（Python 3）如何組合列表的產品？

Question

輸入：

Choc 5
Vani 10
Stra 7
Choc 3
Stra 4
END

def process_input(lst):
    result = []
    for string in lines:
        A=string.split()
        result.append([A[0],int(A[1])])  
    return result

def merge_products(invent):
    # your code here


# DON’T modify the code below
str = input()
lines = []
while str != 'END':
    lines.append(str)
    str = input()
inventory1 = process_input(lines)
merge_products(inventory1)
print(inventory1)

---

從這個輸出

[['Choc', 5], ['Vani', 10], ['Stra', 7], ['Choc', 3], ['Stra', 4]]

我需要得到這個輸出

[['Choc', 8], ['Vani', 10], ['Stra', 11]]

如何在相同的字符串下組合整數？

Answer 1

類似於Kaushik NP的答案，但是使用collections.Counter實現

from collections import Counter
def merge_products(inventory):
    groceries = Counter()
    for item, count in inventory:
        groceries[item] +=count

    return [[item, count] for item, count in groceries.items()]

這在功能上是等效的，但Counter會為您處理默認值。 對於大型項目列表而言，它也可能更具性能，但在您運營的規模上可能並不是什么大不了的事。

Answer 2

使用defaultdict存儲值並添加到它們。

from collections import defaultdict

def merge_products(invent):
    d = defaultdict(int)
    for x,y in invent:
            d[x]+=y
    return [[k,v] for k,v in d.items()]

#driver值：

>>> merge_products([['Choc', 5], ['Vani', 10], ['Stra', 7], ['Choc', 3], ['Stra', 4]])
=> [['Choc', 8], ['Vani', 10], ['Stra', 11]]

---

如果你想保留你找到它們的順序 ，請使用OrderedDict 。

def merge_products(invent):
    d = OrderedDict()
    for x,y in invent:
        if d.get(x,None)==None :
            d[x]=y
        else :
            d[x]+=y

    return [[k,v] for k,v in d.items()]

Answer 3

你可以使用字典：

def merge_products(invent):
    # your code here
    result = {}
    for item in invent:
        if item[0] not in result:
            result[item[0]] = item[1]
        else:
            result[item[0]] += item[1]

    return result

請注意您將使用返回的值，而不是作為參數傳遞的值：

print(merge_products(inventory1))