问题：将字符列表转换为字符串
如果我有一个字符列表：
a = ['a','b','c','d']
如何将其转换为单个字符串？
a = 'abcd'
If I have a list of chars:
a = ['a','b','c','d']
How do I convert it into a single string?
a = 'abcd'
回答 0
使用join空字符串的方法将所有字符串以及中间的空字符串连接在一起，如下所示：
>>> a = ['a', 'b', 'c', 'd']
>>> ''.join(a)
'abcd'
Use the join method of the empty string to join all of the strings together with the empty string in between, like so:
>>> a = ['a', 'b', 'c', 'd']
>>> ''.join(a)
'abcd'
回答 1
这可以在许多流行的语言（例如JavaScript和Ruby）中使用，为什么不能在Python中使用？
>>> ['a', 'b', 'c'].join('')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'join'
奇怪的是，在Python中，join方法在str类上：
# this is the Python way
"".join(['a','b','c','d'])
为什么对象中join的方法list不像JavaScript或其他流行的脚本语言那样？这是Python社区如何思考的一个示例。由于join返回的是字符串，因此应将其放置在字符串类中，而不是列表类中，因此该str.join(list)方法意味着：使用str分隔符将列表连接到新字符串中（本例中str为空字符串）。
过了一段时间，我莫名其妙地爱上了这种思维方式。我可以抱怨Python设计中的很多事情，但不能抱怨它的连贯性。
This works in many popular languages like JavaScript and Ruby, why not in Python?
>>> ['a', 'b', 'c'].join('')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'join'
Strange enough, in Python the join method is on the str class:
# this is the Python way
"".join(['a','b','c','d'])
Why join is not a method in the list object like in JavaScript or other popular script languages? It is one example of how the Python community thinks. Since join is returning a string, it should be placed in the string class, not on the list class, so the str.join(list) method means: join the list into a new string using str as a separator (in this case str is an empty string).
Somehow I got to love this way of thinking after a while. I can complain about a lot of things in Python design, but not about its coherence.
回答 2
如果您的Python解释器较旧（例如，1.5.2在某些较旧的Linux发行版中很常见），则您可能无法join()在任何旧的字符串对象上将其用作方法，而需要使用字符串模块。例：
a = ['a', 'b', 'c', 'd']
try:
b = ''.join(a)
except AttributeError:
import string
b = string.join(a, '')
字符串b将为'abcd'。
If your Python interpreter is old (1.5.2, for example, which is common on some older Linux distributions), you may not have join() available as a method on any old string object, and you will instead need to use the string module. Example:
a = ['a', 'b', 'c', 'd']
try:
b = ''.join(a)
except AttributeError:
import string
b = string.join(a, '')
The string b will be 'abcd'.
回答 3
这可能是最快的方法：
>> from array import array
>> a = ['a','b','c','d']
>> array('B', map(ord,a)).tostring()
'abcd'
This may be the fastest way:
>> from array import array
>> a = ['a','b','c','d']
>> array('B', map(ord,a)).tostring()
'abcd'
回答 4
减少功能也起作用
import operator
h=['a','b','c','d']
reduce(operator.add, h)
'abcd'
The reduce function also works
import operator
h=['a','b','c','d']
reduce(operator.add, h)
'abcd'
回答 5
如果列表包含数字，则可以map()与结合使用join()。
例如：
>>> arr = [3, 30, 34, 5, 9]
>>> ''.join(map(str, arr))
3303459
If the list contains numbers, you can use map() with join().
Eg:
>>> arr = [3, 30, 34, 5, 9]
>>> ''.join(map(str, arr))
3303459
回答 6
h = ['a','b','c','d','e','f']
g = ''
for f in h:
g = g + f
>>> g
'abcdef'
h = ['a','b','c','d','e','f']
g = ''
for f in h:
g = g + f
>>> g
'abcdef'
回答 7
除了str.join这是最自然的方式，一种可能性是使用io.StringIO和滥用一次writelines编写所有元素：
import io
a = ['a','b','c','d']
out = io.StringIO()
out.writelines(a)
print(out.getvalue())
印刷品：
abcd
当将此方法与生成器函数或不是a tuple或a 的可迭代器一起使用时list，它将保存临时创建的列表，该列表join确实可以一次性分配正确的大小（并且1个字符的字符串列表在内存方面非常昂贵） ）。
如果您的内存不足，并且输入的对象是惰性求值，则此方法是最佳解决方案。
besides str.join which is the most natural way, a possibility is to use io.StringIO and abusing writelines to write all elements in one go:
import io
a = ['a','b','c','d']
out = io.StringIO()
out.writelines(a)
print(out.getvalue())
prints:
abcd
When using this approach with a generator function or an iterable which isn’t a tuple or a list, it saves the temporary list creation that join does to allocate the right size in one go (and a list of 1-character strings is very expensive memory-wise).
If you’re low in memory and you have a lazily-evaluated object as input, this approach is the best solution.
回答 8
您也可以operator.concat()这样使用：
>>> from operator import concat
>>> a = ['a', 'b', 'c', 'd']
>>> reduce(concat, a)
'abcd'
如果您使用的是Python 3，则需要先添加：
>>> from functools import reduce
由于内置函数reduce()已从Python 3中删除，现在位于中functools.reduce()。
You could also use operator.concat() like this:
>>> from operator import concat
>>> a = ['a', 'b', 'c', 'd']
>>> reduce(concat, a)
'abcd'
If you’re using Python 3 you need to prepend:
>>> from functools import reduce
since the builtin reduce() has been removed from Python 3 and now lives in functools.reduce().
