{"id":25882,"date":"2022-12-14T03:37:19","date_gmt":"2022-12-13T22:07:19","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-python-3-6-when-a-value-in-a-list-changes-find-the-sum-of-the-values-in-another-list\/"},"modified":"2022-12-14T03:37:19","modified_gmt":"2022-12-13T22:07:19","slug":"solved-python-3-6-when-a-value-in-a-list-changes-find-the-sum-of-the-values-in-another-list","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-python-3-6-when-a-value-in-a-list-changes-find-the-sum-of-the-values-in-another-list\/","title":{"rendered":"[Solved] Python 3.6: when a value in a list changes, find the sum of the values in another list?"},"content":{"rendered":"

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\n<\/p>\n

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<\/div>\n
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Now that I understand that the list A<\/code> itself is not going to change, but you expressed the idea of change meaning that the list has partitions where there are adjacent elements that are different, then a different solution comes to mind:<\/p>\n

Given your original data:<\/p>\n

A = ['1','1','1','3','3','4']\n\nB = ['5','9','3','4','8','1']\n<\/code><\/pre>\n
We can use a defaultdict<\/code> to group this data:<\/p>\n
from collections import defaultdict\nD = defaultdict(list)\n\nfor a, b in zip(A,B):\n    D[a].append(int(b))\n<\/code><\/pre>\n
Just to show what we have collected:<\/p>\n
for a, b in D.items():\n    print(a,b)\n>> 1 [5, 9, 3]\n>> 3 [4, 8]\n>> 4 [1]\n<\/code><\/pre>\n
Now we can compute some sums:<\/p>\n
result = [sum(b) for a,b in D.items()]\nprint(result)\n>> [17, 12, 1]\n<\/code><\/pre>\n
Notes about defaultdict<\/code>:<\/p>\n
When the instance is created:<\/p>\n
D = defaultdict(list)\n<\/code><\/pre>\n
This creates a defaultdict<\/code> which, when it encounters new keys will create an instance of the parameter. In this case it creates a new list<\/code><\/p>\n
So each line:<\/p>\n
D[a].append(int(b))\n<\/code><\/pre>\n
When a<\/code> is a key that D<\/code> has not seen before, a new list<\/code> is created, which is returned, to which the .append(....<\/code> is applied.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            2<\/span> <\/p><\/div>\n<\/div>\n
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solved Python 3.6: when a value in a list changes, find the sum of the values in another list? <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Now that I understand that the list A itself is not going to change, but you expressed the idea of change meaning that the list has partitions where there are adjacent elements that are different, then a different solution comes to mind: Given your original data: A = [‘1′,’1′,’1′,’3′,’3′,’4’] B = [‘5′,’9′,’3′,’4′,’8′,’1’] We can … 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":[540,349,1097],"class_list":["post-25882","post","type-post","status-publish","format-standard","hentry","category-solved","tag-list","tag-python","tag-sum"],"yoast_head":"\n[Solved] Python 3.6: when a value in a list changes, find the sum of the values in another list? - 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-python-3-6-when-a-value-in-a-list-changes-find-the-sum-of-the-values-in-another-list\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Python 3.6: when a value in a list changes, find the sum of the values in another list? - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Now that I understand that the list A itself is not going to change, but you expressed the idea of change meaning that the list has partitions where there are adjacent elements that are different, then a different solution comes to mind: Given your original data: A = ['1','1','1','3','3','4'] B = ['5','9','3','4','8','1'] We can ... 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