CODEWITHSUNDEEP
pythonstring
johndoe12345
johndoe12345

Reputation: 421

Python String function

Looking to put together a simple python function that checks the following

Given two strings, return True if either of the strings appears at the very end of the other string, ignoring upper/lower case differences (in other words, the computation should not be "case sensitive").

end_other('Hiabc', 'abc') → True
end_other('AbC', 'HiaBc') → True
end_other('abc', 'abXabc') → True

Upvotes: 1

Views: 1079

Answers (6)

TigerhawkT3
TigerhawkT3

Reputation: 49318

You can use OOP if you want, subclassing str and writing an appropriate method:

>>> class String(str):
...     def isend(self, other):
...             self = self.lower()
...             other = other.lower()
...             return self.endswith(other) or other.endswith(self)
...
>>> a = String('HiAbC')
>>> b = String('Bc')
>>> a.isend(b)
True
>>> b.isend(a)
True
>>> c = String('Hello')
>>> d = String('el')
>>> c.isend(d)
False
>>> d.isend(c)
False

Upvotes: 0

Laszlowaty
Laszlowaty

Reputation: 1323

You could use slicing.

>>> s1 = "Hiabc"
>>> s2 = "abc"
>>> l = len(s2)
>>> s2 == s1[-l:]
True
>>> s2 = "aBc"
>>> s2 == s1[-l:]
False

You can read more about string and slicing here.

Slicing means to get part of the string based on passed index. General construction look like this: string[starting_index:ending_index].

Let's take a look at this:

string1 = "doggy" string1[0:1] -> "d" string1[0:2] -> "do" string1[0:] = string[:] = "doggy" (from zero to end)

You can also pass index lesser than 0:

string1[-1:] (from -1 to end) -> "y" So in the code above calling s1[-l:] means starting position is the length of s2 sentece. In this case s1[-l:] = 'abc' = s1[-3:]. Because lenght of s2 is 3.

Upvotes: 0

AChampion
AChampion

Reputation: 30258

Here's another approach using zip, which avoids having to do multiple passes of endswith(), re.search() or slicing. It iterates the 2 strings in reverse and returns True if all the letters are equal up to the exhaustion of one of the strings:

def end_other(s1, s2):
    return all(a == b for a, b in zip(reversed(s1.lower()), reversed(s2.lower())))

In Python2 you can use itertools.izip() for a marginal space improvement:

>>> end_other('Hiabc', 'abc')
True
>>> end_other('AbC', 'HiaBc')
True
>>> end_other('abc', 'abXabc')
True
>>> end_other('cbc', 'abXabc')
False

Upvotes: 1

Luis Ávila
Luis Ávila

Reputation: 699

Or if you need to "create" your own function:

def check(str1, str2):
    str1 = str1.lower()
    str2 = str2.lower()
    check = True

    # If string 1 is bigger then string 2:
    if( len(str1) > len(str2) ):
        # For each character of string 2
        for i in range(len(str2)):
            # Compare the character i of string 2 with the character at the end of string 1, keeping the order
            if str2[i] != str1[-(len(str2)-i)]:
            check = False

    # If string 2 is bigger then string 1:
    else:
        # For each character of string 1
        for i in range(len(str1)):
            # Compare the character i of string 1 with the character at the end of string 2, keeping the order
            if str1[i] != str2[-(len(str1)-i)]:
                check = False
    return check

So, basically, if string1 = "ABCD" and string2 = "CD", it will check character 0 of string2 with 2 of string1 and 1 of string2 with 3 of string1

Upvotes: 1

Pruthvi Raj
Pruthvi Raj

Reputation: 3036

You can use regex

def end_other(s1,s2):
    return bool(re.search(s1+'$',s2,re.I)) or bool(re.search(s2+'$',s1,re.I))

Upvotes: 1

Pynchia
Pynchia

Reputation: 11590

Try with

def end_other(s1, s2):
    s1 = s1.lower()
    s2 = s2.lower()
    return s1.endswith(s2) or s2.endswith(s1)

Upvotes: 4

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