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cummean() translates to mean() not avg() #157

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iangow opened this Issue Sep 10, 2018 · 1 comment

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iangow commented Sep 10, 2018

library(dplyr, warn.conflicts = FALSE)
library(DBI)

df <- 
    expand.grid(id = LETTERS[1:6], time = 1:10) %>%
    arrange(id) %>%
    mutate(val = rnorm(n = n()))

df %>%
    group_by(id) %>%
    arrange(time) %>%
    mutate(cumavg = cummean(val)) %>%
    arrange(id, time)
#> # A tibble: 60 x 4
#> # Groups:   id [6]
#>    id     time    val cumavg
#>    <fct> <int>  <dbl>  <dbl>
#>  1 A         1  0.994 0.994 
#>  2 A         2  0.250 0.622 
#>  3 A         3  0.128 0.457 
#>  4 A         4  0.848 0.555 
#>  5 A         5  0.429 0.530 
#>  6 A         6 -2.09  0.0941
#>  7 A         7  0.709 0.182 
#>  8 A         8  2.13  0.426 
#>  9 A         9  0.893 0.478 
#> 10 A        10 -1.03  0.327 
#> # ... with 50 more rows

pg <- dbConnect(RPostgres::Postgres())

df_pg <- copy_to(pg, df)

df_pg %>%
    group_by(id) %>%
    arrange(time) %>%
    mutate(cumavg = cummean(val)) %>%
    arrange(id, time) %>%
    show_query()
#> <SQL>
#> SELECT "id", "time", "val", mean("val") OVER (PARTITION BY "id" ORDER BY "time" ROWS UNBOUNDED PRECEDING) AS "cumavg"
#> FROM (SELECT *
#> FROM "df"
#> ORDER BY "time") "qnxnacfsyq"
#> ORDER BY "id", "time"

Created on 2018-09-10 by the reprex package (v0.2.0).

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iangow commented Sep 11, 2018

Just in case it isn't clear from the above, mean is not valid SQL, but avg is.

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