Question

BIG = { "Brand" : ["Clothes" , "Watch"], "Beauty" : ["Skin", "Hair"] }
SMA = { "Clothes" : ["T-shirts", "pants"], "Watch" : ["gold", "metal"],
"Skin" : ["lotion", "base"] , "Hair" : ["shampoo", "rinse"]}

我想要合并这些数据像这样...

BIG = {"Brand" : [ {"Clothes" : ["T-shirts", "pants"]}, {"Watch" : ["gold", "metal"]} ],...

请告诉我如何解决这个问题。

Answer 1

首先，这些是字典而不是列表。另外，我不知道你在合并两个词典背后的意图。

无论如何，如果你想让输出完全按照你提到的那样，那就是这样做的方法 -

BIG = { "Brand" : ["Clothes" , "Watch"], "Beauty" : ["Skin", "Hair"] }
SMA = { "Clothes" : ["T-shirts", "pants"], "Watch" : ["gold", "metal"],"Skin" : ["lotion", "base"] , "Hair" : ["shampoo", "rinse"]}
for key,values in BIG.items(): #change to BIG.iteritems() in python 2.x
    newValues = []
    for value in values:
        if value in SMA:
            newValues.append({value:SMA[value]})
        else:
            newValues.append(value)
    BIG[key]=newValues

此外，BIG.update(SMA)不会按照您希望的方式为您提供正确的结果。

这是一个测试运行 -

>>> BIG.update(SMA)
>>> BIG
{'Watch': ['gold', 'metal'], 'Brand': ['Clothes', 'Watch'], 'Skin': ['lotion', 'base'], 'Beauty': ['Skin', 'Hair'], 'Clothes': ['T-shirts', 'pants'], 'Hair': ['shampoo', 'rinse']}

Answer 2

首先，您需要迭代第一个字典并在第二个字典中搜索该对密钥。

BIG = { "Brand" : ["Clothes" , "Watch"], "Beauty" : ["Skin", "Hair"] }
SMA = { "Clothes" : ["T-shirts", "pants"], "Watch" : ["gold", "metal"], "Skin" : ["lotion", "base"] , "Hair" : ["shampoo", "rinse"]}

for key_big in BIG:
    for key_sma in BIG[key_big]:
        if key_sma in SMA:
            BIG[key_big][BIG[key_big].index(key_sma)] = {key_sma: SMA.get(key_sma)}

print BIG

代码的结果：

>>> {'Brand': [{'Clothes': ['T-shirts', 'pants']}, {'Watch': ['gold', 'metal']}], 'Beauty': [{'Skin': ['lotion', 'base']}, {'Hair': ['shampoo', 'rinse']}]}

如何将不同的复杂列表与python结合起来

2 个答案: