Access Data Frame Column Using Variable

access data frame column using variable

After creating your data frame, you need to use ?colnames. For example, you would have:

d = data.frame(a=c(1,2,3), b=c(4,5,6))
colnames(d) <- c("col1", "col2")

You can also name your variables when you create the data frame. For example:

d = data.frame(col1=c(1,2,3), col2=c(4,5,6))

Further, if you have the names of columns stored in variables, as in

a <- "col1"

you can't use $ to select a column via d$a. R will look for a column whose name is a. Instead, you can do either d[[a]] or d[,a].

Dynamically select data frame columns using $ and a character value

You can't do that kind of subsetting with $. In the source code (R/src/main/subset.c) it states:

/*The $ subset operator.

We need to be sure to only evaluate the first argument.

The second will be a symbol that needs to be matched, not evaluated.

*/

Second argument? What?! You have to realise that $, like everything else in R, (including for instance ( , + , ^ etc) is a function, that takes arguments and is evaluated. df$V1 could be rewritten as

`$`(df , V1)

or indeed

`$`(df , "V1")

But...

`$`(df , paste0("V1") )

...for instance will never work, nor will anything else that must first be evaluated in the second argument. You may only pass a string which is never evaluated.

Instead use [ (or [[ if you want to extract only a single column as a vector).

For example,

var <- "mpg"
#Doesn't work
mtcars$var
#These both work, but note that what they return is different
# the first is a vector, the second is a data.frame
mtcars[[var]]
mtcars[var]

You can perform the ordering without loops, using do.call to construct the call to order. Here is a reproducible example below:

#  set seed for reproducibility
set.seed(123)
df <- data.frame( col1 = sample(5,10,repl=T) , col2 = sample(5,10,repl=T) , col3 = sample(5,10,repl=T) )

#  We want to sort by 'col3' then by 'col1'
sort_list <- c("col3","col1")

#  Use 'do.call' to call order. Seccond argument in do.call is a list of arguments
#  to pass to the first argument, in this case 'order'.
#  Since  a data.frame is really a list, we just subset the data.frame
#  according to the columns we want to sort in, in that order
df[ do.call( order , df[ , match( sort_list , names(df) ) ]  ) , ]

   col1 col2 col3
10    3    5    1
9     3    2    2
7     3    2    3
8     5    1    3
6     1    5    4
3     3    4    4
2     4    3    4
5     5    1    4
1     2    5    5
4     5    3    5

Access column using variable name in R in if statement, condition has length 1

The way you're selecting columns in fine. Using df[[col_name]] (list context) is the same as df[, col_name] -- each returns a vector copy of column col_name. You can save the column name as a variable instead of using paste0 directly in the selection.

The reason you're getting an error is that if is not vectorized and you're giving it a vector with length > 1. In this case, if uses only the first value in the vector, but warns that it's doing so. ifelse is the vectorized version in base R (there's also dplyr::if_else). If I understand your code, the below should be close to what you're looking for.

t1 <- paste0('Z_in_', trait1)
t2 <- paste0('Z_in_', trait2)

# a single boolean vector indicating if trait1 and trait2 are 
# both positive or both negative
same_sign <- ((leadsnp4[, t1] > 0) & (leadsnp4[, t2] > 0)) | 
  ((leadsnp4[, t1] < 0) & (leadsnp4[, t2] < 0))

leadsnp4$ConcordEffect <- ifelse(same_sign, "Yes", "No")

Note that if trait1 and/or trai2 are equal to 0 they will be assigned false. You'll need to modify the logic if this is not the desired behavior.

How to use a string variable to select a data frame column using $ notation

If you have a variable x with a column name in tmp, tmp[,x] or tmp[[x]] are the correct ways to extract it. You cannot get R to treat tmp$x as tmp$"Q5.3". tmp$x will always refer to the item named "x" in "tmp".

How to access the name of a column which is a number through a variable

You can simply do-

> k=99
> df[[paste0(k)]]
[1] 1 2 3

Just for more clarification on your for loop -

cols <- c(99,66)
for(i in 1:length(cols))
{
   print(df[[paste0(cols[i])]])
}

Output-

[1] 1 2 3
[1] 4 5 6

Select columns using a variable in R

We can use .. before the idx to select the columns in data.table or with = FALSE

library(data.table)
df[, ..idx]
df[, idx, with = FALSE]

How can I Access a Column by Name as a Variable to use the isin() Method

Give this a try.. using the set functionality

Usercol ='Account' #user entry
Common = 
list(set(df1.loc[:Usercol]).intersect(set(df2.loc[:Usercol])))
#fetch index of each data frame using
df1[df1[Usercol].isin(Common)].index
df2[df2[Usercol].isin(Common)].index

using a variable to specify a column in a dataframe

If we want to use an object, then use [[ instead of $ similar to the 'zonename'

mean(zones[[zonename]][[atr]] , na.rm = TRUE) #

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