Select only the first row when merging data frames with multiple matches

Here is another method using dplyr::distinct. It is useful if you want to keep all rows from 'data' even if there is no match.

data = data.frame(id=c(1,2,3,4,5),
                  state=c("KS","MN","AL","FL","CA"))
scores = data.frame(id=c(1,1,1,2,2,3,3,3),
                    score=c(66,75,78,86,85,76,75,90))
data %>% dplyr::left_join(dplyr::distinct(scores, id, .keep_all = T))
# Joining, by = "id"
# id state score
# 1  1    KS    66
# 2  2    MN    86
# 3  3    AL    76
# 4  4    FL    NA
# 5  5    CA    NA

Moreover, if you want to replace the NAs in the new data.frame, try the tidyr::replace_na() function. Example:

data %>% dplyr::left_join(dplyr::distinct(scores, id, .keep_all = T)) %>% tidyr::replace_na(replace = list("score"=0L))
# Joining, by = "id"
# id state score
# 1  1    KS    66
# 2  2    MN    86
# 3  3    AL    76
# 4  4    FL     0
# 5  5    CA     0

Here is a base R method using aggregate and head:

merge(data, aggregate(score ~ id, data=scores, head, 1), by="id") 

The aggregate function breaks up the scores dataframe by id, then head is applied to get the first observation from each id. Since aggregate returns a data.frame, this is directly merged onto the data.frame data.

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Probably more efficient is to subset the scores data.frame using duplicated which will achieve the same result as aggregate, but will reduce the computational overhead.

merge(data, scores[!duplicated(scores$id),], by="id")

Using data.table along with mult = "first" and nomatch = 0L:

require(data.table)
setDT(scores); setDT(data) # convert to data.tables by reference

scores[data, mult = "first", on = "id", nomatch=0L]
#    id score state
# 1:  1    66    KS
# 2:  2    86    MN
# 3:  3    76    AL

For each row on data's id column, the matching rows in scores' id column are found, and the first one alone is retained (because mult = "first"). If there are no matches, they're removed (because of nomatch = 0L).