Question

我有以下方式的列表和字典：

lista = [
    ['Pasta', 'is', 'bad'],
    ['Pasta', 'is', 'good', 'and','is','expensive']
]

dica = {
    'Pasta': 0,
     'is': 0,
     'bad': 0,
     'good': 0,
     'and': 0,
     'expensive': 0
}

我想获取列表中的每个子列表并对其应用字典，创建一个字典列表-每个列表一个-其中每个键的值是该键在子列表中出现的次数。 / p>

fin_dict = {
    {
        'Pasta': 1,
        'is': 1,
        'bad': 1,
        'good': 0,
        'and': 0,
        'expensive': 0
    },
    {
        'Pasta': 1,
        'is': 2,
        'bad': 0,
        'good': 1,
        'and': 1,
        'expensive': 1
    }
}

Answer 1

尝试一下：

def count_items(dic, lst):
    counts = dict(dic)
    for x in lst:
        counts[x] += 1
    return counts

lista=[['Pasta', 'is', 'bad'],['Pasta', 'is', 'good', 'and','is','expensive']]

dica={'Pasta': 0,'is': 0,'bad': 0,'good': 0,'and': 0,'expensive': 0}

print([count_items(dica, lista[0]), count_items(dica, lista[1])])

# prints [{'Pasta': 1, 'is': 1, 'bad': 1, 'good': 0, 'and': 0, 'expensive': 0}, {'Pasta': 1, 'is': 2, 'bad': 0, 'good': 1, 'and': 1, 'expensive': 1}]

（此解决方案假定fin_dict是一个列表，而输入列表中的所有项目都在字典中。）

Answer 2

我在以下脚本中解释了详细信息：

lista=[['Pasta', 'is', 'bad'],['Pasta', 'is', 'good', 'and','is','expensive']]

dica={'Pasta': 0,'is': 0,'bad': 0,'good': 0,'and': 0,'expensive': 0}

# Create an empty array to store dictionaries.
myList = []

for i in range(len(lista)):
    # Append `dica` in each iteration to get a reference of it.
    myList.append(dica)
    # Iterate through `dica` keys.
    for key in dica:
        # Iterate through list elements in lista.
        for listElem in lista[i]:
            # Check if the `key` is equal to the value in list,
            # in other words, count the occurences of list elements.
            if key == listElem:
                myList[i][key] += 1
print(myList)

注意：据我所知，您所请求的输出类型是不可能的，我还认为您并不想使用dictionary，因此我将其输出为list。

输出：

[{'Pasta': 2, 'is': 3, 'bad': 1, 'good': 1, 'and': 1, 'expensive': 1}, {'Pasta': 2, 'is': 3, 'bad': 1, 'good': 1, 'and': 1, 'expensive': 1}]

使用列表和字典的期望输出

2 个答案: