通过以列表为值的python字典进行迭代

Question

I'm trying to iterate through a dictionary that looks like:

    d = {
    "list_one": [
        "hello",
        "two",
        "three"
    ],
    "list_two": [
        "morning",
        "rain"
    ]
}

I'm using the function:

def combine_words(d):
    for k, v in d.items():
        a = {k: ("|".join(v))}
    return a

When I run this with print, my output is just one key, value pair. I'm not sure what is happening here. My ideal out put would be:

{
'list_one': 'hello|two|three',
'list_two': 'morning|rain'
}

Answer 1

def combine_words(d):
    for k, v in d.items():
        a = {k: ("|".join(v))}
    return a

This constantly reassigns the dictionary to a, and isn't combining the results

def combine_words(d):
    a = {}
    for k, v in d.items():
        a[k] =  ("|".join(v))
    return a

Would keep adding new entries to the dictionary

Answer 2

a gets replaced by a new dict each time through your loop. You want a dict comprehension.

def combine_words(d):
    return {k: "|".join(v) for k, v in d.items()}

Answer 3

The problem is that you change the value of a during every iteration of your for loop because you reassign a . You can modify your code to get around this:

for k, v in d.items():
    a[k] = "|".join(v)

Alternatively, you can use a dict comprehension:

return {k: "|".join(v) for k, v in d.items()}