Question

我有像

这样的长元组

(2, 2, 10, 10, 344, 344, 45, 43, 2, 2, 10, 10, 12, 8, 2, 10)

我试图把它分成像

这样的元组元组

((2, 2, 10, 10), (344, 344, 45, 43), (2, 2, 10, 10), (12, 8, 2, 10))

我是python的新手，对元组o（2，2,10,10,344,344,45,43,2,10,10,12,8,2,10）不太好名单。我的朋友说我应该分开它，但我不能得到它-_-

我需要将元组拆分为带有4个元素的元组，稍后我将使用矩形绘制到PIL的图像。

Answer 1

那有一定的成语：

def grouper(n, iterable):
    args = [iter(iterable)] * n
    return zip(*args)

t = (2, 2, 10, 10, 344, 344, 45, 43, 2, 2, 10, 10, 12, 8, 2, 10)
print grouper(4, t)

但它有点复杂的解释。 the itertools receipes中列出了稍微更为通用的版本。

你也可以自己切片元组

parts = (t[0:4], t[4:8], t[8:12], t[12:16])

# or as a function
def grouper2(n, lst):
    return [t[i:i+n] for i in range(0, len(t), n)]

这可能是你朋友的意思。

Answer 2

>>> atup
(2, 2, 10, 10, 344, 344, 45, 43, 2, 2, 10, 10, 12, 8, 2, 10)
>>> [ atup[n:n+4] for n,i in enumerate(atup) if n%4==0 ]
[(2, 2, 10, 10), (344, 344, 45, 43), (2, 2, 10, 10), (12, 8, 2, 10)]

Answer 3

另一个可能的答案（使用发电机）：

 oldTuple = (2, 2, 10, 10, 344, 344, 45, 43, 2, 2, 10, 10, 12, 8, 2, 10)
 newTuple = tuple(oldTuple[x:x+4] for x in range(0, len(oldTuple), 4))

Answer 4

我曾经在http://gist.github.com/616853

提供了这样的功能作为gist

def split(input_list,num_fractions=None,subset_length=None):
   '''                                                                                                                                
   Given a list/tuple split original list based on either one of two parameters given but NOT both,                                   
   Returns generator                                                                                                                  
   num_fractions : Number of subsets original list has to be divided into, of same size to the extent possible.                       
                   In case equilength subsets can't be generated, all but the last subset                                             
                   will have the same number of elements.                                                                             
   subset_length : Split on every subset_length elements till the list is exhausted.                                                  

   '''
   if not input_list:
       yield input_list #For some reason I can't just return from here : return not allowed in generator expression                   
   elif not bool(num_fractions) ^ bool(subset_length): #Will check for both the invalid cases, '0' and 'None'.. Oh Python :)          
       raise Exception("Only one of the params : num_fractions,subset_length to be provided")
   else:
       if num_fractions: #calcluate subset_length in this case                                                                        
           subset_length = max(len(input_list)/num_fractions,1)

       for start in xrange(0,len(input_list),subset_length):
           yield input_list[start:start+subset_length]



>> list(list_split.split((2, 2, 10, 10, 344, 344, 45, 43, 2, 2, 10, 10, 12, 8, 2, 10),subset_length=4))
 [(2, 2, 10, 10), (344, 344, 45, 43), (2, 2, 10, 10), (12, 8, 2, 10)]

代码比上面给出的解决方案更长，但涵盖了所有可能的序列分割条件。

将一个长元组分成较小的元组

4 个答案: