How to Create a Variable of Rownames

Convert row names into first column

Or you can use tibble's rownames_to_column which does the same thing as David's answer:

library(tibble)
df <- tibble::rownames_to_column(df, "VALUE")

Note: The earlier function called add_rownames() has been deprecated and is being replaced by tibble::rownames_to_column()

assign rownames within a loop to constructed variables names

Based on Roland's comment, I come out with this solution:

rows<-c("a","b")
df<-data.frame(var1=c(1,2),var2=c(10,20))
dfs<-list()
for (n in 1:2){
  dfs[[n]]<-df
  rownames(dfs[[n]])<-rows
}

A list is the key!

Convert the values in a column into row names in an existing data frame

This should do:

samp2 <- samp[,-1]
rownames(samp2) <- samp[,1]

So in short, no there is no alternative to reassigning.

Edit: Correcting myself, one can also do it in place: assign rowname attributes, then remove column:

R> df<-data.frame(a=letters[1:10], b=1:10, c=LETTERS[1:10])
R> rownames(df) <- df[,1]
R> df[,1] <- NULL
R> df
   b c
a  1 A
b  2 B
c  3 C
d  4 D
e  5 E
f  6 F
g  7 G
h  8 H
i  9 I
j 10 J
R> 

Convert column/variable to row names for multiple lists in R with sapply

Because in R functions return the last line or explicit return() call, your anonymous function function(x) { rownames(x) <- x$zipcode } returns the result of row.names which per docs:

row.names returns a character vector.

This can be quickly fixed by calling x after row names change as commented, function(x) { rownames(x) <- x$zipcode; x } or function(x) { rownames(x) <- x$zipcode; return(x) }.

However, consider row.names<- (subtle difference with assignment operator embedded):

function(x) { `rownames<-`(x, value= x$zipcode }

which per same docs:

row.names<- returns a data frame with the row names changed.

Altogether, removing redundant curly braces {...} and default USE.NAMES = TRUE:

test_list <- sapply(polygon_nyc_listings, 
                    function(df) `row.names<-`(df, value = df$zipcode),
                    simplify = FALSE)

How to make column 0: rownames a number starting at 1 like a regular dataset in r?

We can use rownames_to_column from tibble

library(tibble)
df1 <- rownames_to_column(df1, "Variable")

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Or using base R

df1$Variable <- row.names(df1)
row.names(df1) <- NULL

set row names of dataset to be equal to another variable

very simple.

row.names(yourdf) <- yourdf$psid

How to convert a column to row.names from a tibble

First, you can see the difference

> str(entrepreneur[, 1])
tibble [5 x 1] (S3: tbl_df/tbl/data.frame)
 $ Factors: chr [1:5] "Competition" "Cultural Support" "Financing" "High Growth" ...      

> str(entrepreneur[[1]])
 chr [1:5] "Competition" "Cultural Support" "Financing" "High Growth" ...

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Try the code below (using entrepreneur[[1]] rather than entrepreneur[,1])

> `rownames<-`(as.data.frame(entrepreneur[-1]), entrepreneur[[1]])
                 Baden-Wⁿrttemberg Bayern Berlin Brandenburg
Competition                   0.71   0.67   1.00        1.00
Cultural Support              0.66   0.66   0.56        0.55
Financing                     0.81   0.83   0.90        0.64
High Growth                   0.62   0.77   0.82        1.00
Human Capital                 0.46   0.49   0.79        0.77

How do I create a variable based on selected characters from row.names?

Specifically, keep all text prior to the first . character:

df$Name <- sub('[.].*','', rownames(df))

In a regular expression, . means to match any character. So to match a literal dot, the character must be escaped. However, within a character class (bounded by [ and ]), the . does represent a literal dot. That is, [.] matches a literal dot only.

Following that is .* which matches any number of any characters.

Thus, the expression matches everything from the first literal . to the end of the string, and replaces it with the empty string.

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