
print(rev_list)
print(''.join(rev_list))
output:

def rev(a):
if a == "":
return ""
else:
z = rev(a[1:]) + a[0]
return z
Reverse string --> gnirts esreveR
def rev(k):
y = rev(k).split()
for i in range(len(y)-1,-1,-1):
print y[i],
-->esreveR gnirts

Remove text between () and []

I have a very long string of text with () and [] in it. I'm trying to remove the characters between the parentheses and brackets but I cannot figure out how.
The list is similar to this:
x = "This is a sentence. (once a day) [twice a day]"
This list isn't what I'm working with but is very similar and a lot shorter.

You can use re.sub function.
>>> import re
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("([\(\[]).*?([\)\]])", "\g<1>\g<2>", x)
'This is a sentence. () []'
If you want to remove the [] and the () you can use this code:
>>> import re
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("[\(\[].*?[\)\]]", "", x)
'This is a sentence. '
Important: This code will not work with nested symbols
Explanation
The first regex groups ( or [ into group 1 (by surrounding it with parentheses) and ) or ] into group 2, matching these groups and all characters that come in between them. After matching, the matched portion is substituted with groups 1 and 2, leaving the final string with nothing inside the brackets. The second regex is self explanatory from this -> match everything and substitute with the empty string.
-- modified from comment by Ajay Thomas

Run this script, it works even with nested brackets.
Uses basic logical tests.
def a(test_str):
ret = ''
skip1c = 0
skip2c = 0
for i in test_str:
if i == '[':
skip1c += 1
elif i == '(':
skip2c += 1
elif i == ']' and skip1c > 0:
skip1c -= 1
elif i == ')'and skip2c > 0:
skip2c -= 1
elif skip1c == 0 and skip2c == 0:
ret += i
return ret
x = "ewq[a [(b] ([c))]] This is a sentence. (once a day) [twice a day]"
x = a(x)
print x
print repr(x)
Just incase you don't run it,
Here's the output:
>>>
ewq This is a sentence.
'ewq This is a sentence. '

Here's a solution similar to #pradyunsg's answer (it works with arbitrary nested brackets):
def remove_text_inside_brackets(text, brackets="()[]"):
count = [0] * (len(brackets) // 2) # count open/close brackets
saved_chars = []
for character in text:
for i, b in enumerate(brackets):
if character == b: # found bracket
kind, is_close = divmod(i, 2)
count[kind] += (-1)**is_close # `+1`: open, `-1`: close
if count[kind] < 0: # unbalanced bracket
count[kind] = 0 # keep it
else: # found bracket to remove
break
else: # character is not a [balanced] bracket
if not any(count): # outside brackets
saved_chars.append(character)
return ''.join(saved_chars)
print(repr(remove_text_inside_brackets(
"This is a sentence. (once a day) [twice a day]")))
# -> 'This is a sentence. '

This should work for parentheses. Regular expressions will "consume" the text it has matched so it won't work for nested parentheses.
import re
regex = re.compile(".*?\((.*?)\)")
result = re.findall(regex, mystring)
or this would find one set of parentheses, simply loop to find more:
start = mystring.find("(")
end = mystring.find(")")
if start != -1 and end != -1:
result = mystring[start+1:end]

You can split, filter, and join the string again. If your brackets are well defined the following code should do.
import re
x = "".join(re.split("\(|\)|\[|\]", x)[::2])

You can try this. Can remove the bracket and the content exist inside it.
import re
x = "This is a sentence. (once a day) [twice a day]"
x = re.sub("\(.*?\)|\[.*?\]","",x)
print(x)
Expected ouput :
This is a sentence.

For anyone who appreciates the simplicity of the accepted answer by jvallver, and is looking for more readability from their code:
>>> import re
>>> x = 'This is a sentence. (once a day) [twice a day]'
>>> opening_braces = '\(\['
>>> closing_braces = '\)\]'
>>> non_greedy_wildcard = '.*?'
>>> re.sub(f'[{opening_braces}]{non_greedy_wildcard}[{closing_braces}]', '', x)
'This is a sentence. '
Most of the explanation for why this regex works is included in the code. Your future self will thank you for the 3 additional lines.
(Replace the f-string with the equivalent string concatenation for Python2 compatibility)

The RegEx \(.*?\)|\[.*?\] removes bracket content by finding pairs, first it remove paranthesis and then square brackets. I also works fine for the nested brackets as it acts in sequence. Ofcourse, it would break in case of bad brackets scenario.
_brackets = re.compile("\(.*?\)|\[.*?\]")
_spaces = re.compile("\s+")
_b = _brackets.sub(" ", "microRNAs (miR) play a role in cancer ([1], [2])")
_s = _spaces.sub(" ", _b.strip())
print(_s)
# OUTPUT: microRNAs play a role in cancer

How to replace the Nth appearance of a needle in a haystack? (Python)

I am trying to replace the Nth appearance of a needle in a haystack. I want to do this simply via re.sub(), but cannot seem to come up with an appropriate regex to solve this. I am trying to adapt: http://docstore.mik.ua/orelly/perl/cookbook/ch06_06.htm but am failing at spanning multilines, I suppose.
My current method is an iterative approach that finds the position of each occurrence from the beginning after each mutation. This is pretty inefficient and I would like to get some input. Thanks!

