unique

问题 How can I keep my duplicates, but remove unique values based on one column(qol)? ID qol Sat A 7 6 A 7 5 B 3 3 B 3 4 B 1 7 C 2 7 c 1 2 But I need this: ID qol Sat A 7 6 A 7 5 B 3 3 B 3 4 What can I do? 回答1: dplyr solution: library(dplyr) ID <- c("A", "A", "B", "B", "B", "C", "c") qol <- c(7,7,3,3,1,2,1) Sat <- c(6,5,3,4,7,7,2) test_df <- data.frame(cbind(ID, qol, Sat)) filtered_df <- test_df %>% group_by(qol) %>% filter(n()>1) Please note that this will return ID qol Sat 1 A 7 6 2 A 7 5 3 B 3

问题 I'm trying to do an upsert to a table that has partial unique indexes create table test ( p text not null, q text, r text, txt text, unique(p,q,r) ); create unique index test_p_idx on test(p) where q is null and r is null; create unique index test_pq_idx on test(p, q) where r IS NULL; create unique index test_pr_idx on test(p, r) where q is NULL; In plain terms, p is not null and only one of q or r can be null. Duplicate inserts throw constraint violations as expected insert into test(p,q,r

问题 I'm trying to do an upsert to a table that has partial unique indexes create table test ( p text not null, q text, r text, txt text, unique(p,q,r) ); create unique index test_p_idx on test(p) where q is null and r is null; create unique index test_pq_idx on test(p, q) where r IS NULL; create unique index test_pr_idx on test(p, r) where q is NULL; In plain terms, p is not null and only one of q or r can be null. Duplicate inserts throw constraint violations as expected insert into test(p,q,r

问题 How do I get the unique values of a column of lists in pandas or numpy such that second column from would result in 'action', 'crime', 'drama' . The closest (but non-functional) solutions I could come up with were: genres = data['Genre'].unique() But this predictably results in a TypeError saying how lists aren't hashable. TypeError: unhashable type: 'list' Set seemed to be a good idea but genres = data.apply(set(), columns=['Genre'], axis=1) but also results in a TypeError: set() takes no

问题 I trying to generate all possible unique combination of items. Ex: item1, item2, item3 Combinations: item1+item2+item3 item1+item2 item1+item3 item2+item3 item1 item2 item3 I am unable to get an idea on how to solve this? for(int i=0;i<size;i++){ for(int j=i+1;j<size;j++){ System.out.println(list.item(i)+list.item(j)); } } The above code certainly works for all unique combination of two elements. But not for 3 element pair and more.. 回答1: If you have N items, count from 1 to 2^N-1. Each

问题 I trying to generate all possible unique combination of items. Ex: item1, item2, item3 Combinations: item1+item2+item3 item1+item2 item1+item3 item2+item3 item1 item2 item3 I am unable to get an idea on how to solve this? for(int i=0;i<size;i++){ for(int j=i+1;j<size;j++){ System.out.println(list.item(i)+list.item(j)); } } The above code certainly works for all unique combination of two elements. But not for 3 element pair and more.. 回答1: If you have N items, count from 1 to 2^N-1. Each

问题 In genomics research, you often have many strings with duplicate gene names. I would like to find an efficient way to only keep the unique gene names in a string. This is an example that works. But, isn't it possible to do this in one step, i.e., without having to split the entire string and then having to past the unique elements back together? genes <- c("GSTP1;GSTP1;APC") a <- unlist(strsplit(genes, ";")) paste(unique(a), collapse=";") [1] "GSTP1;APC" 回答1: An alternative is doing unique

问题 Why is the code below returning $null ? I'm trying to store just unique values. $DailyPathsToDelete = @("C:\temp\IMG000483\","C:\temp\IMG000483\") $DailyPathsToDelete = Select-Object $DailyPathsToDelete -Unique 回答1: You can try : $unique = $DailyPathsToDelete | Get-Unique 回答2: Short answer: To get all unique paths, you should pipe $DailyPathsToDelete to Select-Object and set the Unique switch. $DailyPathsToDelete = $DailyPathsToDelete | Select-Object -Unique Longer answer: 1. Why it's not

问题 I would like to prove something like this lemma in Isabelle lemma assumes "y = (THE x. P x)" shows "P (THE x. P x)" I imagine that the assumption implies that THE x. P x exists and is well-defined. So this lemma ought to be true too lemma assumes "y = (THE x. P x)" shows "∃! x. P x" I'm not sure how to prove this because I've looked through all the theorems that turn up when I type "name: the" into the query box in Isabelle and they don't seem useful. I can't find the definition of THE either

问题 I would like to prove something like this lemma in Isabelle lemma assumes "y = (THE x. P x)" shows "P (THE x. P x)" I imagine that the assumption implies that THE x. P x exists and is well-defined. So this lemma ought to be true too lemma assumes "y = (THE x. P x)" shows "∃! x. P x" I'm not sure how to prove this because I've looked through all the theorems that turn up when I type "name: the" into the query box in Isabelle and they don't seem useful. I can't find the definition of THE either