将两个列表连接到格式化字符串中的最聪明方法

Question

可以说我有两个长度相同的列表：

a = ['a1', 'a2', 'a3']
b = ['b1', 'b2', 'b3']

我想产生以下字符串：

c = 'a1=b1, a2=b2, a3=b3'

实现此目标的最佳方法是什么？

我有以下实现：

import timeit

a = [str(f) for f in range(500)]
b = [str(f) for f in range(500)]

def func1():
    return ', '.join([aa+'='+bb for aa in a for bb in b if a.index(aa) == b.index(bb)])

def func2():
    list = []
    for i in range(len(a)):
        list.append('%s=%s' % (a[i], b[i]))
    return ', '.join(list)

t = timeit.Timer(setup='from __main__ import func1', stmt='func1()')
print 'func1 = ' + t.timeit(10)

t = timeit.Timer(setup='from __main__ import func2', stmt='func2()')
print 'func2 = ' + t.timeit(10)

输出为：

func1 = 32.4704790115
func2 = 0.00529003143311

你有权衡吗？

Answer 1

a = ['a1', 'a2', 'a3']
b = ['b1', 'b2', 'b3']

pat = '%s=%%s, %s=%%s, %s=%%s'

print pat % tuple(a) % tuple(b)

给 a1=b1, a2=b2, a3=b3

。

然后：

from timeit import Timer
from itertools import izip

n = 300

a = [str(f) for f in range(n)]
b = [str(f) for f in range(n)]

def func1():
    return ', '.join([aa+'='+bb for aa in a for bb in b if a.index(aa) == b.index(bb)])

def func2():
    list = []
    for i in range(len(a)):
        list.append('%s=%s' % (a[i], b[i]))
    return ', '.join(list)

def func3():
    return ', '.join('%s=%s' % t for t in zip(a, b))

def func4():
    return ', '.join('%s=%s' % t for t in izip(a, b))

def func5():
    pat = n * '%s=%%s, '
    return pat % tuple(a) % tuple(b)

d = dict(zip((1,2,3,4,5),('heavy','append','zip','izip','% formatting')))
for i in xrange(1,6):
    t = Timer(setup='from __main__ import func%d'%i, stmt='func%d()'%i)
    print 'func%d = %s  %s' % (i,t.timeit(10),d[i])

结果

func1 = 16.2272833558  heavy
func2 = 0.00410247671143  append
func3 = 0.00349569568199  zip
func4 = 0.00301686387516  izip
func5 = 0.00157338432678  % formatting