我在 Python 程序中有一个列表,其中包含一系列数字,这些数字本身就是 ASCII 值。如何将其转换为可以回显到屏幕的“常规”字符串?
Electrons_Ahoy
问问题
191663 次
9 回答
152
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'
于 2008-10-07T21:54:33.040 回答
25
与其他人相同的基本解决方案,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'
于 2008-10-08T01:22:05.087 回答
14
import array
def f7(list):
return array.array('B', list).tostring()
于 2008-10-08T20:22:31.633 回答
8
l = [83, 84, 65, 67, 75]
s = "".join([chr(c) for c in l])
print s
于 2008-10-07T21:55:06.970 回答
7
您可以使用它bytes(list).decode()
来执行此操作 - 并list(string.encode())
取回值。
于 2019-04-04T06:50:34.487 回答
5
也许不像 Pyhtonic 一个解决方案,但对于像我这样的菜鸟来说更容易阅读:
charlist = [34, 38, 49, 67, 89, 45, 103, 105, 119, 125]
mystring = ""
for char in charlist:
mystring = mystring + chr(char)
print mystring
于 2015-12-13T00:35:37.630 回答
3
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()
于 2015-08-04T03:42:20.930 回答
1
我已经计时了现有的答案。要重现的代码如下。TLDR 是bytes(seq).decode()
迄今为止最快的。结果在这里:
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
设置为 CPython 3.8.2(32 位)、Windows 10、i7-2600 3.4GHz
有趣的观察:
- “官方”最快的答案(由 Toni Ruža 转发)现在对于 Python 3 来说已经过时了,但一旦修复,基本上仍然排在第二位
- 加入映射序列的速度几乎是列表理解的两倍
- 列表理解比它的非列表对应物更快
要重现的代码在这里:
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))
于 2020-05-14T06:40:10.113 回答
-1
Question = [67, 121, 98, 101, 114, 71, 105, 114, 108, 122]
print(''.join(chr(number) for number in Question))
于 2018-12-09T10:42:33.723 回答