Google python class [string2.py] exercise F

Question

I've done the exercise and it works. But I want to know if there a smarter way to do it. Thanks.

# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back

My code:

def front_back(a, b):
    if len(a) % 2 == 0:
    a_front = a[0:len(a) / 2]
    else:
    a_front = a[0:(len(a) / 2) + 1]
    if len(a) % 2 == 0:
    a_back = a[len(a) / 2:]
    else:
    a_back = a[(len(a) / 2) + 1:]
    if len(b) % 2 == 0:
    b_front = b[0:len(b) / 2]
    else:
    b_front = b[0:(len(b) / 2) + 1]
    if len(b) % 2 == 0:
    b_back = b[len(b) / 2:]
    else:
    b_back = b[(len(b) / 2) + 1:]
    return a_front + b_front + a_back + b_back

This question would be appropriate at [codereview.se]. – Jeff Mercado Sep 07 '14 at 00:08 — Jeff Mercado, Sep 07 '14 at 00:08

Brian McCrory · Accepted Answer · 2014-09-07T00:11:27.267

2

You probably don't need all the if tests. Using substrings (as you use the slices in python), the following example in Java works.

public static String SplitMixCombine(String a, String b) {

    return a.substring(0, (a.length()+1)/2) + 
            b.substring(0, (b.length()+1)/2) + 
            a.substring((a.length()+1)/2, a.length()) + 
            b.substring((b.length()+1)/2, b.length()); 
}

Just got this to work in python:

>>> def mix_combine(a, b):
...     return a[0:int((len(a)+1)/2)] + b[0:int((len(b)+1)/2)] + a[int((len(a)+1)/2):] +b[int((len(b)+1)/2):]
...

>>> a = "abcd"
>>> b = "wxyz"
>>> mix_combine(a,b)
'abwxcdyz'
>>> a = "abcde"
>>> b = "vwxyz"
>>> mix_combine(a,b)
'abcvwxdeyz'
>>>

edited Sep 07 '14 at 00:11

answered Sep 06 '14 at 23:50

Brian McCrory

91
4

1

Thank you so much! I didn't had noticed that a number pair of digits + 1 divided by 2 is the same as a number odd of digits + 1 divided by 2 (in integer values). – Matheus Martins Sep 07 '14 at 00:36
it would be a lot cleaner code if you did not ram it all into one line – Padraic Cunningham Sep 07 '14 at 00:41

score 1 · Answer 2 · answered Sep 07 '14 at 07:17

This is a lot cleaner code.

def front_back(a, b):
    print a, b
    a_indx = int((len(a)+1)/2)
    b_indx = int((len(b)+1)/2)

    print a[:a_indx] + b[:b_indx] + a[a_indx:] +b[b_indx:]
    print "\n"

front_back("ab", "cd")
front_back("abc", "de")
front_back("ab", "cde")
front_back("abc", "def")

Hope this helps !

score 0 · Answer 3 · edited May 23 '17 at 11:51

0

I wrote another one-liner for this:

def front_back(a, b):
  return a[:len(a) / 2 + len(a) % 2] + b[:len(b) / 2 + len(b) % 2] + a[-(len(a) / 2):] + b[-(len(b) / 2):]

An interesting thing to note is that in Python 5 / 2 yields 2, but -5 / 2 yields -3, you might expect to get -2. That's why I added the parenthesis. See Negative integer division surprising result

edited May 23 '17 at 11:51

Community

1
1

answered Nov 14 '14 at 11:27

Pawel Krakowiak

9,940
3
37
55

score -1 · Answer 4 · edited Jan 07 '16 at 12:49

-1

def front_back(a, b):
return front_string(a)+front_string(b)+back_string(a)+back_string(b)

def front_string(s):
return  s[:int((len(s)+1)/2)]  

def back_string(s):
return  s[int((len(s)+1)/2):] 

# int((len(s)+1)/2) returns the correct index for both odd and even length strings

edited Jan 07 '16 at 12:49

Nazik

8,696
27
77
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answered Jan 07 '16 at 12:25

Burnie

1

Google python class [string2.py] exercise F

4 Answers4