匹配行和列，然后减去一个值

Question

I am reconciling two data sets. A has a list of transactions and a value. B contains several values from after a process. I want to subtract the values in A from a the identified field in B .

library(tidyverse)
A<-tribble(
  ~idA, ~group, ~column, ~value, ~idB,
  1, "x", "t1", 11, 1,
  2, "x", "t1",  22, 3,
  3, "x", "t3",  33, 4,
  4, "x", "t1",  25, 5)

B<-tribble(
  ~idB, ~group, ~t1, ~t2, ~t3,
  1, "x", 11, 0, 0,
  2, "x", 0, 11, 0,
  3, "x", 22, 0, 0 ,
  4, "x", 0, 0, 33,
  5, "x", 50, 50, 50)

Desired output:

Boutput<-tribble(
  ~idB, ~g,~t1, ~t2, ~t3,
  1, "x", 0, 0, 0, 
  2, "x", 0, 11, 0, 
  3, "x", 0, 0, 0,  
  4, "x", 0, 0, 0,  
  5, "x", 25, 50, 50)

I've tried inner_joining then mutating based on rules.

How to mathematically subtract the matches?

Answer 1

I was hesitating about posting this, but thought it might be helpful in looking at some alternative solutions.

I might consider converting A from long to wide first:

Awide <- A %>%
  pivot_wider(names_from = column)

R> Awide
# A tibble: 4 x 5
    idA group   idB    t1    t3
  <dbl> <chr> <dbl> <dbl> <dbl>
1     1 x         1    11    NA
2     2 x         3    22    NA
3     3 x         4    NA    33
4     4 x         5    25    NA

In this case, there are no values for t2 . Before joining A and B , would make sure there are columns for all 3 ( t1 , t2 , t3 ):

cols <- c("idA", "group", "idB", "t1", "t2", "t3")
missing <- setdiff(cols, names(Awide))
Awide[missing] <- NA
Awide <- Awide[cols]

R> Awide
# A tibble: 4 x 6
    idA group   idB    t1 t2       t3
  <dbl> <chr> <dbl> <dbl> <lgl> <dbl>
1     1 x         1    11 NA       NA
2     2 x         3    22 NA       NA
3     3 x         4    NA NA       33
4     4 x         5    25 NA       NA

Then could do a left_join and make sure all the NAs present are zero for subtraction later.

AB <- left_join(B, Awide, by=c("idB", "group")) %>%
  mutate_at(c("t1.y", "t2.y", "t3.y"), ~replace(., is.na(.), 0))

R> AB
# A tibble: 5 x 9
    idB group  t1.x  t2.x  t3.x   idA  t1.y  t2.y  t3.y
  <dbl> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1 x        11     0     0     1    11     0     0
2     2 x         0    11     0    NA     0     0     0
3     3 x        22     0     0     2    22     0     0
4     4 x         0     0    33     3     0     0    33
5     5 x        50    50    50     4    25     0     0

Then would do the subtraction on the columns that match the pattern t*.x and t*.y (alternatives could be used depending on what you need):

tdiff <- AB[,grepl("^t.*\\.x$", names(AB))] - AB[,grepl("^t.*\\.y$", names(AB))]

R> tdiff
  t1.x t2.x t3.x
1    0    0    0
2    0   11    0
3    0    0    0
4    0    0    0
5   25   50   50

Then bind these totals to AB to get final result:

cbind(AB[,1:2,drop=FALSE], tdiff)

  idB group t1.x t2.x t3.x
1   1     x    0    0    0
2   2     x    0   11    0
3   3     x    0    0    0
4   4     x    0    0    0
5   5     x   25   50   50

Answer 2

This is the loop I've come up with

Bout<-B
for (i in A$idA){
  Bout[A$idB[i],A$column[i]] <- (as.numeric(Bout[A$idB[i],A$column[i]])) - A$value[i]
}
Bout