[Tutor] Counting number of occurrence of each character in a string

[Alan Gauld]

On 02/09/2020 11:19, Manprit Singh wrote:

>>>> a = "ABRACADABRA"
>>>> d = {}
>>>> for i in a:
>             d[i] = d.get(i, 0) + 1
> 
> will result in
>>>> d
> {'A': 5, 'B': 2, 'R': 2, 'C': 1, 'D': 1}
> 
> Here keys are characters in the string a, and the values are the counts.
> 
> in an another way the same result can be achieved by replacing d[i] =
> d.get(i, 0) + 1 with  dict .update () method,as the code written below :
> 
> a = "ABRACADABRA"
>>>> d = {}
>>>> for i in a:
>             d.update({i : d.get(i, 0) + 1})
> 
> The result  again is
>>>> d
> {'A': 5, 'B': 2, 'R': 2, 'C': 1, 'D': 1}
> 
> So in this situation which way should be adopted, :

Since the get call appears in both solutions the second one does nothing
but add complexity. Our aim as programmers is to remove unnecessary
complexity therefore the first option is better.

However an arguably simpler solution exists using sets and the built
in methods:

>>> a = "ABRACADABRA"
>>> s = sorted(set(a))
>>> s
['A', 'B', 'C', 'D', 'R']
>>> d = {key:a.count(key) for key in s}
>>> d
{'A': 5, 'B': 2, 'C': 1, 'D': 1, 'R': 2}
>>>

Which could be done in a one-liner as:

d2 = {k:a.count(k) for k in sorted(set(a))}

Although personally I'd say that was overly complex for 1 line.

If you are not yet familiar with the generator style of dictionary
creation you could unwrap it to

d = {}
for ch in s:
   d[ch] = a.count(ch)

-- 
Alan G
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