如何返回列表中元素的轮次？

Question

I want to know if an element is in the list multiple times and if so what is the order of the 2nd one. The 2nd one is important, not the 3rd or the 4th one.

list = [1, 3, 4, 2, 0, 1, 6, 7, 0]

My expected output is:

"1" is in list 2 times and the order of the 2nd "1" is six .

Answer 1

Doing it with one pass over the list, while saving the indexes of the target number and checking their amount in the end:

l = [1, 3, 4, 2, 0, 1, 6, 7, 0]
target = 1

idxs = []
for i, num in enumerate(l):
    if num == target:
        idxs.append(i)

if len(idxs) > 1:
    print(f'"{target}" is in the list {len(idxs)} times and the order of the 2nd "{target}" is {idxs[1]+1}')
else:
    print(f'{target} is in the list 0 or 1 times')

The indexes can also be obtained with a neat list-comprehension:

idxs = [i for i, num in enumerate(l) if num == target]

Answer 2

Pretty rough code, just to convey the logic you can apply to get it done:

list = [1, 3, 4, 2, 0, 1, 6, 1, 7, 0]

flagged = []
for index, elem in enumerate(list):
    elem_count = list[index:].count(elem)
    if elem_count > 1 and elem not in flagged:
        flagged.append(elem)
        temp_index = list[index + 1:].index(elem)
        actual_index = index + temp_index + 1
        print(str(elem) + ' is in the list ' + str(elem_count) + ' times and the order of 2nd is ' + str(actual_index))

# OP
# 1 is in the list 3 times and the order of 2nd is 5
# 0 is in the list 2 times and the order of 2nd is 9

Answer 3

Are you sure it's 6 and not 5? Indexing starts from 0. In any case, you can use index for finding it:

first_idx = list.index(element)
print(list.index(element, first_idx + 1))

Like that you will find the first occurrence of the element and than return the index of the second one. If you want it 6 and not 5 - increment it by 1.