I have a dataframe with multiple NULL values. The class type of the columns are LIST not NUMERIC. Is it possible to replace all the NULL values with the median value of the column? I tried a manual way was to change the NULL value of the column 1 by 1, using the as.numeric() function and subsequently apply the median() function. Is there a more efficient way to do this?
i1 <- sapply(pivot_table_1$`Start Working`, is.null)
pivot_table_1$`Start Working`[i1] <- 0
Output from dput():
structure(list(Day = 1:31, `Start Sleeping` = list(0, 20, 35,
40, 50, 0, 40, 0, 0, 40, 50, 0, 0, 40, 0, 40, 35, 45, 0,
0, 65, 35, 40, 40, 0, 50, 40, 0, 0, 0, 0), `Stop Sleeping` = list(
440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440,
440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440, 440,
440, 440, 440, 440, 440, 440, 440), `Start Working` = list(
490, 490, 490, 490, 0, 0, 490, 490, 490, 490, 490, 0, 0,
490, 490, 490, 490, 490, 0, 0, 490, 490, 490, 490, 490, 0,
0, 490, 490, 490, 490), `Stop Working` = list(1005, 1005,
1005, 1005, NULL, NULL, 965, 965, 965, 965, 965, NULL, NULL,
965, 965, 965, 965, 965, NULL, NULL, 965, 965, 965, 965,
965, NULL, NULL, 965, 965, 965, 965), Breakfast = list(690,
645, 615, 540, NULL, NULL, NULL, NULL, NULL, NULL, NULL,
475, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 475, NULL,
NULL, NULL, NULL, NULL, 475, NULL, NULL, NULL, NULL, NULL),
Dinner = list(1390, 1360, 1285, 1270, 1390, NULL, 1140, 1140,
1130, 1135, 1130, NULL, 1165, 1140, 1130, 1135, 1130,
1140, 1140, 1180, NULL, 1145, 1135, 1140, 1135, 1160,
1140, 1140, NULL, 1140, NULL)), row.names = c(NA, -31L
), class = c("tbl_df", "tbl", "data.frame"))
If you wish to keep the entries as length-one lists you can do:
pivot_table_1[] <- lapply(pivot_table_1, function(x) {
ifelse(lengths(x) == 1, x, list(median(unlist(x))))})
pivot_table_1
#> # A tibble: 31 x 7
#> Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#> <int> <list> <list> <list> <list>
#> 1 1 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 2 2 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 3 3 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 4 4 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 5 5 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 6 6 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 7 7 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 8 8 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 9 9 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> 10 10 <dbl [1]> <dbl [1]> <dbl [1]> <dbl [1]>
#> # ... with 21 more rows, and 2 more variables: Breakfast <list>, Dinner <list>
Or, if you want them as numeric columns, do:
pivot_table_1[] <- lapply(pivot_table_1, function(x) {
unlist(ifelse(lengths(x) == 1, x, list(median(unlist(x)))))})
pivot_table_1
#> # A tibble: 31 x 7
#> Day `Start Sleeping` `Stop Sleeping` `Start Working` `Stop Working`
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0 440 490 1005
#> 2 2 20 440 490 1005
#> 3 3 35 440 490 1005
#> 4 4 40 440 490 1005
#> 5 5 50 440 0 965
#> 6 6 0 440 0 965
#> 7 7 40 440 490 965
#> 8 8 0 440 490 965
#> 9 9 0 440 490 965
#> 10 10 40 440 490 965
#> # ... with 21 more rows, and 2 more variables: Breakfast <dbl>, Dinner <dbl>
Created on 2022-05-22 by the reprex package (v2.0.1)
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