【Pythonノート】join関数の使い方

Pythonでは、関数join()を使用して文字列を接続します.

>>>li = ['my','name','is','bob'] 
>>>' '.join(li) 
'my name is bob' 
 
>>>'_'.join(li) 
'my_name_is_bob' 
 
>>> s = ['my','name','is','bob'] 
>>> ' '.join(s) 
'my name is bob' 
 
>>> '..'.join(s) 
'my..name..is..bob' 

以上王偉のブログから抜粋:http://wangwei007.blog.51cto.com/68019/1100587
また、http://www.codewars.com上のKataには、次のような問題があります.

Description:

Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer. Square all numbers k (0 <= k <= n) between 0 and n. Count the numbers of digits d used in the writing of all the k**2. Call nb_dig (or nbDig or ...) the function taking n and d as parameters and returning this count.

Examples:

n = 10, d = 1, the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in 1, 16, 81, 100. The total count is then 4.

nb_dig(25, 1):
the numbers of interest are
1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared are 1, 16, 81, 100, 121, 144, 169, 196, 361, 441
so there are 11 digits `1` for the squares of numbers between 0 and 25.
Note that 121 has twice the digit 1.

Solution of 
‘zebulan’:

def nb_dig(n, d):
    return ''.join(str(a ** 2) for a in xrange(n + 1)).count(str(d))

の中で、joinとgeneratorの優位性を柔軟に使って、1行のコードだけで私の何行のコードの量を解決して、very clever~