What is the pythonic way to extract values from dict

Question

What is the best way to extract values from dictionary. Let's suppose we have a list of dicts:

projects = [{'project': 'project_name1', 
             'dst-repo': 'some_dst_path', 
             'src-repo': 'some_src_path', 
             'branches': ['*']},
            {...}, 
            {...}]

Now I just iterate through this dictionary and get values, something like:

   for project in projects:
       project_name = project.get('project')
       project_src = ....
       project_dst = ....
       ....
       ....

So the question is: "Are there any more pythonic approaches to extract values by key from dictionary that allow not making so many lines of code for new variable assignment?"

Answer 1

There's nothing wrong with what you're doing, but you can make it more compact by using a list comprehension to extract the values from the current dictionary. Eg,

projects = [
    {
        'project': 'project_name1', 
        'dst-repo': 'some_dst_path', 
        'src-repo': 'some_src_path', 
        'branches': ['*']
    },
]

keys = ['project', 'src-repo', 'dst-repo', 'branches']
for project in projects:
    name, src, dst, branches = [project[k] for k in keys]
    # Do stuff with the values
    print(name, src, dst, branches)

output

project_name1 some_src_path some_dst_path ['*']

However, this approach gets unwieldy if the number of keys is large.

---

If keys are sometimes absent from the dict, then you will need to use the .get method, which returns None for missing keys (unless you pass it a default arg):

name, src, dst, branches = [project.get(k) for k in keys]

If you need specific default for each key, you could put them into a dict, eg

defaults = {
    'project': 'NONAME',
    'src-repo': 'NOSRC',
    'dst-repo': 'NODEST',
    'branches': ['*'],
}

projects = [
    {
        'project': 'project_name1', 
        'src-repo': 'some_src_path', 
    },
]

keys = ['project', 'src-repo', 'dst-repo', 'branches']
for project in projects:
    name, src, dst, branches = [project.get(k, defaults[k]) for k in keys]
    # Do stuff with the values
    print(name, src, dst, branches)

output

project_name1 some_src_path NODEST ['*']

Answer 2

out = [elt.values() for elt in projects]

Answer 3

   for project in projects:
       project_name = project['project']
       project_src = ....
       project_dst = ....
       ....
       ....

I'm not sure you can get less typing

//EDIT: Ok, it looks I misunderstood the question: Assume we have a list of dicts like this:

projects = [ {'project': "proj1", 'other': "value1", 'other2': "value2"},
             {'project': "proj2", 'other': "value3", 'other2': "value4"},
             {'project': "proj2", 'other': "value3", 'other2': "value4"} ]

To extract the list of project fields, you can use the following expression:

projects_names =  [x['project'] for x in projects]

This will iterate over project list, extracting the value of 'project' key from each dictionary.