Question

这里我有字典列表，我的目标是迭代列表，每当有2个或更多列表可用时，我想合并它们并附加在输出列表中，并且只要有一个列表就需要按原样存储。

data = [
            [[{'font-weight': '1'},{'font-weight': '1'}],[{'font-weight': '2'},{'font-weight': '2'}]],
            [{'font-weight': '3'},{'font-weight': '3'},{'font-weight': '3'}],
            [[{'font-weight': '1'},{'font-weight': '1'}],[{'font-weight': '2'},{'font-weight': '2'}]],
            [{'font-weight': '3'},{'font-weight': '3'}]
        ]

我可以为特定元素data[0]

列出扁平化

print([item for sublist in data[0] for item in sublist])
[{'font-weight': '1'}, {'font-weight': '1'}, {'font-weight': '2'}, {'font-weight': '2'}]

预期产出：

data = [
            [{'font-weight': '1'},{'font-weight': '1'},{'font-weight': '2'},{'font-weight': '2'}],
            [{'font-weight': '3'},{'font-weight': '3'},{'font-weight': '3'}],
            [{'font-weight': '1'},{'font-weight': '1'},{'font-weight': '2'},{'font-weight': '2'}]
            [{'font-weight': '3'},{'font-weight': '3'}]
        ]

Answer 1

您可以将conditional list comprehension与itertools.chain一起用于需要展平的元素：

In [54]: import itertools

In [55]: [list(itertools.chain(*l)) if isinstance(l[0], list) else l for l in data]
Out[55]: 
[[{'font-weight': '1'},
  {'font-weight': '1'},
  {'font-weight': '2'},
  {'font-weight': '2'}],
 [{'font-weight': '3'}, {'font-weight': '3'}, {'font-weight': '3'}],
 [{'font-weight': '1'},
  {'font-weight': '1'},
  {'font-weight': '2'},
  {'font-weight': '2'}],
 [{'font-weight': '3'}, {'font-weight': '3'}]]

Answer 2

试试这个，

result = []
for item in data:
    result.append([i for j in item for i in j])

包含列表理解的单行代码，

[[i for j in item for i in j] for item in data]

替代方法，

import numpy as np
[list(np.array(i).flat) for i in data]

<强>结果

[[{'font-weight': '1'},
  {'font-weight': '1'},
  {'font-weight': '2'},
  {'font-weight': '2'}],
 [{'font-weight': '3'}, {'font-weight': '3'}, {'font-weight': '3'}],
 [{'font-weight': '1'},
  {'font-weight': '1'},
  {'font-weight': '2'},
  {'font-weight': '2'}],
 [{'font-weight': '3'}, {'font-weight': '3'}]]

Answer 3

遍历列表并检查每个项目是否为列表列表。如果这样压平它。

data = [
         [[{'font-weight': '1'},{'font-weight': '1'}],[{'font-weight': '2'},{'font-weight': '2'}]],
         [{'font-weight': '3'},{'font-weight': '3'},{'font-weight': '3'}],
         [[{'font-weight': '1'},{'font-weight': '1'}],[{'font-weight': '2'},{'font-weight': '2'}]],
         [{'font-weight': '3'},{'font-weight': '3'}]
       ]
for n, each_item in enumerate(data):
   if any(isinstance(el, list) for el in each_item):
        data[n] = sum(each_item, [])

print data

如何使用列表推导在循环内进行列表展平？

3 个答案: