Python：找到第一个x的索引，然后找到第二个x的索引

Question

I am trying to loop through a list and find the index of the first occurrence of x and the index of the second occurrence of x.

I have the following list:

int_list = [1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 14, 15, 15, 17, 17, 14, 20, 20, 5, 23, 23, 25, 26, 26, 28, 28, 25, 31, 32, 32, 31, 35, 35, 4, 3, 39, 39, 41, 42, 42, 44, 44, 41, 47, 47, 2, 50, 50, 1, 53, 54, 55, 56, 57, 58, 58, 60, 60, 62, 62, 57, 65, 66, 67, 67, 66, 65, 56, 72, 73, 74, 75, 75, 74, 78, 78, 73, 72, 82, 83, 83, 85, 85, 82, 55, 89, 90, 91, 92, 93, 93, 92, 91, 90, 98, 99, 99, 98, 102, 103, 103, 102, 106, 106, 89, 109, 109, 54, 112, 113, 113, 112, 116, 116, 53, 119, 120, 121, 121, 120, 119, 0, 126, 127, 128, 128, 130, 131, 131, 130, 127, 135, 135, 126]

I would like to start from index 0 which is a 1, as this is the first occurrence assign this to first_index, then search the rest of the list until I find another 1 which is at index 1 and so on.

I understand I need to use enumerate() which will loop through my list and allow me to find the index, however I am struggling with how to find the first and second occurrences

for example:

for i, e in enumerate(int_list):
    # if e 'is the first occurrence':
         first_index = i
    # else e is second occurrence:
         second_index = i

I don't know how to determine whether e is the first or second occurrence

I have tried the following code:

for i in int_list:

    first_ind = firstInt_ls.index(i, 0)
    second_ind = firstInt_ls.index(i, first_ind+1)

    print(first_ind)
    print(second_ind)

Which works unless there is only one occurrence. If there is not a second occurrence it produces a ValueError: 10 is not in list

Answer 1

Use list's index() function, where the first argument is the element you are trying to find and the second argument is the index to start from. Maybe something like this:

a = [1, 2, 3, 1, 4, 5, 6, 1, 5]

first_ind_of_one = a.index(1, 0)
second_ind_of_one = a.index(1, first_ind_of_one+1)

print(first_ind_of_one)
print(second_ind_of_one)

Code below for case when we don't know if and how many occurrences of element in list:

occurrences_needed = 2
list_of_inds = []
for ind, elem in enumerate(a):
    if len(list_of_inds) == occurrences_needed:
        break
    if elem == 1:
        list_of_inds.append(ind)

print(list_of_inds)

Just for fun, I have added a third method that is easier to read but may be inefficient for large lists (some testing may be required):

s = ''.join(str(i) for i in a)
first_ind = s.find('1')
second_ind = s.find('1', first_ind + 1)
print(first_ind)
print(second_ind)

This converts the list to a string and use's string's find() method. This method simply returns -1 in the event where the element is not found. So you may have to check for that in your code.