的Python：問題處理字符串

Question

我有一個字符串，如下所示：

names = "name:fred, name:wilma, name:barney, name2:gauss, name2:riemann" 

假設的字符串名稱具有名稱和NAME屬性。

如何編寫函數is_name_attribute（），該函數檢查值是否爲名稱屬性？這是is_name_attribute（'fred'）應該返回True，而is_name_attribute（'gauss'）應該返回False。

此外，如何創建分隔的字符串包含只有名稱即屬性逗號，

"fred, wilma, barney" 

Answer 1

事情是這樣的：

>>> names = "name:fred, name:wilma, name:barney, name2:gauss, name2:riemann" 
>>> pairs = [x.split(':') for x in names.split(", ")] 
>>> attrs = [x[1] for x in pairs if x[0]=='name'] 
>>> attrs 
['fred', 'wilma', 'barney'] 
>>> def is_name_attribute(x): 
...  return x in attrs 
... 
>>> is_name_attribute('fred') 
True 
>>> is_name_attribute('gauss') 
False 

Answer 2

我想在一個字符串寫入ALS這玩意不是最好的解決方案，但是：

import re 

names = "name:fred, name:wilma, name:barney, name2:gauss, name2:riemann" 

def is_name_attribute(names, name): 
    list = names.split() 
    compiler = re.compile('^name:(.*)$') 
    for line in list: 
     line = line.replace(',','') 
     match = compiler.match(line) 
     if match: 
      if name == match.group(1): 
       return True 
    return False 

def commaseperated(names): 
    list = names.split() 
    compiler = re.compile('^name:(.*)$') 
    commasep = "" 
    for line in list: 
     line = line.replace(',','') 
     match = compiler.match(line) 
     if match: 
      commasep += match.group(1) + ', ' 
    return commasep[:-2] 

print is_name_attribute(names, 'fred') 
print is_name_attribute(names, 'gauss') 
print commaseperated(names) 

Answer 3

還有其他方法可以做到這一點（正如您從答案），但也許是時候學習一些Python列表魔術了。

>>> names = "name:fred, name:wilma, name:barney, name2:gauss, name2:riemann" 
>>> names_list = [pair.split(':') for pair in names.split(', ')] 
>>> names_list 
[['name', 'fred'], ['name', 'wilma'], ['name', 'barney'], ['name2', 'gauss'], ['name2', 'riemann']] 

從那裏，這只是一個檢查的情況。如果你正在尋找一個特定的名稱：

for pair in names_list: 
    if pair[0] == 'name' and pair[1] == 'fred': 
     return true 
return false 

並加入只是名字版本：

>>> new_name_list = ','.join([pair[1] for pair in names_list if pair[0] == 'name']) 
>>> new_name_list 
'fred,wilma,barney' 

Answer 4

簡單的正則表達式匹配：

>>> names = re.compile ('name:([^,]+)', 'g') 
>>> names2 = re.compile ('name2:([^,]+)', 'g') 
>>> str = "name:fred, name:wilma, name:barney, name2:gauss, name2:riemann" 
>>> 'fred' in names.findall(str) 
True 
>>> names.findall(str) 
['fred', 'wilma', 'barney']