Find all occurrences of a regex pattern and replace with eval output - python

I have a txt file which contains a lot of strings such as
Chr(101)
Chr(97)
Chr(104)
...
I am using the below code to find all occurrences of such strings using regex. What I'd like to do is to replace each occurrence with its evaluated output. So in this case I'd replace the above with:
e
a
h
The code I have is as follows:
with open(oFile, "r") as f:
for line in f:
# find all occurrences of Chr(\d+\) and put in a list
chrList = [str(s) for s in re.findall(r'Chr\(\d+\)', line)]
# print chrList
for c in chrList:
# print eval(c.lower())
out = re.sub(c, eval(c.lower()), line)
If I print the eval(c.lower()) line then it outputs as expected. However the re.sub line fails with the following error:
raise error, v # invalid expression sre_constants.error: bogus escape (end of line)
Not sure where I'm going wrong here.

You don't have to use distinct search and replace functions. You can invoke eval using the functional form of re.sub:
for line in f:
out = re.sub(r'Chr\(\d+\)', lambda c: eval(c.group(0).lower()), line)
print out

Your going to want to escape your search pattern because parenthesis are special characters in regular expressions. You can easily do this using re.escape.
out = re.sub(re.escape(c), eval(c.lower()), line)
And as an example:
strings = ['Chr(100)', 'Chr(101)', 'Chr(102)']
values = [re.sub(re.escape(c), eval(c.lower()), c) for c in strings]
# ['d', 'e', 'f']
That being said, why not just use replace()?
out = line.replace(c, eval(c.lower())

Same thing but without eval() or imports:
strings = ['Chr(100)', 'Chr(101)', 'Chr(102)']
values = [chr(x) for x in (int(c.replace("Chr(", "").replace(")","")) for c in strings)]

