A reason why I love Pythonic way - Series (Part1)

Problem Statement - 1

Lets say we have to validate whether a string has at least one Alphanumeric or Alphabetical or Digit or Lower Case or Upper Case character in it.  

Pythonic Solution

Like many other programming language , python has also string methods such as  isalnum,  isalpha.  isdigit,  islower, isupper. So to solve the problem with these methods , we need to check each characters of the string against each method. That needs many lines of codes. However, using getattr, any method & list comprehension , we can solve the problem in just three lines of code:

s = input()
my_methods = ['isalnum', 'isalpha', 'isdigit', 'islower', 'isupper']
for i in my_methods:
    print(any([getattr(char, i)() for char in s]))

Is not that beautiful code ?  do let me know in comments if you find it hard to understand. 

Another pythonic approach is  to solve above problem could be by using for-else statements instead of any something like this:

s = input()
my_methods = ['isalnum', 'isalpha', 'isdigit', 'islower', 'isupper']
for i in my_methods:
    for char in s:
        obj = getattr(char, i)
        if obj():
            print(obj())
            break
    else:
        print(False)

Problem Statement - 2

Given the a List of Integers and value of 'k', find the most common 'k' values in the list.  Lets consider the following as example:

input_arr = [1, 3, 4, 5, 6, 7, 8, 4, 4, 8, 10, 20, 17, 15, 1, 3, 5]

k = 5

# Solution - 1 : Using Naive Approach 

def k_most_frequent(k, input_arr):
    """ Given an Array find the Most K-frequent Numbers"""
    results = []
    mapdict = {}
    my_bucket = [0] * len(input_arr)
    for i in input_arr:
        if i in mapdict:
            mapdict[i] += 1
        else:
            mapdict[i] = 1
    for key in mapdict:
        value = mapdict[key]
        if my_bucket[value] == 0:
            my_bucket[value] = [key]
        else:
            my_bucket[value].append(key)
    i = len(my_bucket) - 1
    while i >= 0:
        if my_bucket[i] != 0:
            if len(my_bucket[i]) == k:
                results.extend(my_bucket[i])
                break
            elif len(my_bucket[i]) < k:
                results.extend(my_bucket[i])
                k -= len(my_bucket[i])
                i -= 1
            else:
                results.extend(my_bucket[i][0:k])
                break
        else:
            i -= 1
    return results


print(k_most_frequent(k, input_arr))

# Solution - 2 : Using Python Sort by Values on Dictionary Approach 

from collections import defaultdict

my_dict=defaultdict(int)
for num in input_arr:
    my_dict[num] += 1
my_dict = dict(sorted(my_dict.items(),reverse=True, key= lambda x: x[1]))
print(list(my_dict.keys())[:k] )

# Solution - 3 : Using Python Counter module from collection 

from collections import Counter
Counter = Counter(input_arr)
print(list(dict(Counter.most_common(k)).keys()))

So this is the most pythonic solution and only two lines of code solve the problem stated.

Is not that beautiful code ?  do let me know in comments if you find it hard to understand. 

Happy Python-ing  ✌