Aggregation json elements by sub-string
Clash Royale CLAN TAG#URR8PPP
Aggregation json elements by sub-string
I have the following structure:
[
"name": "a-v1",
"date": "2018-05-08T08:40:35.000Z"
,
"name": "a-v2",
"date": "2018-05-20T08:40:35.000Z"
,
"name": "a-v3",
"date": "2018-05-22T08:40:35.000Z"
,
"name": "b-v1",
"date": "2018-02-08T08:40:35.000Z"
,
"name": "b-v2",
"date": "2018-05-08T08:40:35.000Z"
,
"name": "b-v3",
"date": "2018-05-10T08:40:35.000Z"
,
"name": "c-v1",
"date": "2018-10-08T08:40:35.000Z"
,
"name": "c-v2",
"date": "2018-11-08T08:40:35.000Z"
,
"name": "d-v1",
"date": "2018-08-08T08:40:35.000Z"
]
Each name combines from type and version (In a-v1 for example, a in the name and v1 is the type).
name
a-v1
How can i create a list of all the name which are not the 2 latest versions?
In our case, The output would be:
name
a-v1
b-v1
Any idea how to do that in Python? I've been thinking about counting sub-strings. For example: To use - as a delimiter, And count how many times i find the left side of the string (aa, b, c). Is this possible to implement such as thing in Python? Any better ideas?
-
The output should contain also a-v3,d-v1,... ? why only a-v1 and b-v1?
– newbie
12 mins ago
or you may use something like priority queue with limit size, maybe overkill I think.
– apple apple
7 mins ago
@newbie I have 3 versions of
a, And i want to keep only the 2 latest version, So and output would be a-v1 (Which is the oldest version). Same as for b. As for c and d, I don't have more than 2 versions of each, So the output would be empty for them.– Omri
5 mins ago
a
a-v1
b
c
d
do you sort by postfix like
v1 or you account for dates as well? do you need to check the order of v-somehting is in proper date?– EPo
4 mins ago
v1
2 Answers
2
The problem would be easier with a slightly different data format.
You didn't write any code so I won't give you a complete answer:
data = ['name': 'a-v1', 'date': '2018-05-08T08:40:35.000Z', 'name': 'a-v2', 'date': '2018-05-20T08:40:35.000Z', 'name': 'a-v3', 'date': '2018-05-22T08:40:35.000Z', 'name': 'b-v1', 'date': '2018-02-08T08:40:35.000Z', 'name': 'b-v2', 'date': '2018-05-08T08:40:35.000Z', 'name': 'b-v3', 'date': '2018-05-10T08:40:35.000Z', 'name': 'c-v1', 'date': '2018-10-08T08:40:35.000Z', 'name': 'c-v2', 'date': '2018-11-08T08:40:35.000Z', 'name': 'd-v1', 'date': '2018-08-08T08:40:35.000Z']
temp = [d['name'].split('-') for d in data]
# [['a', 'v1'], ['a', 'v2'], ['a', 'v3'], ['b', 'v1'], ['b', 'v2'], ['b', 'v3'], ['c', 'v1'], ['c', 'v2'], ['d', 'v1']]
versions = [(letter, int(v[1:])) for letter, v in temp]
sorted(versions)
It outputs:
[('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('b', 3),
('c', 1),
('c', 2),
('d', 1)]
You could now try to use itertools.groupby to group the versions by letter and remove every version but the last two ones for each group.
itertools.groupby
Assuming your list L is pre-sorted, you can use itertools.groupby:
L
itertools.groupby
from itertools import groupby
from operator import itemgetter
groups = [list(vals)[:-2] for _, vals in groupby(map(itemgetter('name'), L),
key=lambda x: x.split('-')[0])]
res = list(chain.from_iterable(filter(None, groups)))
# ['a-v1', 'b-v1']
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I don't see any problem with the approach you proposed .
– apple apple
12 mins ago