CODEWITHSUNDEEP
python-3.xlistdictionary
KoalaKey
KoalaKey

Reputation: 242

How do I print out each value for a dictionary item in a list

I have a list of 'id' and their vaules. I am trying to loop through the list and print out the values of each 'id'. It has to be done in one line though. I am struggling to come up with the solution though. The closest I have gotten is below:

ls = [
{'id':'1'},{'id':'2'},{'id':'3'},{'id':'4'},{'id':'5'},
{'id':'6'},{'id':'7'},{'id':'8'},{'id':'9'},{'id':'10'},
{'id':'11'},{'id':'12'},{'id':'13'},{'id':'14'},{'id':'15'},
{'id':'16'},{'id':'17'},{'id':'18'},{'id':'19'},{'id':'20'},
{'id':'21'},{'id':'22'},{'id':'23'},{'id':'24'},{'id':'25'}
]

print([(ls[num]['id']) for num in range(25)])

This gives me the values for each 'id' in the list which is great! But, it returns them all in a list. I need them all on separate lines. Like so:

'1'
'2'
'3'
'4'
'5'

Any ideas how I can go about this? It has to be done with a one-liner.

Upvotes: 0

Views: 27

Answers (1)

Carcigenicate
Carcigenicate

Reputation: 45741

If you really want a one liner, you can make use of unpacking and the sep argument of print:

print(*[(ls[num]['id']) for num in range(25)], sep="\n")

* unpacks the elements into the arguments of print, then sep tells print to add a newline between each printed argument.

I think this would be neater by making use of a full for loop though, with separate prints:

for elem in [(ls[num]['id']) for num in range(25)]:
    print(elem)

I'd personally only use that one-liner for quick debugging.

Upvotes: 1

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