如果value在另一个数据帧中，请用NA替换多个列

Question

I have two data.frames like this:

#df1
ID     a1      a2     a3      b1      b2      b3     Date
3xy    Evan    Greg   Ryan   Ben      Bob     Alex   12/3
4lm    John    Bill   Sue    Randy    Mark    Seth   12/5

#df2
Name
Evan
Mark

If a name from any of the "a" columns appears in df2$Name, I want to replace all of the "a" columns with NA. Same for the "b" columns. My desired output would look like this:

ID     a1      a2     a3      b1      b2      b3     Date
3xy    NA      NA     NA     Ben      Bob     Alex   12/3
4lm    John    Bill   Sue    NA       NA      NA     12/5

I've found several other posts that appear to be on similar topics, but I haven't found a way to get this to work. I've been able to replace the names in df1 that appear in df2 with NA using the code below, but haven't figured out how to replace the other columns that begin with the same letter:

df1[apply(df1, 2, function(df1) df1 %in% df2$Name)] <- NA

Gives me an output like this:

ID     a1      a2     a3      b1      b2      b3     Date
3xy    NA      Greg   Ryan   Ben      Bob     Alex   12/3
4lm    John    Bill   Sue    Randy    NA      Seth   12/5

I also keep trying different ifelse statements, but no success.

Answer 1

We can split the dataset based on the 'a', and 'b' columns, then loop through rows and assign the rows to NA values if there is any match with the 'name' column of 'df2'

nm1 <- names(df1)[c(-1, -8)]
lst <- lapply(split.default(df1[nm1], sub("\\d+", "", nm1)), function(x) {
         x[apply(x, 1, function(y) any(y %in% df2$Name)),] <- NA
     x})
df1[nm1] <- do.call(cbind, unname(lst))
df1
#   ID   a1   a2   a3   b1   b2   b3 Date
#1 3xy <NA> <NA> <NA>  Ben  Bob Alex 12/3
#2 4lm John Bill  Sue <NA> <NA> <NA> 12/5

---

Or another option is melt/dcast from data.table

library(data.table)
dcast(melt(setDT(df1), measure = patterns("^a\\d+", "^b\\d+"),
    value.name = c('a', 'b'))[, c('a', 'b') := lapply(.SD, function(x) 
  replace(x, any(x %in% df2$Name), NA)), ID, .SDcols = a:b][],
        ID + Date ~ variable, value.var = c('a', 'b'), sep='')
#    ID Date   a1   a2  a3  b1  b2   b3
#1: 3xy 12/3   NA   NA  NA Ben Bob Alex
#2: 4lm 12/5 John Bill Sue  NA  NA   NA

Answer 2

library(tidyverse)
df3 <- df1 %>%
  gather(key, value, -ID, -Date) %>%
  mutate(group = substr(key, 1, 1)) %>%
  select(group, ID, value) %>%
  inner_join(df2, by = c("value" = "Name")) %>%
  select(group, ID)

df1 %>%
  gather(key, value, -ID, -Date) %>%
  mutate(group = substr(key, 1, 1)) %>%
  anti_join(df3) %>%
  select(-group) %>%
  spread(key, value) %>%
  select(ID, matches("^a"), matches("^b"), Date)

Output:

# A tibble: 2 x 8
     ID    a1    a2    a3    b1    b2    b3  Date
* <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1   3xy  <NA>  <NA>  <NA>   Ben   Bob  Alex  12/3
2   4lm  John  Bill   Sue  <NA>  <NA>  <NA>  12/5

Answer 3

Here is a dplyr/tidyr approach

library(dplyr)
library(tidyr)

df1= df1%>% gather(Type, Names, -c(ID, Date)) %>%
  mutate(type2 = gsub("\\d", "", Type)) %>%
  group_by(type2, ID) %>%
  mutate(names2 = ifelse(any(Names %in% df2$Name), "", Names),
         Names = ifelse(names2 == "", NA, Names)) %>%
  ungroup() %>%
  select(-type2, -names2) 

which results in a (long format)

       ID   Date  Type Names
   <fctr> <fctr> <chr> <chr>
 1  3xy     12/3    a1  <NA>
 2  4lm     12/5    a1  John
 3  3xy     12/3    a2  <NA>
 4  4lm     12/5    a2  Bill
 5  3xy     12/3    a3  <NA>
 6  4lm     12/5    a3   Sue
 7  3xy     12/3    b1   Ben
 8  4lm     12/5    b1  <NA>
 9  3xy     12/3    b2   Bob
10  4lm     12/5    b2  <NA>
11  3xy     12/3    b3  Alex
12  4lm     12/5    b3  <NA>