How to Use the Row.Names Attribute to Order the Rows of My Dataframe in R

How can I use the row.names attribute to order the rows of my dataframe in R?

This worked for me:

new_df <- df[ order(row.names(df)), ]

How do I order by row.names in dataframe

I've had this problem before. Here's how I solved it. Using some iris sample data from base R,

> dat <- iris[sample(nrow(iris), 10), ]
> rn <- rownames(dat)
> dat[order(as.numeric(rn)), ]

How to order a data.frame based on row.names in another data frame?

Here's one option, using the built-in mtcars data frame for illustration:

# Create new sorted data frame. This is analogous to the second data frame
# by which you want to sort the first one.
mtcars.sorted = mtcars[order(rownames(mtcars)),]

# Sort original data frame by the order of the new data frame
mtcars[match(rownames(mtcars.sorted), rownames(mtcars)),]

So, in your case, the code would be:

DF1[match(rownames(DF2), rownames(DF1)), ]

How can I order the rownames of a dataframe in R?

row_names_test <- data.frame(cbind(genes <- rownames(read.csv("~/row_names_test.csv")),
                                   read.csv("~/row_names_test.csv")))

row_names_test <- row_names_test[order(row_names_test$genes), ]

How to convert row names to column names and bind it by order in r

Assuming your long-form data starts with the rows in the order you want, try this:

month_order = unique(returns)
returns.df <- returns %>%
  spread(key = DATE, value = RETURN) %>% 
  select(c("TICKER", month_order))

If your data doesn't begin in the right order, append a year and convert it to a Date class object. Then you can sort it to the right order and use the method above.

How to sort each row of a data frame WITHOUT losing the column names

Store the names and apply them:

nm = names(df)
sorted_df <- as.data.frame(t(apply(df, 1, sort)))
names(sorted_df) = nm

You could compress this down to a single line if you prefer:

sorted_df = setNames(as.data.frame(t(apply(df, 1, sort))), names(df))

R, How to set row names attribute as numeric from character?

From R's help on ?row.names:

All data frames have a row names attribute, a character vector of length the number of rows with no duplicates nor missing values.

This means that the row names will always be a character vector. You would need to use workarounds as suggested in the comments to make them "usable" as integers, basically always coercing. One suggestion could be that you create an id column of class integer and do not use row.names as id:

df$id <- as.integer(row.names(df)) 
df[order(df$id), ]

Omitting row.names also seems to be the way to go with popular data frame rethinking such as data.table or tibble - none of those use row names.

How to set row names for list of matrix or data frame?

I'm adding @Roland's answer here so the question appears answered.

ismr3 <- lapply(ismr2, function(x){ row.names(x)<-as.character(x$ID); x})

The important part is that the function inside lapply actually returns the value to be added to the list. With just the row.names<- assignment, the value returned from that is just the row names themselves. So because each assignment was returning a character vector, you were not creating an object that had a dim() attribute. (Atomic vectors have length, not dim).

---

Related Topics