python - 通过字典值之一的索引将字典拆分为多个字典

Question

将 data 字典拆分为与 data['legend'] 中的唯一值一样多的字典是一种优雅而有效的方法，每个字典代表传奇条目？所有列表将始终按正确顺序排列。

我目前正在使用下面的方法，但如果有更短的解决方案我很感兴趣:

data = {
    'x':  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
    'y': [10, 20, 30, 40, 50, 60, 70, 80, 90, 100],
    'legend': ["active", "pending", "pending", "active", "completed", "pending", "pending","active", "completed", "active"],
    };

traces = [] 
unique_legend = list(set(data['legend']))
for item in unique_legend:
    trace = {}
    item_indices = [k for k,v in enumerate(data['legend']) if v == item]
    trace['x'] = [data['x'][indice] for indice in item_indices]
    trace['y'] = [data['y'][indice] for indice in item_indices]
    traces.append(trace)

for trace in traces:
    print(trace)

Answer 1

这是我可能采用的一种方法:

result = {}
for x in zip(data['x'], data['y'], data['legend']):
    result.setdefault(x[2], []).append(x[:2])

# result will look like the following:
# {
#    'active': [(1, 10), (4, 40), (8, 80), ...],
#    ...
# }