CODEWITHSUNDEEP
r
Alex
Alex

Reputation: 2788

How to replace row in one data frame with matching rows from another dataframe for all columns

I would like to replace a full row of data in one dataframe with matching rows from another dataframe. I have a reproducible example with only a couple of columns, but in practice I have a dataframe with dozens of columns. 

# main dataframe
df1 <- tibble(id = letters[1:5], v1 = seq(1,5), v2 = seq(1,5), v3 = seq(1,5))

>df1
# A tibble: 5 x 4
  id       v1    v2    v3
  <chr> <int> <int> <int>
1 a         1     1     1
2 b         2     2     2
3 c         3     3     3
4 d         4     4     4
5 e         5     5     5

# values to replace
df2 <- tibble(id = letters[3:4], v1 = rep(0,2), v2 = rep(0,2), v3 = rep(0,2))

> df2
# A tibble: 2 x 4
  id       v1    v2    v3
  <chr> <dbl> <dbl> <dbl>
1 c         0     0     0
2 d         0     0     0

# what the final result should look like
result <- tibble(id = c("a", "b", "c", "d", "e"), v1 = c(1, 2, 0, 0, 5), v2 = c(1, 2, 0, 0, 5), v3 = c(1, 2, 0, 0, 5))

>result
# A tibble: 5 x 4
  id       v1    v2    v3
  <chr> <dbl> <dbl> <dbl>
1 a         1     1     1
2 b         2     2     2
3 c         0     0     0
4 d         0     0     0
5 e         5     5     5

Upvotes: 0

Views: 1509

Answers (3)

Iroha
Iroha

Reputation: 34761

Based on your comment, if you have ids that exist in df2 but not in df1 you can do:

df1[na.omit(match(df2$id, df1$id)),] <- df2[df2$id %in% df1$id,]

Upvotes: 1

user2332849
user2332849

Reputation: 1450

This is a straightforward solution

df3 <- left_join(df1, df2, by = "id", suffix = c("", ".x"))
df3[!is.na(df3$v1.x), 2:4] <- df3[!is.na(df3$v1.x), 5:7]
df3[, 5:7] <- NULL

Upvotes: 0

Jonathan V. Sol&#243;rzano
Jonathan V. Sol&#243;rzano

Reputation: 5640

Here is one solution using tidyverse

library(tidyverse)
df1 %>%
  #Stay with the rows that are not found in df2 according to its id
  filter(! id %in% df2$id) %>%
  #bind rows of df2
  bind_rows(df2) %>%
  #Order data according to id values
  arrange(id)

Upvotes: 2

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