How to find dict value by Key in list

Question

I have following List:

lst = [
    {'Title1': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name'}},
    {'Title2': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name2'}},
]

Now I want to search list by word Title1

How to get values of Link and Episode of certain key

Answer 1

Using a list comprehension to produce all matches:

[d[searchtitle] for d in lst if searchtitle in d]

where searchtitle contains 'Title1' . The result is a list of matching dictionaries.

Finding the first match only:

next((d[searchtitle] for d in lst if searchtitle in d), None)

which returns None if there is no match, or a dictionary with the matching key in it.

Demo:

>>> lst = [
...     {'Title1': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name'}},
...     {'Title2': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name2'}},
... ]
>>> searchtitle = 'Title1'
>>> [d[searchtitle] for d in lst if searchtitle in d]
[{'Episode': 'Episode Name', 'Link': 'ZbELsW_tyWA'}]
>>> next((d[searchtitle] for d in lst if searchtitle in d), None)
{'Episode': 'Episode Name', 'Link': 'ZbELsW_tyWA'}

Instead of storing each title as a separate dictionary in a list, your search would be much simpler if you just stored each title as a key in one dictionary:

titles = {
    'Title1': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name'},
    'Title2': {'Link': 'ZbELsW_tyWA', 'Episode': 'Episode Name2'},
}

as now all you have to do to get the nested dictionary is directly reference the title key:

titles['Title1']

Provided your titles are unique.

Answer 2

You can do something like this

for item in lst:
    for key,value in item.iteritems():
        if key == "Title1":
            link = value["Link"]
            episode = value["Episode"]

Answer 3

You can use the filter function. The first argument to the function is a function which would help filter the list. The function we need here needs to check if 'Title1' exists in the dictionary. So this expression would give you your answer :

filter(lambda x : 'Title1' in x, lst)