将双元组转换为字符串

Question

我想转换双元组，如下所示：

  {tuple}  <type 'list'>: ((u'833',), (u'765',) ,
 (u'11',) , (u'56',) , (u'61',) , (u'84',) )

到这个字符串：

"833, 765, 11, 56, 61, 84"

我这样尝试：

query = double_tuple_from_above
str = ["'{}' ".format(element) for element in query ]
final_str = ', '.join(str)
# aaaa = steamids64.replace("(", "',)") # does not work

但是结果看起来无论哪种方式都不正确，例如：

final_str = '(u'833',)' , '(u'765',)' , '(u'11',)' , '(u'56',)' , '(u'61',)' , '(u'84',)' 

如何弄平正确的？

Answer 1

使用正则表达式，即使用列表理解功能将非数字替换为'' ，然后与,

import re
m = ["'(u'833',)' ", "'(u'765',)' ", "'(u'11',)' ", "'(u'56',)' ", "'(u'61',)' ", "'(u'84',)' "]
new =  ', '.join([re.sub('[^0-9]+' ,'', i) for i in m])

输出：

'833, 765, 11, 56, 61, 84'

Answer 2

您的输入混乱了，可能值得在上游修复。

如果您ast.literal_eval ，请使用ast.literal_eval评估元组（在ast.literal_eval字符串开头和结尾的引号和空格之后），获取第一个项目并重新加入字符串：

l = ["'(u'833',)' ", "'(u'765',)' ",
 "'(u'11',)' ", "'(u'56',)' ", "'(u'61',)' ", "'(u'84',)' "]

import ast

s = ", ".join(y[0] for y in (ast.literal_eval(x.strip("' ")) for x in l))
print(s)

结果：

833, 765, 11, 56, 61, 84

Answer 3

您可以使用联接进行正则表达式：

import re
l = ["'(u'833',)' ", "'(u'765',)' ", "'(u'11',)' ", "'(u'56',)' ", "'(u'61',)' ", "'(u'84',)' "]

', '.join([re.search("\d+", elem).group() for elem in l if re.search("\d+", elem)])

这样产生：

'833，765，11，56，61，84'

Answer 4

import re

l = ["'(u'833',)' ", "'(u'765',)' ", "'(u'11',)' ", "'(u'56',)' ", "'(u'61',)' ", "'(u'84',)' "]
result = []

for elem in l:
    num = re.split(r'([0-9]+)', elem)
    result.append(num[1])

print(", ".join(result))

Answer 5

在ipython中对此进行了测试。 您只需要更深入。

Python 2.7.13 (default, Feb  4 2017, 17:39:59)
Type "copyright", "credits" or "license" for more information.

IPython 5.4.1 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.

In [1]: a = ((u'833',), (u'765',), (u'11',), (u'56',), (u'61',), (u'84',))

In [2]: a
Out[2]: ((u'833',), (u'765',), (u'11',), (u'56',), (u'61',), (u'84',))

In [3]: flatter = (x[0] for x in a)

In [4]: ret = ', '.join(flatter)

In [5]: ret
Out[5]: u'833, 765, 11, 56, 61, 84'

In [6]: ret = ret.encode('utf8')

In [7]: ret
Out[7]: '833, 765, 11, 56, 61, 84'