解壓Python3中的iterables？

Question

這是爲什麼返回in sort_tup_from_list for key, val in tup: ValueError: not enough values to unpack (expected 2, got 1)

# list with tuples 
lst = [("1", "2"), ("3", "4")] 

# sorting list by tuple val key 
def sort_tup_from_list(input_list): 
    tmp = [] 
    print(tup) 
    for tup in input_list: 
     for key, val in tup: 
      tmp.append((val, key)) 
      tmp.sort(reverse=True) 
    return tmp 

print(sort_tup_from_list(lst)) 

當我註釋掉第二for循環，它打印元組：

lst = [("1", "2"), ("3", "4")] 
def sort_tup_from_list(input_list): 
    tmp = [] 
    for tup in input_list: 
     print(tup) 
     # for key, val in tup: 
     #  tmp.append((val, key)) 
     #  tmp.sort(reverse=True) 
    return tmp 

print(sort_tup_from_list(lst)) 

輸出：

('1', '2') 
('3', '4') 
[] 

因此，元組在那裏。他們爲什麼不拆包自己？

Answer 1

你的第二個for循環是遍歷元組中的項目，但是你抓取了它中的兩個項目。我認爲這是你想要的：

# list with tuples 
lst = [("1", "2"), ("3", "4")] 

# sorting list by tuple val key 
def sort_tup_from_list(input_list): 
    tmp = [] 
    print(tmp) 
    for key,val in input_list: 
     tmp.append((val, key)) 
     tmp.sort(reverse=True) 
    return tmp 

print(sort_tup_from_list(lst))