Python pop dictionary list values

Question

I have a dictionary:

{'dict': [['IE', '5', '-5'], ['UK', '3', '-9']]}

I wish to pop the list values that are outside of the UK, therefore taking the first value within the lists and comparing to see if it is equal to 'UK'.

I currently have:

for k,v in insideUK.items():
    for i in v:
            if i[0] == "UK":
                print(x) 
            else:
                k.pop(v)

I know after the else is wrong but need help!

I wish for the dict to look like this once finished popping values that aren't equal to "UK".

{'dict': [['UK', '3', '-9']]}

score 1 · Answer 1 · answered Mar 19 '18 at 15:53

1

You can use a list comprehension to filter out based on the first element

>>> data = {'dict': [['IE', '5', '-5'], ['UK', '3', '-9']]}
>>> {'dict': [i for i in data['dict'] if i[0] == 'UK']}
{'dict': [['UK', '3', '-9']]}

answered Mar 19 '18 at 15:53

Cory Kramer

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Is there a way to do it using my way instead of using list comprehension? – Overtime Mar 19 '18 at 16:02

score 1 · Answer 2 · answered Mar 19 '18 at 15:55

1

You can also do it using the filter function:

d = {'dict': list(filter(lambda i: 'UK' in i, d['dict']))}
print(d)

Output:

{'dict': [['UK', '3', '-9']]}

answered Mar 19 '18 at 15:55

Vasilis G.

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Maarten Fabré · Accepted Answer · 2018-03-19T16:10:39.040

0

A nested dict and list expression can do this:

{k: [i for i in v if i[0] == 'UK'] for k,v in insideUK.items()}

If you really want to do it with a for-loop and change the list in-place, you could do something like this:

for k,v in insideUK.items():
    for i in v:
            if i[0] == "UK":
                print(x) 
            else:
                v.remove(i)

But it is discouraged strongly to change the list you are iterating over during the iteration

edited Mar 19 '18 at 16:10

answered Mar 19 '18 at 15:53

Maarten Fabré

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for a reason why not to take the second approach, check [this SO question](https://stackoverflow.com/q/6260089/1562285) – Maarten Fabré Mar 19 '18 at 16:23

Python pop dictionary list values

3 Answers3