Is there any way to do this? r programming

Question

How can I tie it in the same cust_cd so that L_num is displayed?

If the cust_cd is 1000401, it shows 9,4,11,11,4,4.

example:

cust_cd   L_num
1000301    5,6,2,11,5
1000401    9,4,11,11,4,4
1000601    14,22,9,9,4

I can't do this. Please...

Welcome to [so]! Please read [ask]! [so] is not a code writing service. What have you tried so far? ... edit your question: https://stackoverflow.com/posts/65166740/edit — jogo, Dec 06 '20 at 10:02
https://stackoverflow.com/questions/24862046/cumulatively-paste-concatenate-values-grouped-by-another-variable — jogo, Dec 07 '20 at 08:58

score 1 · Answer 1 · answered Dec 06 '20 at 11:46

You can do:

df_short <- data.frame(cust_cd = unique(df$cust_cd),
                       i_num =  sapply(unique(df$cust_cd), function(i) {
                         paste(unique(df$i_num[which(df$cust_cd == i)]),
                               collapse = ", ")}))

df_short
#>   cust_cd     i_num
#> 1 1000101        18
#> 2 1000301        18
#> 3 1000401  9, 4, 11
#> 4 1000601 14, 22, 9

Data from question

df <- data.frame(cust_cd = c(rep(1000101, 7), 1000301, rep(1000401, 6),
                             rep(1000601, 5)),
                 i_num = c(rep(18, 8), 9, 4, 11, 11, 4, 4, 14, 22, 9, 9, 14))

df
#>    cust_cd i_num
#> 1  1000101    18
#> 2  1000101    18
#> 3  1000101    18
#> 4  1000101    18
#> 5  1000101    18
#> 6  1000101    18
#> 7  1000101    18
#> 8  1000301    18
#> 9  1000401     9
#> 10 1000401     4
#> 11 1000401    11
#> 12 1000401    11
#> 13 1000401     4
#> 14 1000401     4
#> 15 1000601    14
#> 16 1000601    22
#> 17 1000601     9
#> 18 1000601     9
#> 19 1000601    14

Thanks for upvote. I missed that there could be duplicate rows in my solution, so that needs to taken care of before paste(). Your solution on base R is also amazing. Upvoted. — AnilGoyal, Dec 06 '20 at 12:59

score 1 · Answer 2 · answered Dec 06 '20 at 12:06

Since you have not added any data, I have added the following sample data

> df
   custid lnum
1       1    4
2       1    2
3       1    6
4       1    5
5       2    1
6       2    2
7       3    3
8       3    4
9       3    9
10      3    8
11      4    0
12      4    7
13      5    2
14      6    0
15      6    7
16      6    8

Proceed as

df %>% group_by(custid) %>% 
  mutate(d = paste(lnum, collapse = ", ")) %>%
  select(-lnum) %>% slice_head()

  custid d         
   <int> <chr>     
1      1 4, 2, 6, 5
2      2 1, 2      
3      3 3, 4, 9, 8
4      4 0, 7      
5      5 2         
6      6 0, 7, 8

Nice to see a tidyverse solution +1. Feel free to use the data from my answer; I transcribed it from the picture of the data posted by the OP — Allan Cameron, Dec 06 '20 at 12:32

score 0 · Answer 3 · answered Dec 06 '20 at 18:55

We can use aggregate from base R

aggregate(i_num ~ cust_cd, unique(df), FUN = toString)
#  cust_cd     i_num
#1 1000101        18
#2 1000301        18
#3 1000401  9, 4, 11
#4 1000601 14, 22, 9

data

df <- structure(list(cust_cd = c(1000101, 1000101, 1000101, 1000101, 
1000101, 1000101, 1000101, 1000301, 1000401, 1000401, 1000401, 
1000401, 1000401, 1000401, 1000601, 1000601, 1000601, 1000601, 
1000601), i_num = c(18, 18, 18, 18, 18, 18, 18, 18, 9, 4, 11, 
11, 4, 4, 14, 22, 9, 9, 14)), class = "data.frame", row.names = c(NA, 
-19L))

Is there any way to do this? r programming

3 Answers3

data