Related

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

I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
How do I do this properly?
Strings in Python are immutable (can't be changed). Because of this, the effect of line.replace(...) is just to create a new string, rather than changing the old one. You need to rebind (assign) it to line in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use str.translate, (see Python 3 answer below):
line = line.translate(None, '!##$')
or regular expression replacement with re.sub
import re
line = re.sub('[!##$]', '', line)
The characters enclosed in brackets constitute a character class. Any characters in line which are in that class are replaced with the second parameter to sub: an empty string.
Python 3 answer
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a comment on one of the answers, and it's noted in the documentation for str.translate.
When calling the translate method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass None as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling ord on them) to the ordinal values of the characters which should replace them, or—usefully to us—None to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
translation_table = dict.fromkeys(map(ord, '!##$'), None)
unicode_line = unicode_line.translate(translation_table)
Here dict.fromkeys and map are used to succinctly generate a dictionary containing
{ord('!'): None, ord('#'): None, ...}
Even simpler, as another answer puts it, create the translation table in place:
unicode_line = unicode_line.translate({ord(c): None for c in '!##$'})
Or, as brought up by Joseph Lee, create the same translation table with str.maketrans:
unicode_line = unicode_line.translate(str.maketrans('', '', '!##$'))
* for compatibility with earlier Pythons, you can create a "null" translation table to pass in place of None:
import string
line = line.translate(string.maketrans('', ''), '!##$')
Here string.maketrans is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.
Am I missing the point here, or is it just the following:
string = "ab1cd1ef"
string = string.replace("1", "")
print(string)
# result: "abcdef"
Put it in a loop:
a = "a!b#c#d$"
b = "!##$"
for char in b:
a = a.replace(char, "")
print(a)
# result: "abcd"
>>> line = "abc##!?efg12;:?"
>>> ''.join( c for c in line if c not in '?:!/;' )
'abc##efg12'
With re.sub regular expression
Since Python 3.5, substitution using regular expressions re.sub became available:
import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
Example
import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
'QDoIwriteNo'
Explanation
In regular expressions (regex), | is a logical OR and \ escapes spaces and special characters that might be actual regex commands. Whereas sub stands for substitution, in this case with the empty string ''.
The asker almost had it. Like most things in Python, the answer is simpler than you think.
>>> line = "H E?.LL!/;O:: "
>>> for char in ' ?.!/;:':
... line = line.replace(char,'')
...
>>> print line
HELLO
You don't have to do the nested if/for loop thing, but you DO need to check each character individually.
For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator [^ABCabc]. For example, to remove everything except ascii letters, digits, and the hyphen:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
From the python regular expression documentation:
Characters that are not within a range can be matched by complementing
the set. If the first character of the set is '^', all the characters
that are not in the set will be matched. For example, [^5] will match
any character except '5', and [^^] will match any character except
'^'. ^ has no special meaning if it’s not the first character in the
set.
line = line.translate(None, " ?.!/;:")
>>> s = 'a1b2c3'
>>> ''.join(c for c in s if c not in '123')
'abc'
Strings are immutable in Python. The replace method returns a new string after the replacement. Try:
for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
This is identical to your original code, with the addition of an assignment to line inside the loop.
Note that the string replace() method replaces all of the occurrences of the character in the string, so you can do better by using replace() for each character you want to remove, instead of looping over each character in your string.
I was surprised that no one had yet recommended using the builtin filter function.
import operator
import string # only for the example you could use a custom string
s = "1212edjaq"
Say we want to filter out everything that isn't a number. Using the filter builtin method "...is equivalent to the generator expression (item for item in iterable if function(item))" [Python 3 Builtins: Filter]
sList = list(s)
intsList = list(string.digits)
obj = filter(lambda x: operator.contains(intsList, x), sList)))
In Python 3 this returns
>> <filter object # hex>
To get a printed string,
nums = "".join(list(obj))
print(nums)
>> "1212"
I am not sure how filter ranks in terms of efficiency but it is a good thing to know how to use when doing list comprehensions and such.
UPDATE
Logically, since filter works you could also use list comprehension and from what I have read it is supposed to be more efficient because lambdas are the wall street hedge fund managers of the programming function world. Another plus is that it is a one-liner that doesnt require any imports. For example, using the same string 's' defined above,
num = "".join([i for i in s if i.isdigit()])
That's it. The return will be a string of all the characters that are digits in the original string.
If you have a specific list of acceptable/unacceptable characters you need only adjust the 'if' part of the list comprehension.
target_chars = "".join([i for i in s if i in some_list])
or alternatively,
target_chars = "".join([i for i in s if i not in some_list])
Using filter, you'd just need one line
line = filter(lambda char: char not in " ?.!/;:", line)
This treats the string as an iterable and checks every character if the lambda returns True:
>>> help(filter)
Help on built-in function filter in module __builtin__:
filter(...)
filter(function or None, sequence) -> list, tuple, or string
Return those items of sequence for which function(item) is true. If
function is None, return the items that are true. If sequence is a tuple
or string, return the same type, else return a list.
Try this one:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
This method works well in Python 3
Here's some possible ways to achieve this task:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)
PS2: If you're interested on performance you could measure these attempts with a simple code like:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
In my box you'd get:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
So it seems attempt4 is the fastest one for this particular input.
Here's my Python 2/3 compatible version. Since the translate api has changed.
def remove(str_, chars):
"""Removes each char in `chars` from `str_`.
Args:
str_: String to remove characters from
chars: String of to-be removed characters
Returns:
A copy of str_ with `chars` removed
Example:
remove("What?!?: darn;", " ?.!:;") => 'Whatdarn'
"""
try:
# Python2.x
return str_.translate(None, chars)
except TypeError:
# Python 3.x
table = {ord(char): None for char in chars}
return str_.translate(table)
#!/usr/bin/python
import re
strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr
You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.
*NB: works with Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!##$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.
The output:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:
words = "things"
removed = "%s%s" % (words[:3], words[-1:])
This will result in 'removed' holding the word 'this'.
Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d.
Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.
So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle.
If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.
Examples:
words = "control"
removed = "%s%s" % (words[:2], words[-2:])
removed equals 'cool'.
words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
removed equals 'macs'.
In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position).
Remember, Python starts counting at 0, so you will need to as well.
In Python 3.5
e.g.,
os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
To remove all the number from the string
How about this:
def text_cleanup(text):
new = ""
for i in text:
if i not in " ?.!/;:":
new += i
return new
Below one.. with out using regular expression concept..
ipstring ="text with symbols!##$^&*( ends here"
opstring=''
for i in ipstring:
if i.isalnum()==1 or i==' ':
opstring+=i
pass
print opstring
Recursive split:
s=string ; chars=chars to remove
def strip(s,chars):
if len(s)==1:
return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
example:
print(strip("Hello!","lo")) #He!
You could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.
import re
text = "This is absurd!"
text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
print(text)
Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.
# for each file on a directory, rename filename
file_list = os.listdir (r"D:\Dev\Python")
for file_name in file_list:
os.rename(file_name, re.sub(r'\d+','',file_name))
Even the below approach works
line = "a,b,c,d,e"
alpha = list(line)
while ',' in alpha:
alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
output: abcde
The string method replace does not modify the original string. It leaves the original alone and returns a modified copy.
What you want is something like: line = line.replace(char,'')
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
for char in s:
if ord(char) < 96 or ord(char) > 123:
s = s.replace(char, "")
It will remove all the characters beyond a....z even upper cases.

