从字典列表中删除所有元素，并重复一个（键，值）

Question

i have this list of dictionaries, and i want to remove those dictionaries with same genre until have only one, for example if i have two dictionaries with same genre ( Kids) i want only the first one, and the others, should be removed.

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Source

[
     {'genre': 'Kids', 'jpg': 'as.jpg', 'file': '01-26'},
     {'genre': 'Kids', 'jpg': 'la.jpg', 'file': '02-02'},
     {'genre': 'Action', 'jpg': 'na.jpg', 'file': '01-41'},
     {'genre': 'Action', 'jpg': 'lo.jpg', 'file': '00-17'}, 
     {'genre': 'Drama', 'jpg': 'do.jpg', 'file': '01-54'}
]

Output:

[
     {'genre': 'Kids', 'jpg': 'as.jpg', 'file': '01-26'},
     {'genre': 'Action', 'jpg': 'na.jpg', 'file': '01-41'},
     {'genre': 'Drama', 'jpg': 'do.jpg', 'file': '01-54'}
]

Answer 1

You could do:

data = [
     {'genre': 'Kids', 'jpg': 'as.jpg', 'file': '01-26'},
     {'genre': 'Kids', 'jpg': 'la.jpg', 'file': '02-02'},
     {'genre': 'Action', 'jpg': 'na.jpg', 'file': '01-41'},
     {'genre': 'Action', 'jpg': 'lo.jpg', 'file': '00-17'},
     {'genre': 'Drama', 'jpg': 'do.jpg', 'file': '01-54'}
]


seen = set()
result = []
for e in data:
    if e['genre'] not in seen:
        seen.add(e['genre'])
        result.append(e)

print(result)

Output

[{'file': '01-26', 'jpg': 'as.jpg', 'genre': 'Kids'}, {'file': '01-41', 'jpg': 'na.jpg', 'genre': 'Action'}, {'file': '01-54', 'jpg': 'do.jpg', 'genre': 'Drama'}]

Answer 2

Suppose your list is l . Just scan the list by the end, and stock items in a dictionary with genre as keys. Only the first of each genre will stay, hiding the others. Then you just have to drop the key, keeping the values.

stock = {d['genre']:d for d in reversed (l)}
print( [v for v in stock.values()] )

result:

[{'genre': 'Drama', 'jpg': 'do.jpg', 'file': '01-54'}, 
{'genre': 'Action', 'jpg': 'na.jpg', 'file': '01-41'}, 
{'genre': 'Kids', 'jpg': 'as.jpg', 'file': '01-26'}]

Answer 3

As usual there are multiple ways:

data = [
 {'genre': 'Kids', 'jpg': 'as.jpg', 'file': '01-26'},
 {'genre': 'Kids', 'jpg': 'la.jpg', 'file': '02-02'},
 {'genre': 'Action', 'jpg': 'na.jpg', 'file': '01-41'},
 {'genre': 'Action', 'jpg': 'lo.jpg', 'file': '00-17'}, 
 {'genre': 'Drama', 'jpg': 'do.jpg', 'file': '01-54'}
]

Here the solution posted before as a function:

def filter1(dict_list, by='grenre'):
    seen   = []
    result = []
    for e in dict_list:
        if e[by] not in seen:
            seen.append(e[by])
            result.append(e)
    return result

And here a long-ish concatenation of lists:

def filter2(dict_list, by='grenre'):
    result = [[e for e in dict_list if e[by]==key][0] for key in set([e[by] for e in dict_list])]
    return result

And a test:

filter_key = 'genre'
print(filter1(data, by=filter_key))
print(filter2(data, by=filter_key))