Question

我有一个可变长度元组的列表。如何将其转换为字典？

class StringArrayList {

    // make the variable private, to protect it from being modified from outside the class
    private final String[] arr;

    StringArrayList(int size) {
        // constructors are the only method allowed to set final variables
        arr = new String[size];
    }

    String get(int i) {
        // get method doesn't change the state of the object, so is fine
        // However, if the object returned is mutable then there might be issues.
        return arr[i];
    }

    void set(int i, String item) {
        // set changes the state of arr, and so with this method, StringArrayList cannot 
        // be considered immutable
        arr[i] = item;
    }

}

使用理解功能时出现错误

tup = [ ("x", 1), ("x", 2, 4), ("x", 3), ("y", 1), ("y", 2), ("z", 1), ("z", 2, 3) ]

错误：

{key: [i[1:] for i in tup if i[0] == key] for (key, value) in tup}

预期输出：

ValueError
Traceback (most recent call last)
>ipython-input-26-bedcc2e8a704< in module()
----> 1 {key: [i[1] for i in tuplex if i[0] == key] for (key, value) in tuplex}  
>ipython-input-26-bedcc2e8a704< in dictcomp((key, value))
----> 1 {key: [i[1] for i in tuplex if i[0] == key] for (key, value) in tuplex}  
ValueError: too many values to unpack

Answer 1

这似乎不是列表理解的任务。您可以使用字典的setdefault方法通过简单的for循环将密钥的默认值设置为一个空列表，然后使用该元组中的值extend将该列表设置为：

tup = [ ("x", 1), ("x", 2, 4), ("x", 3), ("y", 1), ("y", 2), ("z", 1), ("z", 2, 3) ]

res = {}
for k, *rest in tup:
    res.setdefault(k, []).extend(rest)

print(res)
# {'y': [1, 2], 'x': [1, 2, 4, 3], 'z': [1, 2, 3]}

对于Python 2.7，我认为您无法像这样打开一个元组，因此您可以尝试执行以下操作：

for t in tup:
    res.setdefault(t[0], []).extend(t[1:])

Answer 2

tup = [("x", 1), ("x", 2, 4), ("x", 3), ("y", 1), ("y", 2), ("z", 1), ("z", 2, 3)]
data = {}
for item in tup:
    if item[0] in data:
        data[item[0]] = data[item[0]] + list(item[1:])
    else:
        data[item[0]] = list(item[1:])
print(data)

将可变长度元组的列表转换为字典

2 个答案: