从Python中模式重复的列表中删除重复字符

使用set方法可以从列表中删除重复项

数据=

 ['','a','a','a','a','a','a','','b','b','b','b','b','b','b','b',
 '','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d',
 '','','e','e','e','e','e','e','','','a','a','a','a','a','a',
 '','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c',
 '','','','d','d','d','d','d','d','','','e','e','e','e','e','e',
 '','','a','a','a','a','a','a','','b','b','b','b','b','b','b','b',
 '','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d',
 '','','e','e','e','e','e','e','','','a','a','a','a','a','a',
 '','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c',
 '','','','d','d','d','d','d','d','','','e','e','e','e','e','e','','']

打印(集合(数据))

from itertools import groupby
from operator import itemgetter
# data <- your data
a = [k for k, v in groupby(data) if k] # approach 1
b = list(filter(bool, map(itemgetter(0), groupby(data)))) # approach 2
assert a == b
print(a)

结果:

['a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e']

类似于:

result = []
before = ''
for e in data:
    if e and e != before:
        result.append(e)
        before = e
print(result)

你可以用 itertools.groupby :

如果你的名单是 ll

ll = [i for i in ll if i]
out = []
for k, g in groupby(ll, key=lambda x: ord(x)):
    out.append(chr(k))
print(out)
#prints ['a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', ...

你需要这样的东西

li = ['','a','a','a','a','a','a','','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','','','e','e','e','e','e','e','','']

new_li = []
e_ = ''

for e in li:
    if len(e) > 0 and e_ != e:
        new_li.append(e)
        e_ = e

print(new_li)

输出

['a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e']

您可以在列表上循环并添加适当的连接。对于您期望的响应,您只需要确定前一个字符是否与当前字符不同

    current_sequence = ['','a','a','a','a','a','a','','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','d','d','','','e','e','e','e','e','e','','','a','a','a','a','a','a','','','','b','b','b','b','b','b','b','b','b','','','c','c','c','c','c','c','','','','d','d','d','d','d','d','','','e','e','e','e','e','e','','']

sequence_list = []
for x in range(len(current_sequence)):
    if current_sequence[x]:
        if current_sequence[x] != current_sequence[x-1]:
            sequence_list.append(current_sequence[x])

print(sequence_list)

下面是一个使用列表理解和 itertools.zip_longest :仅当元素不是空字符串且不等于下一个元素时才保留该元素。您可以使用迭代器跳过第一个元素,以避免切片列表的成本。

from itertools import zip_longest

def remove_consecutive_duplicates(lst):
    ahead = iter(lst)
    next(ahead)
    return [ x for x, y in zip_longest(lst, ahead) if x and x != y ]

用法:

>>> remove_consecutive_duplicates([1, 1, 2, 2, 3, 1, 3, 3, 3, 2])
[1, 2, 3, 1, 3, 2]
>>> remove_consecutive_duplicates(my_list)
['a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'd',
 'e', 'a', 'b', 'c', 'd', 'e']

我假设没有由空字符串分隔的重复项(例如。 'a', '', 'a' ),或者不希望删除此类重复项。如果这个假设是错误的,那么您应该首先过滤掉空字符串:

>>> example = ['a', '', 'a']
>>> remove_consecutive_duplicates([ x for x in example if x ])
['a']