Conditionally adding characters to new column based on separate dataset - r

Question

Hello all and thank you in advance.
I would like to add a new column to my pre-existing data frame where the values sourced from a second data frame based on certain conditions. The dataset I wish to add the new column to ("data_melt") has many different sample IDs (sample.#) under the variable column. Using a second dataset ("metadata") I want to add the pond names to the "data_melt" new column based on the sample-ids. The sample IDs are the same in both datasets.
My gut tells me there's an obvious solution but my head is pretty fried. Here is a toy example of my data_melt df (since its 25,000 observations):
> dput(toy)
structure(list(gene = c("serA", "mdh", "fdhB", "fdhA"), process = structure(c(1L,
1L, 1L, 1L), .Label = "energy", class = "factor"), category = structure(c(1L,
1L, 1L, 1L), .Label = "metabolism", class = "factor"), ko = structure(1:4, .Label = c("K00058",
"K00093", "K00125", "K00148"), class = "factor"), variable = structure(c(1L,
2L, 3L, 3L), .Label = c("sample.10", "sample.19", "sample.72"
), class = "factor"), value = c(0.00116, 2.77e-05, 1.84e-05,
0.0125)), row.names = c(NA, -4L), class = "data.frame")
And here is a toy example of my metadata df:
> dput(toy)
structure(list(sample = c("sample.10", "sample.19", "sample.72",
"sample.13"), pond = structure(c(2L, 2L, 1L, 1L), .Label = c("lower",
"upper"), class = "factor")), row.names = c(NA, -4L), class = "data.frame")
Thank you again!

Answer 1

We can use match from base R to create a numeric index to replace the values
toy$pond <- with(toy, out$pond[match(variable, out$sample)])

Answer 2

I believe merge will work here.
sss <- structure(list(gene = c("serA", "mdh", "fdhB", "fdhA"), process = structure(c(1L,
1L, 1L, 1L), .Label = "energy", class = "factor"), category = structure(c(1L,
1L, 1L, 1L), .Label = "metabolism", class = "factor"), ko = structure(1:4, .Label = c("K00058",
"K00093", "K00125", "K00148"), class = "factor"), variable = structure(c(1L,
2L, 3L, 3L), .Label = c("sample.10", "sample.19", "sample.72"
), class = "factor"), value = c(0.00116, 2.77e-05, 1.84e-05,
0.0125)), row.names = c(NA, -4L), class = "data.frame")
ss <- structure(list(sample = c("sample.10", "sample.19", "sample.72",
"sample.13"), pond = structure(c(2L, 2L, 1L, 1L), .Label = c("lower",
"upper"), class = "factor")), row.names = c(NA, -4L), class = "data.frame")
ssss <- merge(sss, ss, by.x = "variable", by.y = "sample")

Answer 3

You can use left_join() from the dplyr package after renaming sample to variable in the metadata data frame.
library(tidyverse)
data_melt <- structure(list(gene = c("serA", "mdh", "fdhB", "fdhA"),
process = structure(c(1L, 1L, 1L, 1L),
.Label = "energy",
class = "factor"),
category = structure(c(1L, 1L, 1L, 1L),
.Label = "metabolism",
class = "factor"),
ko = structure(1:4,
.Label = c("K00058", "K00093", "K00125", "K00148"),
class = "factor"),
variable = structure(c(1L, 2L, 3L, 3L),
.Label = c("sample.10", "sample.19", "sample.72"),
class = "factor"),
value = c(0.00116, 2.77e-05, 1.84e-05, 0.0125)),
row.names = c(NA, -4L),
class = "data.frame")
metadata <- structure(list(sample = c("sample.10", "sample.19", "sample.72", "sample.13"),
pond = structure(c(2L, 2L, 1L, 1L),
.Label = c("lower", "upper"),
class = "factor")),
row.names = c(NA, -4L),
class = "data.frame") %>%
# Renaming the column, so we can join the two data sets together
rename(variable = sample)
data_melt <- data_melt %>%
left_join(metadata, by = "variable")