Question

I really have no idea how to go about this. So far all I have is this:

alphabet = 'abcdefghijklmnopqrstuvwxyz'

I am supposed to go through a text file and count the number of words in alphabetical order.
I can’t do

if word in alphabet:

because that requires the word to actually have all of those letters in order.

To clarify: ‘blow’ would pass the test and ‘suck’ would not.
I have a list of thousands of words, I need to go through the entire list and count the numbers of words that are in alphabetical order. Sorry for any prior confusion.

Answer 1

Very simple:

>>> alphabet = 'abcdefghijklmnopqrstuvwxyz'
>>> list(alphabet) == sorted(alphabet)
True
>>> list('blow') == sorted('blow')
True
>>> list('suck') == sorted('suck')
False

So know we can define the predicate we need:

>>> alphabetical = lambda w: list(w.lower()) == sorted(w.lower())

And apply it to a list:

>>> lst = ['blow', 'suck', 'abc']
>>> filter(alphabetical, lst)
['blow', 'abc']

From there it’s not a large step to counting the results 🙂 There’s some other things to consider:

• The sorting is O(n*log n), although this problem could easily be solved in O(n). This is probably okay because words usually have a bounded number of characters and sorted is implemented in C and thus very fast
• If you really need efficiency, you can even use the sum(1 for w in w if ...) trick, which uses a generator expression instead of building a list.