Question

电影镜头数据具有非规范化类型：

movie1  ACTION
movie1  ADVENTURE
movie2  ACTION
...

是否可以编写一个简单的SQL来获取规范化的电影类型：

library(networkD3)
library(data.table)
set.seed(1999)
links <- data.table(
  src = rep(0:4, times=c(1,1,2,3,5)),
  target = sample(1:11, 12, TRUE),
  value = sample(100, 12)
)[src < target, ]  # no loops
nodes <- data.table(name=LETTERS[1:12])

## Need to hover to get counts
sankeyNetwork(Links=links, Nodes=nodes, Source='src', Target='target',
  Value='value', NodeID='name', fontSize=16)

## Add text to label
txt <- links[, .(total = sum(value)), by=c('target')]
nodes[txt$target+1L, name := paste0(name, ' (', txt$total, ')')]

## Displays the counts as part of the labels
sankeyNetwork(Links=links, Nodes=nodes, Source='src', Target='target',
  Value='value', NodeID='name', fontSize=16, width=600, height=300)

假设我在MySQL或PostgreSQL中这样做。

Answer 1

对于PostgreSQL，您可以将unnest与string_to_array：

一起使用

select name, unnest(string_to_array(genres, '|'))
from movies;

SQL Fiddle Demo

对非规范化数据进行简单的SELECT查询

1 个答案: