Question

Python中每个字母加倍（或重复n次）的最有效方法是什么？

"abcd" -> "aabbccdd"

或

"abcd" -> "aaaabbbbccccdddd"

我有一个需要以这种方式进行变异的长字符串，当前的解决方案涉及一个每个字母都有n个连接的循环，我想这可能更有效。

Answer 1

使用str.join：

>>> strs = "abcd"
>>> "".join([x*2 for x in strs])
'aabbccdd'
>>> "".join([x*4 for x in strs])
'aaaabbbbccccdddd'

来自docs：

s = ""
for substring in list:
    s += substring

请改用s = "".join(list)。在构建大型字符串时，前者是一个非常常见和灾难性的错误。

Answer 2

因为您特别询问了效率：

# drewk's answer, optimized by using from_iterable instead of *
def double_chain(s):
    return ''.join(chain.from_iterable(zip(s, s)))

# Ashwini Chaudhary's answer
def double_mult(s):
    return ''.join([x*2 for x in s])

# Jon Clements' answer, but optimized to take the re.compile and *2 out of the loop.
r = re.compile('(.)')
def double_re(s):
    return r.sub(r'\1\1', s)

现在：

In [499]: %timeit double_chain('abcd')
1000000 loops, best of 3: 1.99 us per loop
In [500]: %timeit double_mult('abcd')
1000000 loops, best of 3: 1.25 us per loop
In [501]: %timeit double_re('abcd')
10000 loops, best of 3: 22.2 us per loop

因此，itertools方法比最简单的方法慢约60％，并且使用正则表达式仍然慢了一个数量级。

但像这样的小字符串可能无法代表较长的字符串，所以：

In [504]: %timeit double_chain('abcd' * 10000)
100 loops, best of 3: 4.92 ms per loop
In [505]: %timeit double_mult('abcd' * 10000)
100 loops, best of 3: 5.57 ms per loop
In [506]: %timeit double_re('abcd' * 10000)
10 loops, best of 3: 91.5 ms per loop

正如所料，itertools方法变得更好（现在胜过简单方法），并且随着字符串变长，正则表达式变得更糟。

所以，没有一种“最有效”的方式。如果你将数十亿的小弦加倍，那么Ashwini的答案是最好的。如果你将数百万的大字符串或成千上万的大字符串加倍，那么drewk是最好的。如果你不做任何事情......首先没有理由对此进行优化。

此外，通常的警告：这个测试是我的Mac上没有负载的64位CPython 3.3.0;不保证您的应用中的Python实现，版本和平台与您的真实数据一样。使用32位2.6的快速测试显示了类似的结果，但如果重要，您需要自己进行更现实和相关的测试。

Answer 3

您可以使用join，izip和chain：

>>> st='abcd'
>>> from itertools import chain,izip
>>> ''.join(chain(*izip(st,st)))
'aabbccdd'

虽然它比列表理解的可读性差，但优点是没有中间列表; izip和chain生成迭代器。

Answer 4

我会选择str.join，所以我会提供re替代选项：

>>> s = "abcd"
>>> import re
>>> re.sub('(.)', r'\1' * 2, s)
'aabbccdd'

Answer 5

每当问题是：“将字符串的每个字符映射到其他字符的最有效方法是什么”结果str.translate是最好的选择...对于足够大的字符串：

def double_translate(s):
    return s.translate({ord(x):2*x for x in set(s)})

Timings反对其他答案：

In [5]: %timeit double_chain('abcd')
The slowest run took 11.03 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 992 ns per loop

In [6]: %timeit double_chain('mult')
The slowest run took 13.61 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1 µs per loop

In [7]: %timeit double_mult('abcd')
The slowest run took 7.59 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 869 ns per loop

In [8]: %timeit double_re('abcd')
The slowest run took 8.63 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 9.4 µs per loop

In [9]: %timeit double_translate('abcd')
The slowest run took 5.80 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.78 µs per loop

In [10]: %%timeit t='abcd'*5000
    ...: double_chain(t)
    ...: 
1000 loops, best of 3: 1.66 ms per loop

In [11]: %%timeit t='abcd'*5000
    ...: double_mult(t)
    ...: 
100 loops, best of 3: 2.35 ms per loop

In [12]: %%timeit t='abcd'*5000
    ...: double_re(t)
    ...: 
10 loops, best of 3: 30 ms per loop

In [13]: %%timeit t='abcd'*5000
    ...: double_translate(t)
    ...: 
1000 loops, best of 3: 1.03 ms per loop

但请注意，此解决方案具有额外的优势，在某些情况下，您可能会避免重新构建要传递给translate的表，例如：

def double_translate_opt(s, table=None):
    if table is None:
        table = {ord(x):2*x for x in set(s)}
    return s.translate(table)

这样可以避免一些开销，使其更快：

In [19]: %%timeit t='abcd'; table={ord(x):2*x for x in t}
    ...: double_translate_opt(t, table)
    ...: 
The slowest run took 17.59 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 452 ns per loop

正如你可以看到的那样，只要您避免每次构建表格，它都是当前答案的两倍。对于长文本，构建表的成本以翻译速度偿还（在这些情况下使用set值得，以避免多次调用ord。

Answer 6

def double_letter(str):
    strReturn = ''
    for chr in str:
        strReturn += chr*n
    return strReturn

将字母串中的每个字母加倍

6 个答案: