python - 将元组分组到列表中

Question

在 Python 中，使用公共(public)索引对元组进行分组的最佳方法是什么？

(2, 3, 'z')
(1, 1, 'abc')
(2, 1, 'stu')
(1, 2, 'def')
(2, 2, 'vxy')

结果将是:

[((1, 1, 'abc'),(1, 2, 'def')]
[((2, 1, 'stu'),(2, 2, 'vxy'), (2, 2, 'vxy')]

目标是将第三个元素连接成单个字符串对象。

这是连接部分，但我不确定分组。

def sort_tuples(list_input):
    new = sorted(list_input)
    str = ''
    for i in range(0, len(new)):
        str = str + new[i][2]
    return str

Answer 1

使用字典进行分组；选择您的分组元素并将您想要连接的内容附加到每个键的列表中:

groups = {}
for first, second, third in list_input:
    groups.setdefault(first, []).append(third)

然后您可以连接每个列表:

for key, group in groups.items():
    print(key, ''.join(group))

由于您只想连接每个元组的第三个元素，因此我没有费心在字典中包含第二个元素，但您也可以自由地将整个元组存储在组列表中。

演示:

>>> list_input = [
...     (2, 3, 'z'),
...     (1, 1, 'abc'),
...     (2, 1, 'stu'),
...     (1, 2, 'def'),
...     (2, 2, 'vxy'),
... ]
>>> groups = {}
>>> for first, second, third in list_input:
...     groups.setdefault(first, []).append(third)
... 
>>> for key, group in groups.items():
...     print(key, ''.join(group))
... 
1 abcdef
2 zstuvxy

如果第二个键被用作排序键，那么您必须在分组时包含它；然后您可以排序并提取第三个:

groups = {}
for first, second, third in list_input:
    groups.setdefault(first, []).append((second, third))

for key, group in groups.items():
    print(key, ''.join([third for second, third in sorted(group)]))

演示:

>>> groups = {}
>>> for first, second, third in list_input:
...     groups.setdefault(first, []).append((second, third))
... 
>>> for key, group in groups.items():
...     print(key, ''.join([third for second, third in sorted(group)]))
... 
1 abcdef
2 stuvxyz

由于这涉及到排序，所以不妨对整个输入列表进行一次排序，并使用 itertools.groupby()排序后对输入进行分组:

from itertools import groupby

for key, group in groupby(sorted(list_input), key=lambda t: t[0]):
    print(key, ''.join([third for first, second, third in group]))

再次演示一下这种方法:

>>> from itertools import groupby
>>> for key, group in groupby(sorted(list_input), key=lambda t: t[0]):
...     print(key, ''.join([third for first, second, third in group]))
... 
1 abcdef
2 stuvxyz

字典分组方法是 O(N) 算法)，一旦添加排序，它就变成 O(NlogN) 算法。