如何为一列中的每个值复制相同的行？

Question

How may I copy rows for each value in one column

Let's say that we have column like:

---------------------
record time     id
---------------------
1      12:00    [1,2,3]
2      12:01    [4,5,6,7]
3      12:07    [8,9]

And I would like to get result like:

---------------------
record time    id
---------------------
1      12:00    1
2      12:00    2
3      12:00    3
4      12:01    4
5      12:01    5
...
9      12:07    9

I need to do this in Postgresql or R

Answer 1

One option is separate_rows if the 'id' is string

library(tidyverse)
df1 %>%
    separate_rows(id) %>%
    filter(id != "") %>%
    mutate(record = row_number())
#  record  time id
#1      1 12:00  1
#2      2 12:00  2
#3      3 12:00  3
#4      4 12:01  4
#5      5 12:01  5
#6      6 12:01  6
#7      7 12:01  7
#8      8 12:07  8
#9      9 12:07  9

---

If the 'id' is a list

df1 %>% 
   unnest

data

 df1 <- structure(list(record = 1:3, time = c("12:00", "12:01", "12:07"
 ), id = c("[1,2,3]", "[4,5,6,7]", "[8,9]")), class = "data.frame", 
 row.names = c(NA, -3L))

Answer 2

You could do it in PostgreSQL like so:

SELECT DISTINCT record, time, unnest(translate(id, '[]', '{}'):: int[]) AS ids
FROM tbl  
ORDER BY record, time, ids;

Basically, you create an array out of your text field, and then use unnest to get your desired result.

 record |   time   | ids 
--------+----------+-----
      1 | 12:00:00 |   1
      1 | 12:00:00 |   2
      1 | 12:00:00 |   3
      2 | 12:01:00 |   4
      2 | 12:01:00 |   5
      2 | 12:01:00 |   6
      2 | 12:01:00 |   7
      3 | 12:07:00 |   8
      3 | 12:07:00 |   9

DEMO

Answer 3

In Postgres, you would just use unnest() :

select t.record, t.time, unnest(t.id) as id
from t;

This assumes that your column is actually stored as an array in Postgres. If it is a string, you can do something similar, but it requires more string manipulation.