Print next element of a list, if it matches the previous element

Question

I have a list with similar elements, and I want to print the next element if I match the previous one. So something like this:

my_list=['one', 'two', 'one', 'one', 'two']
for ml in my_list:
    if ml =='one':
        print (next)

so the output would be:

two

two  

Answer 1

Try this:

my_list=['one', 'two', 'one', 'one', 'two']
i = 0
for ml in my_list:
    if ml =='one' and i!=(len(my_list)-1):
        print (my_list[i+1])
    i+=1

Output:

two
one
two
[Finished in 0.0s]

Answer 2

This code should do what you want:

my_list=['one', 'two', 'one', 'one', 'two']
for i, item in enumerate(my_list):
    if i != 0:
        if item == my_list[i-1]:
            print(my_list[i+1])

Answer 3

I guess you are looking to print only two. So you need to check if the next value is also not one.

my_list=['one', 'two', 'one', 'one', 'two']
for i, ml in enumerate(my_list):
  if ml =='one' and my_list[i+1] != 'one':
     print (my_list[i+1])

The output of this will be:

two
two

Answer 4

I would actually say a more elegant solution to this problem would be to first convert your list of words into a string

' '.join(my_list)

and then split it on the word 'one' like so

my_string.split('one')

as this would give you a list of all the subsequent words after each 'one' .

>>> ['', ' two ', ' ', ' two']

In summary, a simple one-line solution would be:

' '.join(my_list).split('one')

Answer 5

You can use the zip function to get adjacent pairs:

for a, b in zip(my_list[:-1], my_list[1:]):
  if a == "one":
    print(b)