a = range(100000)
%timeit -n 10 sum(a)
10 loops, best of 3: 3.15 ms per loop
%%timeit
  ...: s = 0
  ...: for i in a:
  ...:  s += i
  ...:
100 loops, best of 3: 6.93 ms per loop

%timeit -n 1000 s = a + b + c + d + e + f + g + h + i + j + k # 数据量较小时直接累加更快
1000 loops, best of 3: 571 ns per loop
%timeit -n 1000 s = sum([a,b,c,d,e,f,g,h,i,j,k]) # 小数据量调用 sum 函数，空间效率降低
1000 loops, best of 3: 669 ns per loop

for lst in [(1, 2, 3), (4, 5, 6)]: # lst 索引需要额外开销
  pass

for a, b, c in [(1, 2, 3), (4, 5, 6)]: # better
  pass

def force():
 lst = range(4)
 for a1 in [1, 2]:
   for a2 in lst:
     for a3 in lst:
       for b1 in lst:
         for b2 in lst:
           for b3 in lst:
             for c1 in lst:
               for c2 in lst:
                 for c3 in lst:
                   for d1 in lst:
                     yield (a1, a2, a3, b1, b2, b3, c1, c2, c3, d1)
                      
%%timeit -n 10
for t in force():
  sum([t[0], t[1], t[2], t[3], t[4], t[5], t[6], t[7], t[8], t[9]])
10 loops, best of 3: 465 ms per loop
%%timeit -n 10
for a1, a2, a3, b1, b2, b3, c1, c2, c3, d1 in force():
  sum([a1, a2, a3, b1, b2, b3, c1, c2, c3, d1])
10 loops, best of 3: 360 ms per loop

def force(start, end): # 用于密码暴力破解程序
  for i in range(start, end):
    now = i
    sublst = []
    for j in range(10):
      sublst.append(i % 10) # 除法运算开销较大，比乘法大
      i //= 10
    sublst.reverse()
    yield(tuple(sublst), now)

def force(): # better
 lst = range(5)
 for a1 in [1]:
   for a2 in lst:
     for a3 in lst:
       for b1 in lst:
         for b2 in lst:
           for b3 in lst:
             for c1 in lst:
               for c2 in lst:
                 for c3 in lst:
                   for d1 in lst:
                     yield (a1, a2, a3, b1, b2, b3, c1, c2, c3, d1)
　　

r0 = [1, 2] # 可读性与灵活性
r1 = range(10)
r2 = r3 = r4 = r5 = r6 = r7 = r8 = r9 = r1
force = ((a0, a1, a2, a3, a4, a5, a6, a7, a8, a9)
      for a0 in r0 for a1 in r1 for a2 in r2 for a3 in r3 for a4 in r4
      for a5 in r5 for a6 in r6 for a7 in r7 for a8 in r8 for a9 in r9)

def isprime(n):
  if n & 1 == 0:
    return False
  k, q = find_kq(n)
  a = randint(1, n - 1)
  if pow(a, q, n) == 1: # 比使用 a ** q % n 运算优化数倍
    return True
  for j in range(k):
    if pow(a, pow(2, j) * q, n) == n - 1: # a **((2 ** j) * q) % n
      return True
  return False

In [1]: from random import getrandbits
 
In [2]: x = getrandbits(4096)
 
In [3]: y = getrandbits(2048)
 
In [4]: %timeit -n 10000 q, r = divmod(x, y)
10000 loops, best of 3: 10.7 us per loop
 
In [5]: %timeit -n 10000 q, r = x//y, x % y
10000 loops, best of 3: 21.2 us per loop

In [23]: import copy
In [24]: %timeit -n 10 copy.copy(a)
10 loops, best of 3: 606 ns per loop
In [25]: %timeit -n 10 copy.deepcopy(a)
10 loops, best of 3: 1.17 us per loop

>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]

In [1]: r = range(10**7)
In [2]: s = set(r) # 占用 588MB 内存
In [3]: d = dict((i, 1) for i in r) # 占用 716MB 内存
In [4]: %timeit -n 10000 (10**7) - 1 in r
10000 loops, best of 3: 291 ns per loop
In [5]: %timeit -n 10000 (10**7) - 1 in s
10000 loops, best of 3: 121 ns per loop
In [6]: %timeit -n 10000 (10**7) - 1 in d
10000 loops, best of 3: 111 ns per loop

In [1]: %timeit -n 10 a = (i for i in range(10**7)) # 生成器通常遍历更高效
10 loops, best of 3: 933 ns per loop
In [2]: %timeit -n 10 a = [i for i in range(10**7)]
10 loops, best of 3: 916 ms per loop
In [1]: %timeit -n 10 for x in (i for i in range(10**7)): pass
10 loops, best of 3: 749 ms per loop
In [2]: %timeit -n 10 for x in [i for i in range(10**7)]: pass
10 loops, best of 3: 1.05 s per loop