CodeWars練習問題ノート(一)

IQ Test

Introduction
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given numbers differs from the others . Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given numbers finds one that is different in evenness, and return a position of this number.
! Keep in mind that your task is to help Bob solve a real IQ test , which means indexes of the elements start from 1 (not 0)
Examples :

iq_test("2 4 7 8 10") => 3 // Third number is odd, while the rest of the numbers are even

iq_test("1 2 1 1") => 2 // Second number is even, while the rest of the numbers are odd

My submit

def iq_test(numbers):
    ns = [int(x) for x in numbers.split(' ')]
    odd_count = 0
    even_count = 0
    for i in range(0,3):
        if ns[i] %2 == 0:
            even_count += 1
        else:
            odd_count += 1
    standard = int(odd_count > even_count)
    index = 0
    for i in range(0, len(ns)):
        if ns[i] % 2 is not standard:
            index = i + 1
    return index   

Best Practices

def iq_test(numbers):
    e = [int(i) % 2 == 0 for i in numbers.split()]

    return e.index(True) + 1 if e.count(True) == 1 else e.index(False) + 1

Summary
パラメータを扱うときはTrue(双数)、False(単数)の[Bool]と処理し、柔軟に使えるlist methodですが、countを使うと性能に問題があるかもしれません

Replace With Alphabet Position

Introduction
Welcome. In this kata you are required to, given a string, replace every letter with its position in the alphabet. If anything in the text isn't a letter, ignore it and don't return it. a being 1, b being 2, etc. As an example:

alphabet_position("The sunset sets at twelve o' clock.")

Should return "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11" (As a string.)
My Submit

def alphabet_position(text):
    letters = []
    for s in text:
        c = ord(s)
        if c >= 65 and c <= 90:
            c -= 64
        elif c >= 97 and c <= 122:
            c -= 96
        else:
            continue
        letters.append(str(c))
    return ' '.join(letters)

Best Practices

def alphabet_position(text):
    return ' '.join(str(ord(c) - 96) for c in text.lower() if c.isalpha())

Summary
問題を間違えてアルファベット以外の文字を残すために、比較的便利な書き方を犠牲にし、最適な答えから一歩しか離れていない.lower()は巧みに使われています

Count the number of Duplicates

Introduction
Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.
Example "abcde"-> 0 # no characters repeats more than once "aabbcde"-> 2 # 'a' and 'b' "aabBcde"-> 2 # 'a' occurs twice and 'b' twice (bandB) "indivisibility"-> 1 # 'i' occurs six times "Indivisibilities"-> 2 # 'i' occurs seven times and 's' occurs twice "aA11"-> 2 # 'a' and '1' "ABBA"-> 2 # 'A' and 'B' each occur twice
My Submit

def duplicate_count(text):
    letters = [c for c in text.lower()]
    exists = []
    count = 0
    for c in letters:
        if c not in exists and letters.count(c) > 1:
            count += 1
            exists.append(c)
    return count

Best Practices

def duplicate_count(s):
  return len([c for c in set(s.lower()) if s.lower().count(c)>1])

Summary
簡潔に見えますが、実際には良い答えではありませんが、「いいね」をお勧めするのは特に高い案ではありません.

def duplicate_count(text):
    seen = set()
    dupes = set()
    for char in text:
        char = char.lower()
        if char in seen:
            dupes.add(char)
        seen.add(char)
    return len(dupes)