使用Python列出列表中重复值的索引

Question

我想修改这个列出重复项的定义，使其列出重复值的索引。另外，我希望它列出所有的重复项，这意味着对于a = [1,2,3,2,1,5,6,5,5,5]，结果应为duplicate_indexes = [3,4,7,8,9]。以下是定义：

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]

- konrad

Answer 1

使用列表推导式打印重复项的索引。它会将列表切片到所选索引处，并且如果该项已经存在于切片列表中，则返回索引值。

a= [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result=[idx for idx, item in enumerate(a) if item in a[:idx]]
print result #[3, 4, 7, 8, 9]

- Karthik

---

与其他答案相比，这个规范表述最简洁明了，因此获得加一分。 - MarnixKlooster ReinstateMonica

Answer 2

a, seen, result = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5], set(), []
for idx, item in enumerate(a):
    if item not in seen:
        seen.add(item)          # First time seeing the element
    else:
        result.append(idx)      # Already seen, add the index to the result
print result
# [3, 4, 7, 8, 9]

编辑：在该函数中，您可以只使用列表推导式，例如：

def list_duplicates(seq):
    seen = set()
    seen_add = seen.add
    return [idx for idx,item in enumerate(seq) if item in seen or seen_add(item)]

print list_duplicates([1, 2, 3, 2, 1, 5, 6, 5, 5, 5])
# [3, 4, 7, 8, 9]

- thefourtheye

---

你正在使用一个集合seen来加快成员测试的速度？ - wwii

Answer 3

def list_duplicates_index(seq):
    return [i for (i,x) in enumerate(a) if x in list_duplicates(a)]

- jspurim

Answer 4

def list_duplicates(seq):
    d = {}
    for i in seq:
        if i in d:
            d[i] += 1
        else:
            d[i] = 1
    dups = []
    for i in d:
        if d[i] > 1:
            dups.append(i)
    lst = []
    for i in dups:
        l = []
        for index in range(len(seq)):
            if seq[index] == i:
                l.append(index)
        lst.append(l[1:])
    new = []
    for i in lst:
        for index in i:
            new.append(index)   
    return new

- user3552506