CODEWITHSUNDEEP
pythonpython-3.x
Highstaker
Highstaker

Reputation: 1085

Getting the last half of string written on backwards

I have a certain string, or several strings. For example:

a = '12345678'
b = '123456789'

I'm using Python 3.2. I'm trying to get the last half of a string written on backwards. If a string has an odd amount of characters, the middle character is discarded. So, what I'm trying to achieve is:

a_out = '8765'
b_out = '9876'

The way I do it is the following:

a_back = a[::-1]
a_out = a_back[0:len(a_back)//2]

The question is: is there a shorter way to do this? Can it be done in one operation instead of two?

Upvotes: 2

Views: 787

Answers (3)

Saullo G. P. Castro
Saullo G. P. Castro

Reputation: 58985

You can just make it a one-liner:

a_out = a[::-1][0:len(a)//2]

It would be nice if you could do fancy indexing in a string, like we do in a numpy array, then another solution would be:

a_out = a[ range(len(a)-1, len(a)//2-1, -1) ] # NOT POSSIBLE

@jamylak suggested to use 'itemgetter' to accomplish an equivalent for fancy indexing in an efficient way:

from operator import itemgetter
items = [0,3,2,1,0,0,2,3,]
a_out = ''.join( itemgetter( *items )(a) )
#14321134

Upvotes: 1

Inbar Rose
Inbar Rose

Reputation: 43477

Using a method:

a = '12345678'
b = '123456789'

def method(x):  
    return x[:-(len(x)+1)//2:-1]

print method(a)
print method(b)

>>> 
8765
9876

Upvotes: 1

Nolen Royalty
Nolen Royalty

Reputation: 18653

You can calculate the index and reverse the string at the same time:

>>> a[:-(len(a)+1)//2:-1]
'8765'
>>> b[:-(len(b)+1)//2:-1]
'9876'

Upvotes: 6

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