轉換雙元組串

我想轉換雙元組，它看起來像這樣：轉換雙元組串

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

這個字符串：

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

我試着這樣說：

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

但結果看起來不合適，例如這樣：

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

如何修平它？

來源

2017-09-21 John

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))

來源

2017-09-21 12:43:21

你可以做一個正則表達式與聯接：

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)])

這產生了：

'833，765，11，56，61，84'

來源

2017-09-21 12:39:50

使用正則表達式，即替代的非數字用''使用list理解然後加入,

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])

輸出：

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

來源

2017-09-21 12:41:23 Dark

您的輸入被搞砸了，可能值得修復上游。

如果你堅持與使用ast.literal_eval評估元組（從字符串的開始&結尾剝離報價&空格之後），採取的第一個項目，並加入回字符串：

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)

結果：

833, 765, 11, 56, 61, 84

來源

2017-09-21 12:42:15

在IPython中測試了這個。你應該只需要更深入。

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'

來源

2017-09-21 15:57:14

轉換雙元組串

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