python - 将每个键的值存储为字典中的数组

Question

我想规范化字典 data 中的所有值，并将它们再次存储在另一个具有相同键的字典中，对于每个键，值应该存储在一维数组中，所以我做了以下操作:

>>> data = {1: [0.6065306597126334], 2: [0.6065306597126334, 0.6065306597126334, 0.1353352832366127], 3: [0.6065306597126334, 0.6065306597126334, 0.1353352832366127], 4: [0.6065306597126334, 0.6065306597126334]}

>>> norm = {k: [v / sum(vals) for v in vals] for k, vals in data.items()} 

>>> norm
{1: [1], 2: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 3: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 4: [0.5, 0.5]}

现在假设字典 data 只包含其中一个键的零值，例如第一个键 1 的值:

>>> data = {1: [0.0], 2: [0.6065306597126334, 0.6065306597126334, 0.1353352832366127], 3: [0.6065306597126334, 0.6065306597126334, 0.1353352832366127], 4: [0.6065306597126334, 0.6065306597126334]}

然后归一化这个字典的值将得到 [nan] 值，因为除以零

>>> norm = {k: [v / sum(vals) for v in vals] for k, vals in data.items()}

__main__:1: RuntimeWarning: invalid value encountered in double_scalars
>>> norm
{1: [nan], 2: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 3: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 4: [0.5, 0.5]}

所以我插入了一个 if 语句 来解决这个问题，但我无法将每个键的值存储为 ID 数组

代码

>>> norm = {}
>>> for k, vals in data.items():
...     values = []
...     if sum(vals) == 0:
...        values.append(list(vals))
...     else:
...          for v in vals:
...              values.append(list([v/sum(vals)]))
...     norm[k]=values
... 
>>> norm
{1: [[1.0]], 2: [[0.4498162176582741], [0.4498162176582741], [0.10036756468345168]], 3: [[0.4498162176582741], [0.4498162176582741], [0.10036756468345168]], 4: [[0.5], [0.5]]}

我想获取 norm 字典作为

norm = {1: [1.0], 2: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 3: [0.4498162176582741, 0.4498162176582741, 0.10036756468345168], 4: [0.5, 0.5]}

此外，对于字典 data，虽然它包含一个零值，如果它是键，是否有更好的解决方案来规范化它，因为我认为我的解决方案效率不高!

P.S:我在 for 循环的末尾尝试 norm[k]= np.array(values) 而不是 norm[k]=values 但结果不符合要求。

Answer 1

如回答中所述，extend 可用于解决您的问题。如果您确实想使用 append，您可以使用列表的第一个元素。

norm = {}
for k, vals in data.items():
    values = []
    if sum(vals) == 0:
        values.append(vals[0])
    else:
        for v in vals:
            values.append([v / sum(vals)][0])
    norm[k] = values

参见 difference between append vs extend list methods in python以追加与扩展为例

至于优化。完全删除 for 循环是不可能的，但您可以缩短您的解决方案，同时仍保持可读性:

norm = {}
for k, vals in data.items():
    if sum(vals) == 0:
        norm[k] = vals
    else:
        norm[k] = [x / sum(vals) for x in vals]