把物品放入字典不改變爲了

Question

from collections import OrderedDict 

l = [('Monkey', 71), ('Monkey', 78), ('Ostrich', 80), ('Ostrich', 96), ('Ant', 98)] 

d = OrderedDict() 
for i, j in l: 
    d[i] = j 
print d 
OrderedDict([('Monkey', 78), ('Ostrich', 96), ('Ant', 98)]) 

預期「d」應該是：

OrderedDict([('Monkey', (71,78)), ('Ostrich', (80,96)), ('Ant', 98)]) 

如果所有值都tupled或上市沒有問題。

Answer 1

不是每次都來取代值，將其添加到元組：

>>> l = [('Monkey', 71), ('Monkey', 78), ('Ostrich', 80), ('Ostrich', 96), ('Ant', 98)] 
>>> d = OrderedDict() 
>>> for i, j in l: 
...  if i in d: 
...   d[i] += (j,) 
...  else: 
...   d[i] = (j,) 
... 
>>> d 
OrderedDict([('Monkey', (71, 78)), ('Ostrich', (80, 96)), ('Ant', (98,))]) 

順便說一句，因爲tuple s爲不可變的，每次追加創建一個新對象。如果您使用list s，這將更有效率。

Answer 2

下面是一個使用groupby的方法：

from itertools import groupby 

l = [('Monkey', 71), ('Monkey', 78), ('Ostrich', 80), ('Ostrich', 96), ('Ant', 98)] 
       # Key is the animal, value is a list of the available integers obtained by 
d = OrderedDict((animal, [i for _, i in vals]) 
       for (animal, vals) in 
       # Grouping your list by the first value inside animalAndInt, which is the animal 
       groupby(l, lambda animalAndInt: animalAndInt[0])) 
# If you want a tuple, instead of [i for _, i in vals] use tuple(i for _, i in vals) 
print(d) 
>>> OrderedDict([('Monkey', [71, 78]), ('Ostrich', [80, 96]), ('Ant', [98])]) 

Answer 3

for i, j in l: 
    if i in d: 
     #d[i] = (lambda x: x if type(x) is tuple else (x,))(d[i]) 
     #Eugene's version: 
     if not isinstance(d[i], tuple): 
      d[i] = (d[i],) 
     d[i] += (j,) 
    else: 
     d[i] = j 

給出以下。請注意，'Ant'中的98不像原始問題中提到的那樣是「tupled」。

OrderedDict([('Monkey', (71, 78)), ('Ostrich', (80, 96)), ('Ant', 98)])