从列表列表到列表列表列表

Question

我有这样一份清单：

first_list = [[ 1.        , 45.4,  9.1],
              [ 2.        , 45.5,  9.1],
              [ 2.        , 45.4,  9.2],
              [ 2.        , 45.4,  9.2],
              [ 3.        , 45.4,  9.1],
              [ 3.        , 45.4,  9.1],
              [ 3.        , 45.4,  9.1] ]

我想使用folio函数HeatMapWithTime，为此我需要根据每个子列表（1,2,3.ecc）的第一项对上面的数据进行分组：

new_list = [ [ [45.4, 9.1] ],                               # All coords for 1.
             [ [45.5, 9.1], [45.4, 9.2], [45.4, 9.2] ],     # All coords for 2.
             [ [45.4, 9.1], [45.4, 9.1], [45.4, 9.2] ] ]    # All coords for 3.

我该怎么做

Answer 1

可以在字典中收集所有坐标：

res = {}
for entry in first_list:
    res.setdefault(entry[0], []).append(entry[1:])

这将为您提供：

>>> res
{1.0: [[45.4, 9.1]],
 2.0: [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]],
 3.0: [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]}

如果列表已排序，请将值转换为列表（仅限Python 3.6+）：

>>> list(res.values())
[[[45.4, 9.1]],
 [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]],
 [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]]

否则，您需要首先对它们进行排序：

>>> [res[key] for key in sorted(res.keys())]
[[[45.4, 9.1]],
 [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]],
 [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]]

Answer 2

我会使用dict，如果您需要它作为一个列表，您可能希望将它带回一个列表，但是使用dict进行分组通常是有帮助的：

first_list = [[ 1.        , 45.4,  9.1],
              [ 2.        , 45.5,  9.1],
              [ 2.        , 45.4,  9.2],
              [ 2.        , 45.4,  9.2],
              [ 3.        , 45.4,  9.1],
              [ 3.        , 45.4,  9.1],
              [ 3.        , 45.4,  9.1] ]
result = dict()
for group, *values in first_list:
    if group not in result:
        result[group] = [values]
    else:
        result[group].append(values)
print(result)
### if you want it back as a list:
result_list = [v for k,v in result.items()]
print(result_list)

输出：

#dict:
{1.0: [[45.4, 9.1]], 2.0: [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]], 3.0: [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]}
#list:
[[[45.4, 9.1]], [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]], [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]]

Answer 3

假设列表按第一个元素排序，您可以使用^{}：

from itertools import groupby
from operator import itemgetter
[[i[1:] for i in v] for k,v in groupby(first_list, itemgetter(0))]

#[[[45.4, 9.1]],
# [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]],
# [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]]