基于某些过滤将列中的值分配给其他两列中的值

Question

I am currently working on a programming puzzle that sounds straightforward, but apparently it is pretty difficult if I want to do this efficiently in R without having to use for loop to go through a column with 100k+ rows within a data-frame. I am trying to apply dplyr (particularly group_by and mutate ) or data.table , and -apply family, but it's quite tough. Could anyone give some help?

The problem is as follows: given a data-frame df with columns key ("string" data type) x , y , and z ("numeric" data type). Some elements within column key are repeated. Among rows with the same element in key column, check if the corresponding value in column x is smaller than the sum of corresponding elements in column y (row-wise). If it is, then turn that value in column x to 0, while distributing the element in column x to elements in column y based on the ordering of corresponding values in column z . How do we effectively do this given that we need to go through all distinct elements in column key ?

Input

df <- data.frame(key = c('aa_bb_1, aa_bb_0, ab_ca_0, abc_bbb_1, abbbc_aa_1, aaa_ccc_1, aa_bb_1, aa_bb_1, ab_ca_0, abc_bbb_1, abbbc_aa_1, aaa_ccc_1, aa_bb_0, aa_bb_1, ab_ca_0, abc_bbb_0, abbbc_aa_0, aaa_ccc_1, aa_bb_0, aa_bb_1, ab_ca_1, abc_bbb_1, abbbc_aa_1, aaa_ccc_1, aa_bb_1, aa_bb_0, ab_ca_0, abc_bbb_1, abbbc_aa_1, aaa_ccc_1),
                 x = c(10, 19, 30, 25, 37, 13, 30, 40, 100, 53, 11, 27, 89, 21, 30, 30, 17, 9, 5, 57, 10, 19, 30, 25, 37, 13, 30, 40, 100, 53, 11, 27, 89, 21, 30, 30, 17, 9, 5, 57, 10, 19, 30, 25, 37, 13, 30, 40, 100, 53), 
                 y = (3, 10, 18, 15, 32, 4, 6, 29, 71, 92, 11, 7, 21, 19, 13, 26,28,11,8, 8, 5, 23, 3, 12, 19, 7, 9, 11, 7, 12, 9, 3, 20, 13, 7, 2, 9, 3, 6, 13, 11, 8, 8, 5, 21, 5, 21,11, 25, 40),
                 z = (8,13,15,16,10,10,25,21,32,15,45,8,10,50,12,10,0,0,10,12,2,40,9,8,13,15,16,10,10,25,21,32,15,45,8,10,50,12,10,0,0,10,12,2,40,9,12,10,10,20)

          key   x  y  z
1     aa_bb_1  10  3  8
2     aa_bb_0  19 10 13
3     ab_ca_0  30 18 15
4   abc_bbb_1  25 15 16
5  abbbc_aa_1  37 32 10
6   aaa_ccc_1  13  4 10
7     aa_bb_1  30  6 25
8     aa_bb_1  40 29 21
9     ab_ca_0 100 71 32
10  abc_bbb_1  53 92 15
11 abbbc_aa_1  11 11 45
12  aaa_ccc_1  27  7  8
13    aa_bb_0  89 21 10
14    aa_bb_1  21 19 50
15    ab_ca_0  30 13 12
16  abc_bbb_0  30 26 10
17 abbbc_aa_0  17 28  0
18  aaa_ccc_1   9 11  0
....
25    aa_bb_1  37 19 13
26    aa_bb_0  13  7 15
27    ab_ca_0  30  9 16
28  abc_bbb_1  40 11 10
29 abbbc_aa_1 100  7 10
30  aaa_ccc_1  53 12 25


          

Answer 1

Not sure what exactly your outcome looks like

With dplyr you can do something like this. I'm pretty, sure this doesn't exactly solve your issue because of ambiguity of your description. But you can use this as a template to solve your issue.

 df |> 
  group_by(key) |> 
  mutate(x = ifelse(n() > 1 & (x < sum(y)), 0, x)) |> 
  ungroup() |> 
  mutate( y = df |> group_by(key) |> mutate(y= x[order(z)]) |> pull(y))

   key            x     y     z
   <chr>      <dbl> <int> <int>
 1 aa_bb_1        0    10     8
 2 aa_bb_0        0    89    13
 3 ab_ca_0        0    30    15
 4 abc_bbb_1      0    53    16
 5 abbbc_aa_1     0    37    10
 6 aaa_ccc_1      0     9    10
 7 aa_bb_1        0    40    25
 8 aa_bb_1        0    30    21
 9 ab_ca_0        0    30    32
10 abc_bbb_1      0    25    15
11 abbbc_aa_1     0    11    45
12 aaa_ccc_1     27    27     8
13 aa_bb_0       89    19    10
14 aa_bb_1        0    21    50
15 ab_ca_0        0   100    12
16 abc_bbb_0     30    30    10
17 abbbc_aa_0    17    17     0
18 aaa_ccc_1      0    13     0