[英]remove/list duplicates by comparing multiple lists in python
我知道已经要求删除/列出列表中的重复项。 我在使它同时比较多个列表时遇到问题。
lst = [item1, item2, item3, item4, item5]
a = [1,2,1,5,1]
b = [2,0,2,5,2]
c = [0,1,0,1,5]
如果这些是我的列表,我想像使用zip函数一样比较它们。 我想检查列表中的索引0、2和4是否重复,如果那些相同的索引是其他列表的重复,那么例如列表b中的0、2和4也是重复的,但列表c中的0和2是唯一的重复因此,我只希望从结果列表中首先列出索引0和2 [item1,item3]
我将如何采用此定义来做到这一点?
def list_duplicates(seq):
seen = set()
seen_add = seen.add
# adds all elements it doesn't know yet to seen and all other to seen_twice
seen_twice = set( x for x in seq if x in seen or seen_add(x) )
# turn the set into a list (as requested)
return list( seen_twice )
a = [1,2,3,2,1,5,6,5,5,5]
list_duplicates(a) # yields [1, 2, 5]
您正在尝试确定哪些公共索引包含多个列表中的重复值,而不是跟踪重复值本身。 这意味着,除了跟踪给定seq
重复的项目外,我们还需要跟踪找到重复项目的索引。 这很容易添加到现有方法中:
from collections import defaultdict
def list_duplicates(seq):
seen = set()
seen_twice = set()
seen_indices = defaultdict(list) # To keep track of seen indices
for index, x in enumerate(seq): # Can't use a comprehension now, too much logic in there.
seen_indices[x].append(index)
if x in seen:
seen_twice.add(val)
else:
seen.add(val)
print seen_indices
return list( seen_twice )
if __name__ == "__main__":
a = [1,2,3,2,1,5,6,5,5,5]
duped_items = list_duplicates(a)
print duped_items
输出:
defaultdict(<type 'list'>, {1: [0, 4], 2: [1, 3], 3: [2], 5: [5, 7, 8, 9], 6: [6]})
[1, 2, 5]
因此,现在除了追踪值本身之外,我们还将追踪所有重复值的索引。
下一步是以某种方式将其应用于多个列表。 我们可以利用以下事实:遍历一个列表之后,我们将消除一堆我们不指向重复值的索引,而仅对后续列表中已知重复的索引进行迭代。 这需要稍微修改一下逻辑,以遍历“可能重复的索引”而不是遍历整个列表:
def list_duplicates2(*seqs):
val_range = range(0, len(seqs[0])) # At first, all indices could be duplicates.
for seq in seqs:
# Set up is the same as before.
seen_items = set()
seen_twice = set()
seen_indices = defaultdict(list)
for index in val_range: # Iterate over the possibly duplicated indices, not the whole sequence
val = seq[index]
seen_indices[val].append(index)
if val in seen_items:
seen_twice.add(val)
else:
seen_items.add(val)
# Now that we've gone over the current valid_range, we can create a
# new valid_range for the next iteration by only including the indices
# in seq which contained values that we found at least twice in the
# current valid_range.
val_range = [duped_index for seen_val in seen_twice for duped_index in seen_indices[seen_val]]
print "new val_range is %s" % val_range
return val_range
if __name__ == "__main__":
a = [1,2,1,5,1]
b = [2,0,2,5,2]
c = [0,1,0,1,5]
duped_indices = list_duplicates2(a, b, c)
print "duped_indices is %s" % duped_indices
输出:
new val_range is [0, 2, 4]
new val_range is [0, 2, 4]
new val_range is [0, 2]
duped_indices is [0, 2]
正是您想要的。
在此列表中搜索重复项
l = [[a[i],b[i],c[i]] for i in range(len(a))]
对于您的示例,它将产生以下列表:
[[1, 2, 0], [2, 0, 1], [1, 2, 0], [5, 5, 1], [1, 2, 5]]
然后:
result = [lst[i] for (i,x) in enumerate(l) if x in list_duplicates(l)]
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