Hi, I have the following dataset:

```
# A tibble: 134,644 x 3
| `1y`| `5y | date |
| <dbl>| <dbl> | <date> |
| 1 | 3 | 6 | 2009-01-01 |
| 2 | 2 | NA | 2009-01-01 |
| 3 | -1 | NA | 2009-01-01 |
| 4 | 3 | NA | 2009-01-01 |
| 5 | -5 | NA | 2009-01-01 |
| 6 | 3 | -2 | 2009-01-01 |
| 7 | NA | NA | 2009-01-01 |
| 8 | 5 | NA | 2009-01-01 |
| 9 | 0 | 5 | 2009-01-01 |
| 10 | 0 | NA | 2009-01-01 |
# ... with 134,634 more rows
```

It is quarterly data, so the dates are 2009-01-01, 2009-04-01, 2009-07-01, 2009-10-01, 2010-01-01 and so on and so forth. The data has the following class:

sapply(uk_data, class)

```
sapply(uk_data, class)
1y 5y date
"numeric" "numeric" "Date"
```

For each variable, I am trying to get some descriptive statistics per period (mean, median, sd, skewness etc.) ignoring the NAs. I will later substitute the mean of each period for the NAs, hence I was thinking of adding two columns with the period means to the dataset, substitute them for the NAs, and finally find the remaining descriptive statistics. I have run the following code:

uk_data %>%

group_by(date) %>%

summarise_at(vars("1y"),

list(name = mean("1y")))

Yet, I have the following error:

```
Error in `FUN()`:
! expecting a one sided formula, a function, or a function name.
Run `rlang::last_error()` to see where the error occurred.
Warning message:
In mean.default("1y") : argument is not numeric or logical: returning NA
<error/rlang_error>
Error in `FUN()`:
! expecting a one sided formula, a function, or a function name.
Backtrace:
1. uk_data %>% group_by(date) %>% ...
2. dplyr::summarise_at(., vars("1y"), list(name = mean("1y")))
3. dplyr:::manip_at(...)
4. dplyr:::as_fun_list(.funs, .env, ..., .caller = .caller)
5. dplyr:::map(...)
6. base::lapply(.x, .f, ...)
7. dplyr FUN(X[[i]], ...)
Run `rlang::last_trace()` to see the full context.
<error/rlang_error>
Error in `FUN()`:
! expecting a one sided formula, a function, or a function name.
Backtrace:
x
1. +-uk_data %>% group_by(date) %>% ...
2. \-dplyr::summarise_at(., vars("1y"), list(name = mean("1y")))
3. \-dplyr:::manip_at(...)
4. \-dplyr:::as_fun_list(.funs, .env, ..., .caller = .caller)
5. \-dplyr:::map(...)
6. \-base::lapply(.x, .f, ...)
7. \-dplyr FUN(X[[i]], ...)
8. \-rlang::abort("expecting a one sided formula, a function, or a function name.")
```

Is anyone able to help? Thanks in advance!