I think you mean re.sub. You could pass a function and keep track of how often it was called so far:
def replaceNthWith(n, replacement):
def replace(match, c=[0]):
c[0] += 1
return replacement if c[0] == n else match.group(0)
return replace
Usage:
re.sub(pattern, replaceNthWith(n, replacement), str)
But this approach feels a bit hacky, maybe there are more elegant ways.
DEMO

Something like this regex should help you. Though I'm not sure how efficient it is:
#N=3
re.sub(
r'^((?:.*?mytexttoreplace){2}.*?)mytexttoreplace',
'\1yourreplacementtext.',
'mystring',
flags=re.DOTALL
)
The DOTALL flag is important.

I've been struggling for a while with this, but I found a solution that I think is pretty pythonic:
>>> def nth_matcher(n, replacement):
... def alternate(n):
... i=0
... while True:
... i += 1
... yield i%n == 0
... gen = alternate(n)
... def match(m):
... replace = gen.next()
... if replace:
... return replacement
... else:
... return m.group(0)
... return match
...
...
>>> re.sub("([0-9])", nth_matcher(3, "X"), "1234567890")
'12X45X78X0'
EDIT: the matcher consists of two parts:
the alternate(n) function. This returns a generator that returns an infinite sequence True/False, where every nth value is True. Think of it like list(alternate(3)) == [False, False, True, False, False, True, False, ...].
The match(m) function. This is the function that gets passed to re.sub: it gets the next value in alternate(n) (gen.next()) and if it's True it replaces the matched value; otherwise, it keeps it unchanged (replaces it with itself).
I hope this is clear enough. If my explanation is hazy, please say so and I'll improve it.

Could you do it using re.findall with MatchObject.start() and MatchObject.end()?
find all occurences of pattern in string with .findall, get indices of Nth occurrence with .start/.end, make new string with replacement value using the indices?

If the pattern ("needle") or replacement is a complex regular expression, you can't assume anything. The function "nth_occurrence_sub" is what I came up with as a more general solution:
def nth_match_end(pattern, string, n, flags):
for i, match_object in enumerate(re.finditer(pattern, string, flags)):
if i + 1 == n:
return match_object.end()
def nth_occurrence_sub(pattern, repl, string, n=0, flags=0):
max_n = len(re.findall(pattern, string, flags))
if abs(n) > max_n or n == 0:
return string
if n < 0:
n = max_n + n + 1
sub_n_times = re.sub(pattern, repl, string, n, flags)
if n == 1:
return sub_n_times
nm1_end = nth_match_end(pattern, string, n - 1, flags)
sub_nm1_times = re.sub(pattern, repl, string, n - 1, flags)
sub_nm1_change = sub_nm1_times[:-1 * len(string[nm1_end:])]
components = [
string[:nm1_end],
sub_n_times[len(sub_nm1_change):]
]
return ''.join(components)

I have a similar function I wrote to do this. I was trying to replicate SQL REGEXP_REPLACE() functionality. I ended up with:
def sql_regexp_replace( txt, pattern, replacement='', position=1, occurrence=0, regexp_modifier='c'):
class ReplWrapper(object):
def __init__(self, replacement, occurrence):
self.count = 0
self.replacement = replacement
self.occurrence = occurrence
def repl(self, match):
self.count += 1
if self.occurrence == 0 or self.occurrence == self.count:
return match.expand(self.replacement)
else:
try:
return match.group(0)
except IndexError:
return match.group(0)
occurrence = 0 if occurrence < 0 else occurrence
flags = regexp_flags(regexp_modifier)
rx = re.compile(pattern, flags)
replw = ReplWrapper(replacement, occurrence)
return txt[0:position-1] + rx.sub(replw.repl, txt[position-1:])
One important note that I haven't seen mentioned is that you need to return match.expand() otherwise it won't expand the \1 templates properly and will treat them as literals.
If you want this to work you'll need to handle the flags differently (or take it from my github, it's simple to implement and you can dummy it for a test by setting it to 0 and ignoring my call to regexp_flags()).