CODEWITHSUNDEEP
pythonlistpython-2.7dictionary
Ishan Bhatt
Ishan Bhatt

Reputation: 10269

How to get a value from a list of dictionary Python?

I have a list(stop_list) of dictionary. the structure of dictionary is 

stop_id : ""
stop_name : ""

Now, I have a name string which is to be matched against stop_name, and get the stop_id corresponding to that.

Is there any efficient way of doing it? I could come up with only for loop.

for entry in stop_list:
            if name = entry['name']:
                id = entry['id']
                break

Upvotes: 0

Views: 111

Answers (2)

Azurtree
Azurtree

Reputation: 379

After looking at the code it seems that you get the dictionary having the same name. If your name are unique you should consider using a dict for your dicts where the key would be the name.

1 - this will allow you not to browse the list (it is costly compared to dict lookup)

2 - this is more readable

3 - you call entry['name'] only once

Let's say that your stopdict would look like that

stopdict= {
    'stop1' : {'name' : 'stop1', 'id' : 1}
    'stop2' : {'name' : 'stop2', 'id' : 2}
}

accessing the id would look like that : 

stopdict[entry['name']]['id']

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239653

You can use generator expression and next function, like this

next(entry['id'] for entry in stop_list if entry['name'] == name)

This will iterate through the stop_list and when it finds a match, it will yield entry['id']. This will be better because this doesn't have to iterate the entire list.

Another advantage is, if there are more than one matches, then you can use the same expression to get the next id also, like this

>>> ids = next(entry['id'] for entry in stop_list if entry['name'] == name)
>>> next(ids)
# you will get the first matching id here
>>> next(ids)
# you will get the second matching id here

If there is going to be more than one lookups, and given the names are unique, then preprocess the list and create a dictionary, like this

lookup = {entry['name']: entry['id'] for entry in stop_list}

then you can do lookups in constant time, with lookup[name]. This would be the most efficient way if the names are unique and if there are more than one lookups

Upvotes: 4

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