tidyverse: Comparing each row of a data.frame with a single row from another data.frame

Question

I want to compare each row of df1 with a single row of df2 in tidy way. Any hint please.

df1 <-
  structure(
    list(
        Q1 = c("a", "a")
      , Q2 = c("b", "a")
      , Q3 = c("a", "a")
      , Q4 = c("b", "a")
      )
    , class = "data.frame"
    , row.names = c(NA, -2L)
    )

df2 <-
  structure(
    list(
        Q1 = "a"
      , Q2 = "a"
      , Q3 = "b"
      , Q4 = "c"
      )
    , class = "data.frame"
    , row.names = c(NA, -1L)
    )

library(tidyverse)


sum(df1[1, ] == df2)
[1] 1
sum(df1[2, ] == df2)
[1] 2

Answer 1

In Base

apply(df1,1, function(x) sum(x == df2))

[1] 1 2

Answer 2

Using purrr package:

unlist_df2 <- unlist(df2)
    seq_len(nrow(df1)) %>%
      map_lgl(~identical(unlist(df1[.x,]), unlist_df2))

For edit: change map_lgl to map_dbl and identical to sum & ==

unlist_df2 <- unlist(df2)
seq_len(nrow(df1)) %>%
  map_dbl(~sum(unlist(df1[.x,]) == unlist_df2))

Answer 3

An option with base R is rowSums

rowSums(df1 == unlist(df2)[col(df1)])
#[1] 1 2

---

In tidyverse, we can also use c_across

library(dplyr)
df1 %>% 
    rowwise %>%
    mutate(new = sum(c_across(everything()) == df2)) 
# A tibble: 2 x 5
# Rowwise: 
#  Q1    Q2    Q3    Q4      new
#  <chr> <chr> <chr> <chr> <int>
#1 a     b     a     b         1
#2 a     a     a     a         2

Answer 4

Either split it first and check identity:

library(purrr)
asplit(df1,1) %>% map_dbl(~sum(.==df2))

Or just map the row numbers:

1:nrow(df1) %>% map_dbl(function(i)sum(df1[i,]==df2))
[1] 1 2

Answer 5

A base R solution.

Compare and sum by rows:

rowSums(mapply(`==`, df1, df2))
#[1] 1 2

Edit.

Above is a new version of this post. The original summed by columns. Here is the code.

The return value is a list of logical vectors, then *apply function sum.

Map(`==`, df1, df2)
#$Q1
#[1] TRUE TRUE
#
#$Q2
#[1] FALSE  TRUE
#
#$Q3
#[1] FALSE FALSE
#
#$Q4
#[1] FALSE FALSE

res <- Map(`==`, df1, df2)
sapply(res, sum)
#Q1 Q2 Q3 Q4 
# 2  1  0  0

A one-liner would be

sapply(Map(`==`, df1, df2), sum)

Another one, faster.

colSums(mapply(`==`, df1, df2))
#Q1 Q2 Q3 Q4 
# 2  1  0  0