在Python中，连接字符串的首选方法是什么？

Question

由于Python的string无法更改，我想知道如何更有效地连接字符串？

我可以这样写：

s += stringfromelsewhere

或者像这样：

s = []
s.append(somestring)

later

s = ''.join(s)

在写这个问题的时候，我发现一篇关于这个话题的好文章。

http://www.skymind.com/~ocrow/python_string/

但这是在Python2.x中，所以问题是Python3中有什么变化吗？

Answer 1

如果要连接许多值，则两者都不需要。附加一份清单是很贵的。你可以用StringIO来做。尤其是当你在大量的操作中积累起来的时候。

from cStringIO import StringIO
# python3:  from io import StringIO

buf = StringIO()

buf.write('foo')
buf.write('foo')
buf.write('foo')

buf.getvalue()
# 'foofoofoo'

如果您已经从其他操作返回了完整的列表，那么只需使用''.join(aList)

来自python常见问题解答：What is the most efficient way to concatenate many strings together?

str and bytes objects are immutable, therefore concatenating many strings together is inefficient as each concatenation creates a new object. In the general case, the total runtime cost is quadratic in the total string length.

To accumulate many str objects, the recommended idiom is to place them into a list and call str.join() at the end:

chunks = []
for s in my_strings:
    chunks.append(s)
result = ''.join(chunks)

(another reasonably efficient idiom is to use io.StringIO)

To accumulate many bytes objects, the recommended idiom is to extend a bytearray object using in-place concatenation (the += operator):

result = bytearray()
for b in my_bytes_objects:
    result += b

---

编辑：我很傻，结果被向后粘贴，使其看起来像是添加到列表比cStringIO快。我还添加了bytearray/str concat的测试，以及使用带有更大字符串的更大列表的第二轮测试。（python 2.7.3）

ipython大型字符串列表测试示例

try:
    from cStringIO import StringIO
except:
    from io import StringIO

source = ['foo']*1000

%%timeit buf = StringIO()
for i in source:
    buf.write(i)
final = buf.getvalue()
# 1000 loops, best of 3: 1.27 ms per loop

%%timeit out = []
for i in source:
    out.append(i)
final = ''.join(out)
# 1000 loops, best of 3: 9.89 ms per loop

%%timeit out = bytearray()
for i in source:
    out += i
# 10000 loops, best of 3: 98.5 µs per loop

%%timeit out = ""
for i in source:
    out += i
# 10000 loops, best of 3: 161 µs per loop

## Repeat the tests with a larger list, containing
## strings that are bigger than the small string caching 
## done by the Python
source = ['foo']*1000

# cStringIO
# 10 loops, best of 3: 19.2 ms per loop

# list append and join
# 100 loops, best of 3: 144 ms per loop

# bytearray() +=
# 100 loops, best of 3: 3.8 ms per loop

# str() +=
# 100 loops, best of 3: 5.11 ms per loop