python - 如果重复的项目按特定顺序出现，则从列表中删除它们

Question

如果项目以特定顺序出现在列表中，我会尝试从列表中删除它们

我尝试过以下代码；

 a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]

 b = ["ijk", "123", "456","123", "rst", "xyz" ]
 counter=0
 for i in b[:]:
      print(i)
      counter=counter+1
      print(counter)
      if i in a and i in a[counter+2]:
            print(a)
            print(">>>>>",a[counter+2])
            b.remove(i)

  print(b)

我正在寻找以下输出

b = ["ijk", "123", "456","123"]

从 b 中删除了 ["rst", "xyz"]，因为它们在 a 中处于后 2 个后序列中。

Answer 1

这里有一种方法，即利用成对的查找来满足“序列”标准。

这个想法是预先构建“a”或“lookup_list”中所有对的集合。之后，使用对来迭代b。如果找到一对，则设置一个标志以跳过两个元素(当前元素和下一个元素)。否则，附加第一个项目，因为可以保证它不会与顺序中的下一个项目一起出现。

演示:

from itertools import zip_longest

def remove_dupes_in_seq(lookup_list, b):
    ''' lookup_list: list from which you need to check for sequences
    b: list from which you need to get the output that 
    has all elements of b that do not occur in a sequence in lookup_list.
    '''
    pair_set_lookup = set(zip(lookup_list, lookup_list[1:]))
    #make a set of all pairs to check for sequences
    result = []
    skip_next_pair = False #boolean used to indicate that elements need to be skipped
    for pair in zip_longest(b, b[1:]):
        if skip_next_pair:
            #set the boolean according to current pair, then perform a skip
            skip_next_pair = pair in pair_set_lookup
            continue
        if pair in pair_set_lookup:
            #pair found. set flag to skip next element
            skip_next_pair = True
        else:
            #the first item is guaranteed to not occur in a sequence. Append it to output.
            result.append(pair[0])
    return result

a = ["abc", "def", "ijk", "lmn", "opq", "rst", "xyz"]
b = ["ijk", "123", "456","123", "rst", "xyz" ]
out = remove_dupes_in_seq(a, b) #['ijk', '123', '456', '123']
b2 = ["ijk","lmn","456","123","rst","xyz"]
out2 = remove_dupes_in_seq(a, b2) #['456', '123']