[Solved] Loop through data frame and variable names

I am looking for a way to automate some diagrams in R using a FOR loop:

dflist <- c("dataframe1", "dataframe2", "dataframe3", "dataframe4")

for (i in dflist) {
  plot(i$var1, i$var2)
}

All dataframes have the same variables, i.e. var1, var2.

It seems for loops are not the most elegant solution here, but I don’t understand how to use the apply functions for diagrams.

EDIT:

My original example using mean() didn’t help in the original question, so I changed it to a plot function.

Enquirer: Timm S.

||

Solution #1:

To further add to Beasterfield’s answer, it seems like you want to do some number of complex operations on each of the data frames.

It is possible to have complex functions within an apply statement. So where you now have:

for (i in dflist) {
  # Do some complex things
}

This can be translated to:

lapply(dflist, function(df) {
  # Do some complex operations on each data frame, df
  # More steps

  # Make sure the last thing is NULL. The last statement within the function will be
  # returned to lapply, which will try to combine these as a list across all data frames.
  # You don't actually care about this, you just want to run the function.
  NULL
})

A more concrete example using plot:

# Assuming we have a data frame with our points on the x, and y axes,
lapply(dflist, function(df) {
  x2 <- df$x^2
  log_y <- log(df$y)
  plot(x,y)
  NULL
})

You can also write complex functions which take multiple arguments:

lapply(dflist, function(df, arg1, arg2) {
  # Do something on each data.frame, df
  # arg1 == 1, arg2 == 2 (see next line)
}, 1, 2) # extra arguments are passed in here

Hope this helps you out!

Respondent: Timm S.

Solution #2:

Concerning your actual question you should learn how to access cells, rows and columns of data.frames, matrixs or lists. From your code I guess you want to access the j‘th columns of the data.frame i, so it should read:

mean( i[,j] )
# or
mean( i[[ j ]] )

The $ operator can be only used if you want to access a particular variable in your data.frame, e.g. i$var1. Additionally, it is less performant than accessing by [, ] or [[]].

However, although it’s not wrong, usage of for loops it is not very R’ish. You should read about vectorized functions and the apply family. So your code could be easily rewritten as:

set.seed(42)
dflist <- vector( "list", 5 )
for( i in 1:5 ){
  dflist[[i]] <- data.frame( A = rnorm(100), B = rnorm(100), C = rnorm(100) )
}
varlist <- c("A", "B")

lapply( dflist, function(x){ colMeans(x[varlist]) } )
Respondent: Scott Ritchie

Solution #3:

set.seed(42)
dflist <- list(data.frame(x=runif(10),y=rnorm(10)),
               data.frame(x=rnorm(10),y=runif(10)))

par(mfrow=c(1,2))
for (i in dflist) {
  plot(y~x, data=i)
}
Respondent: Beasterfield

Solution #4:

Using the example of @Roland, I wanted to show you the ggplot2 equivalent. First we have to change the datset a bit:

First the original data:

> dflist
[[1]]
           x           y
1  0.9148060 -0.10612452
2  0.9370754  1.51152200
3  0.2861395 -0.09465904
4  0.8304476  2.01842371
5  0.6417455 -0.06271410
6  0.5190959  1.30486965
7  0.7365883  2.28664539
8  0.1346666 -1.38886070
9  0.6569923 -0.27878877
10 0.7050648 -0.13332134

[[2]]
            x          y
1   0.6359504 0.33342721
2  -0.2842529 0.34674825
3  -2.6564554 0.39848541
4  -2.4404669 0.78469278
5   1.3201133 0.03893649
6  -0.3066386 0.74879539
7  -1.7813084 0.67727683
8  -0.1719174 0.17126433
9   1.2146747 0.26108796
10  1.8951935 0.51441293

and put the data into one data.frame, with an id column

require(reshape2)
one_df = melt(dflist, id.vars = c("x","y"))
> one_df
            x           y L1
1   0.9148060 -0.10612452  1
2   0.9370754  1.51152200  1
3   0.2861395 -0.09465904  1
4   0.8304476  2.01842371  1
5   0.6417455 -0.06271410  1
6   0.5190959  1.30486965  1
7   0.7365883  2.28664539  1
8   0.1346666 -1.38886070  1
9   0.6569923 -0.27878877  1
10  0.7050648 -0.13332134  1
11  0.6359504  0.33342721  2
12 -0.2842529  0.34674825  2
13 -2.6564554  0.39848541  2
14 -2.4404669  0.78469278  2
15  1.3201133  0.03893649  2
16 -0.3066386  0.74879539  2
17 -1.7813084  0.67727683  2
18 -0.1719174  0.17126433  2
19  1.2146747  0.26108796  2
20  1.8951935  0.51441293  2

and make the plot:

require(ggplot2)
ggplot(one_df, aes(x = x, y = y)) + geom_point() + facet_wrap(~ L1)

enter image description here

Respondent: Roland

Solution #5:

Based on Scott Ritchi solution, this would the reproducible example, hiding also the feedback message from lapply:

# split dataframe by condition on cars hp
f <- function() trunc(signif(mtcars$hp, 2) / 100)
dflist <- lapply(unique(f()), function(x) subset(mtcars, f() == x ))

This splits the mtcars dataframe is subsets based on the hp variable classification (0 for hp lower than 100, 1 for those in the 100’s, 2 for 200’s, and so on.)

And, plot it:

# use invisible to prevent the feedback message from lapply
invisible(
    lapply(dflist, function(df) {
    x2 <- df$mpg^2
    log_y <- log(df$hp)
    plot(x2, log_y)
    NULL
}))

invisible() will prevent the lapply() message:

16 
9 
6 
1 
[[1]]
NULL

[[2]]
NULL

[[3]]
NULL

[[4]]
NULL

enter image description here

Respondent: Paul Hiemstra

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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