我有一个嵌套的OrderedDict数据:
OrderedDict([('6',
OrderedDict([('idx2', ['6', '6', '6']),
('val2', ['A', 'T', 'T']),
('val1', ['C', 'A', 'T']),
('pos', ['11', '15', '23']),
('idx1', ['4', '4', '4'])])),
('3',
OrderedDict([('idx2', ['3', '3']),
('val2', ['G', 'C']),
('val1', ['A', 'G']),
('pos', ['28', '34']),
('idx1', ['4', '4'])])),
('4',
OrderedDict([('idx2', ['4', '4']),
('val2', ['T', 'C']),
('val1', ['C', 'C']),
('pos', ['41', '51']),
('idx1', ['4', '4'])]))])
我可以为每个键访问key, value
对,例如:
for k, v in grouped_data.items():
print(k, v)
并使用以下内容进一步筑巢:
for k, v in grouped_data.items():
print(k, v)
for x in v.items():
print(x)
for z in x:
print(z)
但是,要处理我的数据,我需要在for循环中一次访问两个连续的键:首先是6-3然后是3-4,依此类推,直到结束。
我在想是否可以为密钥编制索引并获取密钥后面的值:
for k, v in grouped_data.items():
print(k, v)
ind = grouped_data.keys().index(k)
或者,
for k, v in grouped_data.items():
print(k, v)
print(next(k))