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I have two tables: A and B with a one-to-many relationship between them.
create table A ( a_id serial primary key ) create table B ()
I have the following data frame:
items = data_frame( id = 1:4, due = c(1,1,1,2), size = c(5,4,3,2), day = NA )
capacity = 8
items
# # A tibble: 4 × 4
# id due size day
# <int> <dbl> <dbl> <lgl>
# 1 1 1 5 NA
# 2 2 1 4 NA
# 3 3 1 3 NA
# 4 4 2 2 NA
My goal is to assign a day to each item such that the daily capacity 8 is not exceeded. The items are sorted and assigned by their due date.
So the target data frame that I want to obtain for this example is this:
goal = data_frame( id = 1:4, due = c(1,1,1,2), size = c(5,4,3,2), day = c(1,2,1,2) )
goal
# # A tibble: 4 × 4
# id due size day
# <int> <dbl> <dbl> <dbl>
# 1 1 1 5 1
# 2 2 1 4 2
# 3 3 1 3 1
# 4 4 2 2 2
I can solve this problem using two for loops and one if condition like that:
result = data_frame( id = integer(), due = integer(), size = integer(), day = integer() )
df = items
will_drop = integer()
for (day in 1:2) {
cum_size = 0
for (i in 1:nrow(df)) {
item = df[i, ]
if (cum_size + item$size <= capacity) {
cum_size = cum_size + item$size
item$day = day
result = rbind(result, item)
will_drop = c(will_drop, i)
}
}
df = df[-will_drop, ]
}
But I don't like this solution because it is error prone and hard to modify.
I would like to ask if you might come up with a solution without using any for loop and if branching using dplyr or data.table or higher order functions such as lapply or Reduce?
I want to drop some rows from some dataframe using numeric indices of the rows. But sometimes the indices vector that I am going to drop becomes zero length vector. In this case, I expect that nothing should be dropped from the original data frame. But instead of nothing everything is dropped.
For example, here drop works as expected
df = data_frame( a = 10:12 )
drop = c(1,2)
df[ -drop, ]
# # A tibble: 1 × 1
# a
# <int>
# 1 12
But when drop is zero length vector, then removing those rows doesn't work as I expect.
drop = integer()
df[ -drop, ]
# A tibble: 0 × 1
# ... with 1 variables: a <int>
How to remove rows from a data frame using row indices safely?
Solutions:
df %>% filter(!row_number() %in% drop)
df[!seq_len(nrow(df)) %in% drop, ]
df[ setdiff(1:nrow(df), drop), ]
https://stackoverflow.com/questions/44304547/mutate-an-empty-data-frames-column-safely
I have an empty data frame. I try to assign some value to a column of this data frame. Since the data frame is empty, I expect to obtain an empty data frame in return as follows:
df = data_frame(a = integer())
df %>%
mutate(a = 1)
# A tibble: 0 × 1
# ... with 1 variables: a <dbl>
The problem occurs when assigned value is a vector. Then I get the following error:
df %>%
mutate(a = 1:2)
# Error: wrong result size (2), expected 0 or 1
When I try to assign a value using $ operator, then even assigning a single value produces error:
df$a = 1
# Error in `$<-.data.frame`(`*tmp*`, "a", value = 1) :
# replacement has 1 row, data has 0
What is the correct and safe way to try to mutate an empty dataframe's column?
I have a data frame of some items to produce with certain sizes. There is a capacity limit for total size of items to produce. I want to obtain list of items such that they just surpass the capacity limit.
For example, this is the data frame of items:
df = data_frame( id = 1:4, size = 100 ) %>%
mutate( cum_size = cumsum(size) )
# # A tibble: 4 × 3
# id size cum_size
# <int> <dbl> <dbl>
# 1 1 100 100
# 2 2 100 200
# 3 3 100 300
# 4 4 100 400
capacity = 250
I can obtain the list of items such that they don't exceed the capacity limit as such:
df2 = df %>%
filter( cum_size < capacity )
df2
# id size cum_size
# <int> <dbl> <dbl>
# 1 1 100 100
# 2 2 100 200
But I want to include the row that can be partially produced as well. To find that row, I use the following code:
partial_item = df %>%
mutate( remains = cum_size - capacity ) %>%
filter( remains < size & remains > 0 ) %>%
select( -remains )
target = bind_rows(df2, partial_item)
target
# id size cum_size
# <int> <dbl> <dbl>
# 1 1 100 100
# 2 2 100 200
# 3 3 100 300
Solution:
df3 = df %>%
filter( cum_size < capacity + size)