Python 将字符列表转换为字符串
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Convert a list of characters into a string
提问by nos
If I have a list of chars:
如果我有一个字符列表：
a = ['a','b','c','d']
How do I convert it into a single string?
如何将其转换为单个字符串？
a = 'abcd'
采纳答案by Daniel Stutzbach
Use the joinmethod of the empty string to join all of the strings together with the empty string in between, like so:
使用join空字符串的方法将所有字符串连接在一起，中间有空字符串，如下所示：
>>> a = ['a', 'b', 'c', 'd']
>>> ''.join(a)
'abcd'
回答by Paulo Scardine
This works in many popular languages like JavaScript and Ruby, why not in Python?
这适用于许多流行的语言，如 JavaScript 和 Ruby，为什么不适用于 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 joinmethod is on the strclass:
奇怪的是，在 Python 中，该join方法在str类上：
# this is the Python way
"".join(['a','b','c','d'])
Why joinis not a method in the listobject 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 stras a separator (in this case stris an empty string).
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例子。由于 join 返回一个字符串，它应该放在字符串类中，而不是放在列表类中，因此该str.join(list)方法的意思是：使用str作为分隔符（在本例中str为空字符串）将列表连接到一个新字符串中。
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.
不知何故，我在一段时间后开始喜欢这种思维方式。我可以抱怨 Python 设计中的很多东西，但不能抱怨它的连贯性。
回答by bigeagle
This may be the fastest way:
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'
回答by Bill
h = ['a','b','c','d','e','f']
g = ''
for f in h:
g = g + f
>>> g
'abcdef'
回答by Kyle
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:
如果您的 Python 解释器很旧（例如 1.5.2，这在一些较旧的 Linux 发行版中很常见），您可能无法join()作为任何旧字符串对象的方法，而您将需要使用 string 模块。例子：
a = ['a', 'b', 'c', 'd']
try:
b = ''.join(a)
except AttributeError:
import string
b = string.join(a, '')
The string bwill be 'abcd'.
字符串b将为'abcd'.
回答by cce
The reduce function also works
减少功能也有效
import operator
h=['a','b','c','d']
reduce(operator.add, h)
'abcd'
回答by Tonechas
You could also use operator.concat()like this:
你也可以operator.concat()这样使用：
>>> from operator import concat
>>> a = ['a', 'b', 'c', 'd']
>>> reduce(concat, a)
'abcd'
If you're using Python 3 you need to prepend:
如果您使用的是 Python 3，则需要预先添加：
>>> from functools import reduce
since the builtin reduce()has been removed from Python 3 and now lives in functools.reduce().
因为内置函数reduce()已从 Python 3 中删除，现在位于functools.reduce().
回答by yask
If the list contains numbers, you can use map()with join().
如果列表包含数字，您可以使用map()with join()。
Eg:
例如：
>>> arr = [3, 30, 34, 5, 9]
>>> ''.join(map(str, arr))
3303459
回答by Jean-Fran?ois Fabre
besides str.joinwhich is the most natural way, a possibility is to use io.StringIOand abusing writelinesto write all elements in one go:
除了str.join这是最自然的方式之外，一种可能性是使用io.StringIO和滥用writelines一次性编写所有元素：
import io
a = ['a','b','c','d']
out = io.StringIO()
out.writelines(a)
print(out.getvalue())
prints:
印刷：
abcd
When using this approach with a generator function or an iterable which isn't a tupleor a list, it saves the temporary list creation that joindoes to allocate the right size in one go (and a list of 1-character strings is very expensive memory-wise).
当将此方法与生成器函数或不是 atuple或 a的可迭代对象一起使用时list，它会保存临时列表创建，join从而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）。
If you're low in memory and you have a lazily-evaluated object as input, this approach is the best solution.
如果您的内存不足并且您有一个延迟评估的对象作为输入，则此方法是最佳解决方案。
