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Regex Substitution

Create Date: March 31, 2019 at 04:04 PM         Tag: PYTHON         Author Name: Sun, Charles

The re.sub() tool (sub stands for substitution) evaluates a pattern and, for each valid match, it calls a method (or lambda). 
The method is called for all matches and can be used to modify strings in different ways. 
The re.sub() method returns the modified string as an output.

Learn more about .

Transformation of Strings

Code

import re

#Squaring numbers
def square(match):
    number = int(match.group(0))
    return str(number**2)

print re.sub(r"\d+", square, "1 2 3 4 5 6 7 8 9")

Output

1 4 9 16 25 36 49 64 81


Replacements in Strings

Code

import re

html = """
<head>
<title>HTML</title>
</head>
<object type="application/x-flash" 
  data="your-file.swf" 
  width="0" height="0">
  <!-- <param name="movie"  value="your-file.swf" /> -->
  <param name="quality" value="high"/>
</object>
"""

print re.sub("(<!--.*?-->)", "", html) #remove comment

Output

<head>
<title>HTML</title>
</head>
<object type="application/x-flash" 
  data="your-file.swf" 
  width="0" height="0">

  <param name="quality" value="high"/>
</object>

Task

You are given a text of  lines. The text contains && and || symbols. 
Your task is to modify those symbols to the following:

&& → and
|| → or

Both && and || should have a space " " on both sides.

Input Format

The first line contains the integer, . 
The next  lines each contain a line of the text.

Constraints

Neither && nor || occur in the start or end of each line.

Output Format

Output the modified text.

Sample Input

11
a = 1;
b = input();

if a + b > 0 && a - b < 0:
    start()
elif a*b > 10 || a/b < 1:
    stop()
print set(list(a)) | set(list(b)) 
#Note do not change &&& or ||| or & or |
#Only change those '&&' which have space on both sides.
#Only change those '|| which have space on both sides.

Sample Output

a = 1;
b = input();

if a + b > 0 and a - b < 0:
    start()
elif a*b > 10 or a/b < 1:
    stop()
print set(list(a)) | set(list(b)) 
#Note do not change &&& or ||| or & or |
#Only change those '&&' which have space on both sides.
#Only change those '|| which have space on both sides.

 

import re
for i in range(int(input())):
    print(re.sub(r'(?<= )(&&|\|\|)(?= )', lambda x: 'and' if x.group() == '&&' else 'or', input()))

 

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Re.start() & Re.end()

Create Date: March 31, 2019 at 03:57 PM         Tag: PYTHON         Author Name: Sun, Charles

start() & end()

These expressions return the indices of the start and end of the substring matched by the group.

Code

>>> import re
>>> m = re.search(r'\d+','1234')
>>> m.end()
4
>>> m.start()
0

Task 
You are given a string . 
Your task is to find the indices of the start and end of string  in .

Input Format

The first line contains the string . 
The second line contains the string .

Constraints

 

Output Format

Print the tuple in this format: (start _index, end _index). 
If no match is found, print (-1, -1).

Sample Input

aaadaa
aa

Sample Output

(0, 1)  
(1, 2)
(4, 5)
import re
s, k = input(), input()
pattern = re.compile(k)
m = pattern.search(s)
if not m: print("(-1, -1)")
while m:
    print("({0}, {1})".format(m.start(), m.end() - 1))
    m = pattern.search(s, m.start() + 1)

 

New Comment

Re.findall() & Re.finditer()

Create Date: March 31, 2019 at 03:50 PM         Tag: PYTHON         Author Name: Sun, Charles

re.findall()

The expression re.findall() returns all the non-overlapping matches of patterns in a string as a list of strings. 
Code

>>> import re
>>> re.findall(r'\w','http://www.hackerrank.com/')
['h', 't', 't', 'p', 'w', 'w', 'w', 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k', 'c', 'o', 'm']

re.finditer()

The expression re.finditer() returns an iterator yielding MatchObject instances over all non-overlapping matches for the repattern in the string. 
Code

>>> import re
>>> re.finditer(r'\w','http://www.hackerrank.com/')
<callable-iterator object at 0x0266C790>
>>> map(lambda x: x.group(),re.finditer(r'\w','http://www.hackerrank.com/'))
['h', 't', 't', 'p', 'w', 'w', 'w', 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k', 'c', 'o', 'm']

Task 
You are given a string . It consists of alphanumeric characters, spaces and symbols(+,-). 
Your task is to find all the substrings of  that contains  or more vowels. 
Also, these substrings must lie in between  consonants and should contain vowels only.

Note : 
Vowels are defined as: AEIOU and aeiou. 
Consonants are defined as: QWRTYPSDFGHJKLZXCVBNM and qwrtypsdfghjklzxcvbnm.

Input Format

A single line of input containing string .

Constraints

Output Format

Print the matched substrings in their order of occurrence on separate lines. 
If no match is found, print -1.

Sample Input

rabcdeefgyYhFjkIoomnpOeorteeeeet

Sample Output

ee
Ioo
Oeo
eeeee

Explanation

ee is located between consonant  and . 
Ioo is located between consonant  and . 
Oeo is located between consonant  and . 
eeeee is located between consonant  and .

method 1:

import re
v = "aeiou"
c = "qwrtypsdfghjklzxcvbnm"
m = re.findall(r"(?<=[%s])([%s]{2,})[%s]" % (c, v, c), input(), flags = re.I)
print("\n".join(m or ["-1"]))

(?=...)

Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.

(?<=...)

Matches if the current position in the string is preceded by a match for ... that ends at the current position. This is called a positive lookbehind assertion. (?<=abc)def will find a match in abcdef, since the lookbehind will back up 3 characters and check if the contained pattern matches.

method 2:

import re
S = input()

regex = r"[^AEIOUaeiou]?[AEIOUaeiou]{2,}[^AEIOUaeiou]{1}"
pattern = re.compile(regex)
matches = list(map(lambda x: x.group(), pattern.finditer(S))) 

if matches:
        for match in matches:
                vowels = r"[AEIOUaeiou]{2,}"
                result = re.compile(vowels).search(match).group() 
                print(result)
else:
        print('-1')

 

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