How To Check If Characters In A String Are Alphabetically Ordered

I have been trying these code but there is something wrong. I simply want to know if the first string is alphabetical. def alp(s1): s2=sorted(s1) if s2 is s1: retur

Solution 1:

is is identity testing which compares the object IDs, == is the equality testing:

In [1]: s1 = "Hello World"
In [2]: s2 = "Hello World"

In [3]: s1 == s2
Out[3]: True

In [4]: s1 is s2
Out[4]: False

Also note that sorted returns a list, so change it to:

if ''.join(s2) == s1:

Or

if ''.join(sorted(s2)) == s1:

Solution 2:

You could see this answer and use something which works for any sequence:

all(s1[i] <= s1[i+1] for i in xrange(len(s1) - 1))

Example:

>>> def alp(s1):
...     return all(s1[i] <= s1[i+1] for i in xrange(len(s1) - 1))
...
>>> alp("test")
False
>>> alp("abcd")
True

Solution 3:

I would do it using iter to nicely get the previous element:

def is_ordered(ss):
   ss_iterable = iter(ss)
   try:
       current_item = next(ss_iterable)
   except StopIteration:
       #replace next line to handle the empty string case as desired.
       #This is how *I* would do it, but others would prefer `return True`
       #as indicated in the comments :)
       #I suppose the question is "Is an empty sequence ordered or not?"
       raise ValueError("Undefined result.  Cannot accept empty iterable")

   for next_item in ss_iterable:
       if next_item < current_item:
           return False
       current_item = next_item
   return True

This answer has complexity O(n) in the absolute worst case as opposed to the answers which rely on sort which is O(nlogn).

Solution 4:

Make sure that you are comparing strings with strings:

In [8]: s = 'abcdef'

In [9]: s == ''.join(sorted(s))
Out[9]: True

In [10]: s2 = 'zxyw'

In [11]: s2 == ''.join(sorted(s2))
Out[11]: False

If s1 or s2 is a string, sorted will return a list, and you will then be comparing a string to a list. In order to do the comparison you want, using ''.join() will take the list and join all the elements together, essentially creating a string representing the sorted elements.

Solution 5:

use something like this:

sorted() returns a list and you're trying to compare a list to a string, so change that list to a string first:

In [21]: "abcd"=="".join(sorted("abcd"))
Out[21]: True

In [22]: "qwerty"=="".join(sorted("qwerty"))
Out[22]: False

#comparsion of list and a string is False
In [25]: "abcd"==sorted("abcd")
Out[25]: False