如何根据python中的条件在列表中组合或保留字符串? [英] how to combine or leave strings in the lists depending on the condition in python?
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问题描述
我有三个列表:
li1 = ["a", "a", "a", "a", "b", "b", "a", "a", "b"]
li2 = ["a", "a", "a", "b", "a,", "b", "a", "a"]
li3 = ["b", "b", "a", "a", "b"]
我想按"b"
结果应该看起来像这样:
The result is supposed to look like this:
li1 = ["aaaa", "b", "b", "aa", "b"]
li2 = ["aaa", "b", "a", "b", "aa"]
li3 = ["b", "b", "aa", "b"]
但是我不知道该如何处理...请帮助我!
But I don't know how to approach this... please help me!
推荐答案
If you want to join groups not belonging to a certain key
from itertools import groupby
def join_except_key(iterable, key='b'):
groups = groupby(iterable)
for k, group in groups:
if k != key:
yield ''.join(group) # more general: ''.join(map(str, group))
else:
yield from group
演示:
>>> li1 = ["a", "a", "a", "a", "b", "b", "a", "a", "b", "c", "c", "b", "c", "c"]
>>> list(join_except_key(li1))
['aaaa', 'b', 'b', 'aa', 'b', 'cc', 'b', 'cc']
如果要加入属于某个键的组
from itertools import groupby
def join_by_key(iterable, key='a'):
groups = groupby(iterable)
for k, group in groups:
if k == key:
yield ''.join(group) # more general: ''.join(map(str, group))
else:
yield from group
演示:
>>> li1 = ["a", "a", "a", "a", "b", "b", "a", "a", "b", "c", "c", "b", "c", "c"]
>>> list(join_by_key(li1))
['aaaa', 'b', 'b', 'aa', 'b', 'c', 'c', 'b', 'c', 'c']
groupby
产生的内容的详细信息(join_except_key
的非生成器方法)
Details on what groupby
produces (non generator approach for join_except_key
)
>>> li1 = ["a", "a", "a", "a", "b", "b", "a", "a", "b", "c", "c", "b", "c", "c"]
>>> groups = [(k, list(group)) for k, group in groupby(li1)]
>>> groups
[('a', ['a', 'a', 'a', 'a']),
('b', ['b', 'b']),
('a', ['a', 'a']),
('b', ['b']),
('c', ['c', 'c']),
('b', ['b']),
('c', ['c', 'c'])]
>>>
>>> result = []
>>> for k, group in groups:
...: if k != 'b':
...: result.append(''.join(group))
...: else:
...: result.extend(group)
...:
>>> result
['aaaa', 'b', 'b', 'aa', 'b', 'cc', 'b', 'cc']
第二行中的列表理解groups = [...
仅用于检查分组操作的元素,仅使用groups = groupby(li1)
即可正常工作.
The list comprehension groups = [...
in the second line was only needed for inspecting the elements of the grouping operation, it works fine with just groups = groupby(li1)
.
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