在 Python 中使用 enumerate() 访问下一个索引 (+1)

Question

First post here, so any protocol/style things please let me know. I'm only about 2 mos into programming so still figuring out the ropes.

I'm trying to take a string (ex.'AAAABBBCCDAABBB') and return only discrete occurrences of a character while preserving the original order. If a letter repeats after another new letter has been introduced that is ok. So a successful return would be ['A', 'B', 'C', 'D', 'A', 'B'].

After many different ways, I'm trying to use enumerate() such as:

#If j doesn't match the j with an index of + 1, assign it to the list.

def unique_in_order(iterable):
    output= []
    for i, j in enumerate(iterable):
        if j != j[i + 1]:  #Pretty sure the issue is in this line
            output.append(j)
    return output
        

I'm not 100% on the syntax of accessing the index + 1 of j. I'm also getting a string index out of range error here, which I'm assuming is when the iteration gets to the last j that means the next index is off the edge of the cliff and returns the error. I haven't been able to find/generate a good solution to this. Something with len(j) I'm guessing but coming up short.

thanks!

Answer 1

You could wrap everything in a try-catch block. If an IndexError exception is raised, we know we must have hit the end

def unique_in_order(iterable):
    output = []
    for index, element in enumerate(iterable):
        try:
            if element != iterable[index+1]:
                output.append(element)
        except IndexError:
            output.append(element)
            break
    return output

print(unique_in_order("AABCCCDDDDDEE"))

Output:

['A', 'B', 'C', 'D', 'E']
>>> 

Or, like the comments suggest, just use itertools.groupby :

def unique_in_order(iterable):
    from itertools import groupby

    return [key for key, _ in groupby(iterable)]

print(unique_in_order("AABCCCDDDDDEE"))

Output:

['A', 'B', 'C', 'D', 'E']
>>> 

Answer 2

You could just check that the last item in output is not the same as the current value

def unique_in_order(iterable):
    output= []
    for j in iterable:
        # add to list if output empty
        #  or current value not equal to last value
        # in output
        if not output or j != output[-1]:
            output.append(j)
    return output

print(unique_in_order('AAAABBBCCDAABBB')) # ['A', 'B', 'C', 'D', 'A', 'B']

Or Correcting Posted Approach

Rather than this

if j != j[i + 1]:  

You probably meant

if i < len(iterable) -1 and j != iterable[i + 1]: 

But, this is still incorrect. It won't pick up the last B in the end group '...ABBB'

Better

def unique_in_order(iterable):
    output= []
    for i, j in enumerate(iterable):
        # On first index or
        # Adds first element of group
        # so even if group starts on last item in string
        # it still gets added
        if i == 0 or j != iterable[i-1]:
            output.append(j)
    return output

 print(unique_in_order('AAAABBBCCDAABBB')) # ['A', 'B', 'C', 'D', 'A', 'B']