Return row of Data Frame based on value in a column - R

Question

My R data.frame df looks like:

     Name     Amount
1    \"A\"      150
2    \"B\"      120
3    \"C\"      \"NA\"
4    \"D\"      160
.
.
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Answer 1

You could use dplyr:

df %>% group_by("Amount") %>% slice(which.min(x))

Answer 2

Based on the syntax provided

 Select * Where Amount = min(Amount)

You could do using:

 library(sqldf)

Using @Kara Woo's example df

  sqldf("select * from df where Amount in (select min(Amount) from df)")
  #Name Amount
 #1    B    120
 #2    E    120

Answer 3

Use which.min:

df <- data.frame(Name=c('A','B','C','D'), Amount=c(150,120,175,160))
df[which.min(df$Amount),]

> df[which.min(df$Amount),]
  Name Amount
2    B    120

From the help docs:

Determines the location, i.e., index of the (first) minimum or maximum of a numeric (or logical) vector.

Answer 4

@Zelazny7's answer works, but if you want to keep ties you could do:

df[which(df$Amount == min(df$Amount)), ]

For example with the following data frame:

df <- data.frame(Name = c("A", "B", "C", "D", "E"), 
                 Amount = c(150, 120, 175, 160, 120))

df[which.min(df$Amount), ]
#   Name Amount
# 2    B    120

df[which(df$Amount == min(df$Amount)), ]
#   Name Amount
# 2    B    120
# 5    E    120

Edit: If there are NAs in the Amount column you can do:

df[which(df$Amount == min(df$Amount, na.rm = TRUE)), ]