Python regular expression to check start string is present or not

I am trying to write a regular expression which should check start string of the line and count some strings present in the line.
Example:
File.txt
# Compute
[ checking
a = b
a
a=b>c=d
Iterate this file and ignore the line with below condition
My Condition:
(line.startswith("[") or line.startswith("#") or line.count("=") > 1 or '=' not in line)
I need to re write the above condition in regex.
Trying the below,
re.search("^#",line)
re.search("^/[",line)
How to write this regex checking line starts with "#" or "[" and other conditions
If you actually wish to use a singular regular expression, you can use the following pattern;
^[^#\[][^=]*?=[^=]*?$
Which will match everything that does not fit the logic you specified in your answer - and so will extract only things that don't fit the logic you provided, and so will ignore everything all lines with the conditions specified. This single pattern would save you mixing python logic with regular expressions, which may be more consistent.
Demo here
Explanation:
^ anchors to the start of the string
[^#\[] Makes sure there is not a [ or a # at the start of the line
[^=]*? lazily match any number of anything except an =
= match exactly one =
[^=]*? lazily match any number of anything except an =
$ end of string anchor.
You could use this, for example, with grep if you're running bash to extract all the matching lines, and so ignore all desired lines, or use a simple python script as follows;
import re
pattern = re.compile('^[^#[][^=]?=[^=]?$')
# For loop solution
with open('test.txt') as f:
for line in f:
if pattern.match(line):
print(line)
# Alternative one-line generator expression;
with open('test.txt') as f:
print('\n'.join((line for line in f if pattern.match(line))))
For your given output file, both will print out;
a = b
For the first set of startswith conditions you can use re.match:
if re.match(r'[\[#]', text):
...
For the second condition, you can use re.findall (if you want the count):
if len(re.findall('=', text)) != 1:
...
You can combine the two above with an and, like this:
if re.match(r'[\[#]', text) and len(re.findall('=', text)) != 1:
...

