如何在python中将ascii值列表转换为字符串

Question

我在Python程序中得到了一个列表，其中包含一系列数字，这些数字本身就是ASCII值。如何将其转换为"常规"字符串，以便在屏幕上进行回显？

Answer 1

您可能正在寻找"chr（）"：

>>> L = [104, 101, 108, 108, 111, 44, 32, 119, 111, 114, 108, 100]
>>> ''.join(chr(i) for i in L)
'hello, world'

Answer 2

与其他解决方案相同，但我个人更喜欢使用map而不是列表理解：

>>> L = [104, 101, 108, 108, 111, 44, 32, 119, 111, 114, 108, 100]
>>> ''.join(map(chr,L))
'hello, world'

Answer 3

import array
def f7(list):
    return array.array('B', list).tostring()

来自Python模式-优化轶事

Answer 4

l = [83, 84, 65, 67, 75]
s = "".join([chr(c) for c in l])
print s

Answer 5

您可以使用bytes(list).decode()来执行此操作，使用list(string.encode())来获取值。

Answer 6

也许不是Pyhtonic的解决方案，但对于像我这样的人来说更容易阅读：

charlist = [34, 38, 49, 67, 89, 45, 103, 105, 119, 125]
mystring = ""
for char in charlist:
    mystring = mystring + chr(char)
print mystring

Answer 7

def working_ascii():
    """
        G    r   e    e    t    i     n   g    s    !
        71, 114, 101, 101, 116, 105, 110, 103, 115, 33
    """
    hello = [71, 114, 101, 101, 116, 105, 110, 103, 115, 33]
    pmsg = ''.join(chr(i) for i in hello)
    print(pmsg)
    for i in range(33, 256):
        print(" ascii: {0} char: {1}".format(i, chr(i)))
working_ascii()

Answer 8

我已经对现有答案进行了计时。下面是要复制的代码。TLDR是CCD_ 3是迄今为止速度最快的。此处的结果：

 test_bytes_decode : 12.8046 μs/rep
     test_join_map : 62.1697 μs/rep
test_array_library : 63.7088 μs/rep
    test_join_list : 112.021 μs/rep
test_join_iterator : 171.331 μs/rep
    test_naive_add : 286.632 μs/rep

安装程序为CPython 3.8.2（32位），Windows 10，i7-2600 3.4GHz

有趣的观察：

• Python 3的"官方"最快答案（由托尼·鲁扎转发）现在已经过时了，但一旦确定，基本上仍然并列第二
• 加入映射序列的速度几乎是列表理解的两倍
• 列表理解比非列表理解更快

要复制的代码在这里：

import array, string, timeit, random
from collections import namedtuple
# Thomas Wouters (https://stackoverflow.com/a/180615/13528444)
def test_join_iterator(seq):
    return ''.join(chr(c) for c in seq)
# community wiki (https://stackoverflow.com/a/181057/13528444)
def test_join_map(seq):
    return ''.join(map(chr, seq))
# Thomas Vander Stichele (https://stackoverflow.com/a/180617/13528444)
def test_join_list(seq):
    return ''.join([chr(c) for c in seq])
# Toni Ruža (https://stackoverflow.com/a/184708/13528444)
# Also from https://www.python.org/doc/essays/list2str/
def test_array_library(seq):
    return array.array('b', seq).tobytes().decode()  # Updated from tostring() for Python 3
# David White (https://stackoverflow.com/a/34246694/13528444)
def test_naive_add(seq):
    output = ''
    for c in seq:
        output += chr(c)
    return output
# Timo Herngreen (https://stackoverflow.com/a/55509509/13528444)
def test_bytes_decode(seq):
    return bytes(seq).decode()
RESULT = ''.join(random.choices(string.printable, None, k=1000))
INT_SEQ = [ord(c) for c in RESULT]
REPS=10000
if __name__ == '__main__':
    tests = {
        name: test
        for (name, test) in globals().items()
        if name.startswith('test_')
    }
    Result = namedtuple('Result', ['name', 'passed', 'time', 'reps'])
    results = [
        Result(
            name=name,
            passed=test(INT_SEQ) == RESULT,
            time=timeit.Timer(
                stmt=f'{name}(INT_SEQ)',
                setup=f'from __main__ import INT_SEQ, {name}'
                ).timeit(REPS) / REPS,
            reps=REPS)
        for name, test in tests.items()
    ]
    results.sort(key=lambda r: r.time if r.passed else float('inf'))
    def seconds_per_rep(secs):
        (unit, amount) = (
            ('s', secs) if secs > 1
            else ('ms', secs * 10 ** 3) if secs > (10 ** -3)
            else ('μs', secs * 10 ** 6) if secs > (10 ** -6)
            else ('ns', secs * 10 ** 9))
        return f'{amount:.6} {unit}/rep'
    max_name_length = max(len(name) for name in tests)
    for r in results:
        print(
            r.name.rjust(max_name_length),
            ':',
            'failed' if not r.passed else seconds_per_rep(r.time))

Answer 9

Question = [67, 121, 98, 101, 114, 71, 105, 114, 108, 122]
print(''.join(chr(number) for number in Question))