rank

问题 I have a dataframe: import pandas as pd df = pd.DataFrame([[1, 'a'], [1, 'a'], [1, 'b'], [1, 'a'], [2, 'a'], [2, 'b'], [2, 'a'], [2, 'b'], [3, 'b'], [3, 'a'], [3, 'b'], ], columns=['session', 'issue']) df I would like to rank issues within sessions. I tried with: df.groupby(['session', 'issue']).size().rank(ascending=False, method='dense') session issue 1 a 1.0 b 3.0 2 a 2.0 b 2.0 3 a 3.0 b 2.0 dtype: float64 What I need is result like this one: for group session=1, there are three a issues

问题 How could this (Oracle) SQL: select a.*, rank() over (partition by a.field1 order by a.field2 desc) field_rank from table_a a order by a.field1, a.field2 be translated into MySQL? This question seems to be similar but there is no Order By in the end of the base query. Also, does it matter that it is ordered by the fields of partition? 回答1: According to the link you gave it should look like this: SELECT a.*, ( CASE a.field1 WHEN @curType THEN @curRow := @curRow + 1 ELSE @curRow := 1 AND

问题 How could this (Oracle) SQL: select a.*, rank() over (partition by a.field1 order by a.field2 desc) field_rank from table_a a order by a.field1, a.field2 be translated into MySQL? This question seems to be similar but there is no Order By in the end of the base query. Also, does it matter that it is ordered by the fields of partition? 回答1: According to the link you gave it should look like this: SELECT a.*, ( CASE a.field1 WHEN @curType THEN @curRow := @curRow + 1 ELSE @curRow := 1 AND

问题 I have the following dataset: id | date | state ----------------------- 1 | 01/01/17 | high 1 | 02/01/17 | high 1 | 03/01/17 | high 1 | 04/01/17 | miss 1 | 05/01/17 | high 2 | 01/01/17 | miss 2 | 02/01/17 | high 2 | 03/01/17 | high 2 | 04/01/17 | miss 2 | 05/01/17 | miss 2 | 06/01/17 | high I want to create a column rank_state which ranks, within groups of id , the entries as per increasing date (starting from rank 0) which do not have the state of "miss". Furthermore, the rank repeats itself

问题 I have the following dataset: id | date | state ----------------------- 1 | 01/01/17 | high 1 | 02/01/17 | high 1 | 03/01/17 | high 1 | 04/01/17 | miss 1 | 05/01/17 | high 2 | 01/01/17 | miss 2 | 02/01/17 | high 2 | 03/01/17 | high 2 | 04/01/17 | miss 2 | 05/01/17 | miss 2 | 06/01/17 | high I want to create a column rank_state which ranks, within groups of id , the entries as per increasing date (starting from rank 0) which do not have the state of "miss". Furthermore, the rank repeats itself

问题 Does anyone know how to use the rank by distance search option that is mentioned here? https://developers.google.com/maps/documentation/javascript/places#place_search_requests Listing this in the request options doesn't seem to work. Here's my portion of code relative to this: var request = { location: coords, //radius: 30000, keyword: ['puma, retail'], types: ['store'], rankBy: google.maps.places.RankBy.DISTANCE }; service = new google.maps.places.PlacesService(map); service.search(request,

问题 I have a data frame 'test' that look like this: session_id seller_feedback_score 1 1 282470 2 1 275258 3 1 275258 4 1 275258 5 1 37831 6 1 282470 7 1 26 8 1 138351 9 1 321350 10 1 841 11 1 138351 12 1 17263 13 1 282470 14 1 396900 15 1 282470 16 1 282470 17 1 321350 18 1 321350 19 1 321350 20 1 0 21 1 1596 22 7 282505 23 7 275283 24 7 275283 25 7 275283 26 7 37834 27 7 282505 28 7 26 29 7 138359 30 7 321360 and a code (using package dplyr) that apparently should rank the 'seller_feedback

问题 In which SQL standard was RANK() first introduced? List of SQL standards: SQL-86 SQL-89 SQL-92 SQL:1999 SQL:2003 SQL:2008 SQL Rank function: http://en.wikipedia.org/wiki/Select_(SQL)#RANK.28.29_window_function References would be most appreciated. 回答1: The analytic features are defined as part of the ANSI SQL 1999 standard Reference: SS64.com 来源： https://stackoverflow.com/questions/2376290/in-which-sql-dialect-was-rank-first-introduced

问题 I have a vector with doubles which I want to rank (actually it's a vector with objects with a double member called costs ). If there are only unique values or I ignore the nonunique values then there is no problem. However, I want to use the average rank for nonunique values. Furthermore, I have found some question at SO about ranks, however they ignore the non-unique values. Example, say we have (1, 5, 4, 5, 5) then the corresponding ranks should be (1, 4, 2, 4, 4). When we ignore the non

问题 I currently use the following query to get the details of every user. SELECT u.*, sums.total_votes, sums.no_of_events FROM user u LEFT JOIN ( SELECT us.user_uid, count(ev.event_vote_id) AS total_votes count(distinct ue.event_uid) AS no_of_events FROM user_event ue LEFT JOIN event_vote ev ON ev.event_uid = ue.event_uid GROUP BY ue.user_uid ) sums ON sums.user_uid = u.user_uid However, I wish to also return the rank of their highest voted event (out of all events - not just their own). USER |