Extract variables using python regex

Input file contains following lines:
a=b*c;
d=a+2;
c=0;
b=a;
Now for each line I want to extract variables that has been used.For example, for line 1, the output should be [a,b,c].Currently I am doing as follows :
var=[a,b,c,d] # list of variables
for line in file_ptr :
if '=' in line :
temp=line.split('=') :
ans=list(temp[0])
if '+' in temp[1] :
# do something
elif '*' in temp[1] :
# do something
else :
# single variable as line 4 OR constant as line 3
Is it possible to do this using regex?
EDIT:
Expected output for above file :
[a,b,c]
[d,a]
[c]
[a,b]
I would use re.findall() with whatever pattern matches variable names in the example's programming language. Assuming a typical language, this might work for you:
import re
lines = '''a=b*c;
d=a+2;
c=0;
b=a;'''
for line in lines.splitlines():
print re.findall('[_a-z][_a-z0-9]*', line, re.I)
I'd use some shorter pattern for matching variable names:
import re
strs = ['a=b*c;', 'd=a+2;', 'c=0;', 'b=a;']
print([re.findall(r'[_a-z]\w*', x, re.I) for x in strs])
See the Python demo
Pattern matches:
[_a-z] - a _ or an ASCII letter (any upper or lowercase due to the case insensitive modifier use re.I)
\w* - 0 or more alphanumeric or underscore characters.
See the regex demo
If you want just the variables, then do this:
answer = []
for line in file_ptr :
temp = []
for char in line:
if char.isalpha():
temp.append(char)
answer.append(temp)
A word of caution though: this would work only with variables that are exactly 1 character in length. More details about isalpha() can be found here or here.
I'm not entirely sure what you're after, but you can do something like this:
re.split(r'[^\w]', line)
to give a list of the alphabetic characters in the line:
>>> re.split(r'[^\w]', 'a=b*c;')
['a', 'b', 'c', '']
This is how I did :
l=re.split(r'[^A-Za-z]', 'a=b*2;')
l=filter(None,l)

Using parentheses as delimiter in re or str.split() python

I am trying to split a string such as: add(ten)sub(one) into add(ten) sub(one).
I can't figure out how to match the close parentheses. I have used re.sub(r'\\)', '\\) ') and every variation of escaping the parentheses,I can think of. It is hard to tell in this font but I am trying to add a space between these commands so I can split it into a list later.
There's no need to escape ) in the replacement string, ) has a special a special meaning only in the regex pattern so it needs to be escaped there in order to match it in the string, but in normal string it can be used as is.
>>> strs = "add(ten)sub(one)"
>>> re.sub(r'\)(?=\S)',r') ', strs)
'add(ten) sub(one)'
As #StevenRumbalski pointed out in comments the above operation can be simply done using str.replace and str.rstrip:
>>> strs.replace(')',') ').strip()
'add(ten) sub(one)'
d = ')'
my_str = 'add(ten)sub(one)'
result = [t+d for t in my_str.split(d) if len(t) > 0]
result = ['add(ten)','sub(one)']
Create a list of all substrings
import re
a = 'add(ten)sub(one)'
print [ b for b in re.findall('(.+?\(.+?\))', a) ]
Output:
['add(ten)', 'sub(one)']

Python: Regex to find but not include an alphanumeric

Is there an regular expression to find, for example, ">ab" but do not include ">" in the result?
I want to replace some strings using re.sub, and I want to find strings starting with ">" without remove the ">".
You want a positive lookbehind assertion. See the docs.
r'(?<=>)ab'
It needs to be a fixed length expression, it can't be a variable number of characters. Basically, do
r'(?<=stringiwanttobebeforethematch)stringiwanttomatch'
So, an example:
import re
# replace 'ab' with 'e' if it has '>' before it
#here we've got '>ab' so we'll get '>ecd'
print re.sub(r'(?<=>)ab', 'e', '>abcd')
#here we've got 'ab' but no '>' so we'll get 'abcd'
print re.sub(r'(?<=>)ab', 'e', 'abcd')
You can use a back reference in sub:
import re
test = """
>word
>word2
don't replace
"""
print re.sub('(>).*', r'\1replace!', test)
Outputs:
>replace!
>replace!
don't replace
I believe this accomplishes what you actually want when you say "I want to replace some strings using re.sub, and I want to find strings starting with '>' without remove the '>'."
if you want to avoid using the re module you can also use the startswith() string method.
>>> foo = [ '>12', '>54', '34' ]
>>> for line in foo:
... if line.startswith('>'):
... line = line.strip('>')
... print line
...
12
54
34
>>>

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