Issue
I am currently working with a list of dictionaries that is stored in rest_list variable after calling the get_rest(rest) function. I would like to return a list of dictionaries where availability == available (all of them in this case). This is the list of dictionaries in the rest_list variable.
[{'close_time': '5:00 pm',
'open_time': '8:00 am',
'name': 'Pizza DeVille',
'city': 'Las Vegas',
'availability': 'available'},
{'close_time': '10:00 pm',
'open_time': '9:00 am',
'name': 'Cecil’s',
'city': 'Philadelphia',
'availability': 'available'}
The output of the new function should return this (example output):
[{'availability': 'available', 'close_time': '5:00 pm', 'city': 'Las Vegas', 'open_time': '8:00 am', 'name': 'Pizza DeVille'}, {'availability': 'available', 'close_time': '10:00 pm', 'city': 'Philadelphia', 'open_time': '9:00 am', 'name': 'Cecil’s'}
As you can see it seems to be grabbing the last key-value pair and returning it first and then returning to the top. I can basically get it to print in the same order that it's going in but I have no idea how I would grab the last part first, etc. and not do it in an orderly way.
def filter_rest(rest):
rest_list = get_rest(rest) #function that gives me first code chunk above
new_list = []
for x in rest_list:
new_list.append(x)
return new_list
Lastly, the code above is missing an if statement of sorts so that the list of dictionaries returned only includes those where availability == 'available'. Any tips on including an if statement and getting the function to return in the order shown in the example ouput?
Solution
l = [{'close_time': '5:00 pm',
'open_time': '8:00 am',
'name': 'Pizza DeVille',
'city': 'Las Vegas',
'availability': 'available'},
{'close_time': '10:00 pm',
'open_time': '9:00 am',
'name': 'Cecil’s',
'city': 'Philadelphia',
'availability': 'available'}
]
keys=['availability','close_time','city','open_time','name']
l = [ {k:d[k] for k in keys} for d in l]
print(l)
Answered By - islam abdelmoumen
0 comments:
Post a Comment
Note: Only a member of this blog may post a comment.