Change the name of one column
colnames(df)[colnames(df)=="old_name"] <- "new_name"
df in the name of the data frame.
old_name has to match the original name.
Change the names of all columns at once
colnames(df) <- c("col1", "col2")
Be sure to have the same number of items in
c() and columns in
Change the type of a variable
var <- as.data.frame(var)
Change a string to a float
var is a string that contains floats,
as.numeric(var) won’t work. Why? I have absolutely no idea. But basically, you well lose the decimal part of
var, and even the main part will be messed up.
To avoid that, you have to use
Why? No idea ¯\_(ツ)_/¯
Check if a variable is
To check if a variable is empty, do not use
df$var == "NA" or something similar, but
Create an empty data frame
df <- data.frame(var1 = NA, var2 = NA)
Each variable will be a new column.
Drop one or more columns
df <- NULL
This code will drop the second column.
To drop more than one column, use:
df[2:5] <- NULL
The columns from 2 to 5 will be dropped.
Source (with even more options).
df = select(df, -col1., -col23, -col47)
col1, use the actual name of the column in the data frame.
Extract data of a given cell
Two ways to do so.
The first one is the most elegant. It is:
x is the row number.
The second one is by knowing the numbers of the row and the column of the cell:
x is the row number and
y is the column number. Omitting
y will return the values of the row
x, and conversely if omitting
Extract data of a given cell based on the content of another cell
Let’s say you want to know the value of the cell
price when the cell
EUR. The previous code could be tricky, especially because it requires you to look at the data frame, something you don’t want to.
price when currency is
EUR, use the following code:
df$price[df$currency == "EUR"]
Also, see this for more details on the conditions
Isolate part of a data frame based on certain values
Let say you have a data frame
df with two prices:
USD. And you want to extract the observations with only one of those currencies. Use this:
df <- df[ df$currency %in% c("EUR"), ]
subset() might also help (more).
scales package, use the
percent(x, accuracy = NULL, scale = 100, prefix = "", suffix = "%", big.mark = " ", decimal.mark = ".", trim = TRUE)
This exemple will produce a percent with 1 decimal digit:
percent(x = (var1_2018 - var1_2017)/var1_2017, accuracy = .1)
Arguments (copied from source):
accuracy: Number to round to, NULL for automatic guess.
scale: A scaling factor: x will be multiply by scale before formating (useful if the un- derlying data is on another scale, e.g. for computing percentages or thousands).
suffix: Symbols to display before and after value.
big.mark: Character used between every 3 digits to separate thousands.
decimal.mark: The character to be used to indicate the numeric decimal point.
trim: Logical, if FALSE, values are right-justified to a common width (see base::format())
...: Other arguments passed on to base::format().
x: A numeric vector to format.
digits: Deprecated, use accuracy instead.
unit: The units to append.
sep: The separator between the number and the unit label.
Remove empty rows/lines
Sometime, because of formatting or for others reasons, some rows may be empty while still present in the data frame. To remove them, it’s simple:
df <- read.csv2("file.csv", sep = ";", header = TRUE, na.strings = c("", "NA")) df <- remove_empty(df, which = c("rows", "cols"))
na.strings = c("", "NA") ensures that every blank value will be turned into
NA. This is needed for
remove_empty to work (needs
janitor package, see here).
remove_empty removes any row or column entirely made of
NA (so this is why we needed the first line).
remove_empty_columns are deprecated.
Let’s say you want to remove all observations that have an empty value for a given column/variable. Here it is:
df <- dplyr::filter(df, !is.na(var))
It requires the
dplyr package (source).
Reorder columns in a dataframe
Two solutions are available:
# reorder by column name df <- df[c("A", "B", "C")] # reorder by column index df <- df[c(1,3,2)]
df <- select(df, col2, col1)
df will be reorder from
col1, col2 to
With 0 (zeros)
To replace all the
NA in a data frame with
df[is.na(df)] <- 0
To replace only the
NA in a column (variable):
df$var[is.na(df$var)] <- 0
With a value contained in another column
You want to replace
var1 with the value of
NA. Intuitively, you want to write:
df$var1[is.na(df$var1)] <- df$var2
But this code will run into problems. Basically, if only certain values of
var1 are empty, you will have a discrepancy between
var2. For instance, if the 9th row in
var1 are empty, this code will replace
var1 not with the 9th row in
var2, but with the 1st – so the discrepancy. You will get the error
Number of items to replace is not a multiple of replacement length.
To avoid that, you need to use an
df$var1 <- ifelse(is.na(df$var1), df$var2, df$var1)
What this code does is simple: if
df$var1 is empty (
NA), it will be replaced by
var2. If not, it will be replaced by itself – nothing will change.
For whatever reason, this code avoids the discrepancy issue mentioned earlier.
Retrieve the numbers and the names of all columns and all rows
ncol(df) colnames(df) nrow(df) rnames(df)
Let say you want to sort a data frame after one specific column/variable. Here is the code to do that:
df <- arrange(df, desc(var))
var is the variable used to sort
df. To sort
df in the opposite direction, remove