排序元組列表與去年元素

Question

需要用最後一個元素

def sort_last(tuples): 
    sorted_list = [] 
    i = 0 
    i2 = 1 
    while True: 
     if len(tuples) == 2: 
      if tuples[i][-1] < tuples[i2][-1]: 
       sorted_list.append(tuples[i]) 
       sorted_list.append(tuples[i2]) 
       break 
      else: 
       sorted_list.append(tuples[i2]) 
       sorted_list.append(tuples[i]) 
       break 
     elif tuples[i][-1] < tuples[i2][-1]: 
      i2 += 1 
      print 1 
     elif i == i2: 
      i2 += 1 
     elif i2 == len(tuples)-1: 
      if tuples[i][-1] < tuples[i2][-1]: 
       sorted_list.append(tuples[i]) 
       del tuples[i] 
       i = 0 
       i2 = 1 
      else: 
       sorted_list.append(tuples[i2]) 
       del tuples[i2] 
       i = 0 
       i2 = 1 
     elif len(tuples) <= 1: 
      return tuples 
     else: 
      i += 1 
    return sorted_list 

排序給出元組列表隨着元組列表的像([(1, 3), (3, 2), (2, 1)])或([(2, 3), (1, 2), (3, 1)])返回正確：([(2, 1), (3, 2), (1, 3)])和([(3, 1), (1, 2), (2, 3)]），但有3個以上元素的元組或4元組列表它寫入"list index out of range"

Answer 1

def sort_last(tuples): 
    return sorted(tuples, key=lambda i: i[-1]) 

>>> sort_last([(1, 3), (3, 2), (2, 1)]) 
[(2, 1), (3, 2), (1, 3)] 
>>> sort_last([(2, 3), (1, 2), (3, 1)]) 
[(3, 1), (1, 2), (2, 3)] 

Answer 2

a=([(1, 3), (3, 2), (2, 1)]) 
def foo(elem): 
    return elem[-1] 
a.sort(key=foo) 
print(a)