Question

Apologies for the somewhat vague title, I’ll try explain more here.

Currently, I have the following code, which counts the number of times the values “y” and “n” show up in the list called “results”.

NumberOfA = results.count("y")
NumberOfB = results.count("n")

Is there a way of making it so that, for example, values such as “yes” are also counted towards NumberOfA? I was thinking something along the lines of the following:

NumberOfA = results.count("y" and "yes" and "Yes")
NumberOfB = results.count("n" and "no" and "No")

But that doesn’t work. This is probably a pretty easy one to solve, but hey. Thanks in advance!

Answer 1

As for why your answer above does not work, it is because Python will just take the final value of the expression you pass in:

>>> 'Yes' and 'y' and 'yes'
'yes'

Therefore your count will be off because it is just looking for that final value:

>>> results.count('yes' and 'y')
1
>>> results.count('yes' and '???')
0

Would something like this work? Note that this depends on their being only yes/no-esque answers in the list (will be wrong if things like ‘Yeah….um no’ are in there):

In [1]: results = ['yes', 'y', 'Yes', 'no', 'NO', 'n']

In [2]: yes = sum(1 for x in results if x.lower().startswith('y'))

In [3]: no = sum(1 for x in results if x.lower().startswith('n'))

In [4]: print yes, no
3 3

The general idea is to take your results list and then iterate through each item, lowercasing it and then taking the first letter (startswith) – if the letter is a y, we know it is a yes; otherwise, it will be no.

You can also combine the steps above if you want by doing something like this (note this requires Python 2.7):

>>> from collections import Counter
>>> results = ['yes', 'y', 'Yes', 'no', 'NO', 'n']
>>> Counter((x.lower()[0] for x in results))
Counter({'y': 3, 'n': 3})

Counter objects can be treated just like dictionaries, so you would now essentially have a dictionary that contained the counts of yes‘s and no‘s.