{"id":29212,"date":"2023-01-06T07:11:09","date_gmt":"2023-01-06T01:41:09","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-remove-some-duplicates-from-list-in-python\/"},"modified":"2023-01-06T07:11:09","modified_gmt":"2023-01-06T01:41:09","slug":"solved-remove-some-duplicates-from-list-in-python","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-remove-some-duplicates-from-list-in-python\/","title":{"rendered":"[Solved] Remove some duplicates from list in python"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

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\n
<\/div>\n
\n
\n

this is where itertools.groupby<\/code><\/a> may come in handy:<\/p>\n

from itertools import groupby\n\na = [\"a\", \"a\", \"b\", \"b\", \"a\", \"a\", \"c\", \"c\"]\n\nres = [key for key, _group in groupby(a)]\nprint(res)  # ['a', 'b', 'a', 'c']\n<\/code><\/pre>\n
\n
this is a version where you could ‘scale’ down the unique keys (but are guaranteed to have at leas one in the result):<\/p>\n
from itertools import groupby, repeat, chain\n\na = ['a', 'a', 'a', 'a', 'b', 'b', 'c', 'c', 'c', 'c', 'c', 'a', 'a',  \n     'd', 'd', 'a', 'a']\nscale = 0.4\n\nkey_count = tuple((key, sum(1 for _item in group)) for key, group in groupby(a))\n# (('a', 4), ('b', 2), ('c', 5), ('a', 2), ('d', 2), ('a', 2))\n\nres = tuple(\n    chain.from_iterable(\n        (repeat(key, round(scale * count) or 1)) for key, count in key_count\n    )\n)\n# ('a', 'a', 'b', 'c', 'c', 'a', 'd', 'a')\n<\/code><\/pre>\n
there may be smarter ways to determine the scale<\/code> (probably based on the length of the input list a<\/code> and the average group<\/code> length).<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            11<\/span> <\/p><\/div>\n<\/div>\n
[ad_2]<\/p>\n
solved Remove some duplicates from list in python <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] this is where itertools.groupby may come in handy: from itertools import groupby a = [“a”, “a”, “b”, “b”, “a”, “a”, “c”, “c”] res = [key for key, _group in groupby(a)] print(res) # [‘a’, ‘b’, ‘a’, ‘c’] this is a version where you could ‘scale’ down the unique keys (but are guaranteed to have at … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[349],"class_list":["post-29212","post","type-post","status-publish","format-standard","hentry","category-solved","tag-python"],"yoast_head":"\n[Solved] Remove some duplicates from list in python - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-remove-some-duplicates-from-list-in-python\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Remove some duplicates from list in python - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] this is where itertools.groupby may come in handy: from itertools import groupby a = ["a", "a", "b", "b", "a", "a", "c", "c"] res = [key for key, _group in groupby(a)] print(res) # ['a', 'b', 'a', 'c'] this is a version where you could ‘scale’ down the unique keys (but are guaranteed to have at ... 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