Same Python code returns different results for same input string

Question

Below code is supposed to return the most common letter in the TEXT string in the format:

always lowercase
ignoring punctuation and spaces
in the case of words such as "One" - where there is no 2 letters the same - return the first letter in the alphabet

Each time I run the code using the same string, e.g. "One" the result cycles through the letters...weirdly though, only from the third try (in this "One" example).

text=input('Insert String: ')
def mwl(text):
    from string import punctuation
    from collections import Counter
    for l in punctuation:
        if l in text:
            text = text.replace(l,'')
    text = text.lower()
    text=''.join(text.split())
    text= sorted(text)
    collist=Counter(text).most_common(1)
    print(collist[0][0])
mwl(text)

Can you show a sample input? along with the expected output and the result that you actually get when you run this code. — Pankaj Parashar
– Pankaj Parashar, Commented Dec 31, 2014 at 13:19
I don't think the values in Counter are sorted, even if you give it sorted input. — Kevin
– Kevin, Commented Dec 31, 2014 at 13:19
@Pankaj e.g. "One" returns "o", then "o" again, then "e" and then "n". Should always be "e". Everyone is correct on Counter using dictionary, how should I sort then? — Dom13
– Dom13, Commented Dec 31, 2014 at 13:31

fredtantini · Accepted Answer · 2014-12-31 13:19:59Z

10

Counter uses a dictionary:

>>> Counter('one')
Counter({'e': 1, 'o': 1, 'n': 1})

Dictionaries are not ordered, hence the behavior.

answered Dec 31, 2014 at 13:19

fredtantini

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Comments

Irshad Bhat · Accepted Answer · 2014-12-31 15:30:57Z

3

You can get the desired output with OrderedDict replacing the below two lines:

text= sorted(text)
collist=Counter(text).most_common(1)

with:

collist = OrderedDict([(i,text.count(i)) for i in text])
collist = sorted(collist.items(), key=lambda x:x[1], reverse=True)

You also need to import OrderedDict for this.

Demo:

>>> from collections import Counter, OrderedDict
>>> text = 'One'
>>> collist = OrderedDict([(i,text.count(i)) for i in text])
>>> print(sorted(collist.items(), key=lambda x:x[1], reverse=True)[0][0])
O  
>>> print(sorted(collist.items(), key=lambda x:x[1], reverse=True)[0][0])
O    # it will always return O
>>> text = 'hello'
>>> collist = OrderedDict([(i,text.count(i)) for i in text])
>>> print(sorted(collist.items(), key=lambda x:x[1], reverse=True)[0][0])
l    # l returned because it is most frequent

edited Dec 31, 2014 at 15:30

answered Dec 31, 2014 at 13:36

Irshad Bhat

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1 Comment

Dom13 Over a year ago

Txs used your suggestion, did some research, and modified between key sort and value sort depending on my need - it worked. The best and easiest solution, which worked was from @xnx. txs

xnx · Accepted Answer · 2014-12-31 14:42:57Z

1

This can also be done without Counter or OrderedDict:

In [1]: s = 'Find the most common letter in THIS sentence!'
In [2]: letters = [letter.lower() for letter in s if letter.isalpha()]
In [3]: max(set(letters), key=letters.count)
Out[3]: 'e'

answered Dec 31, 2014 at 14:42

xnx

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