Plot one column in every matrix in a list of matrices in R

Question

I'm quite new to programming as well as data analysis, please bear with me here. My data currently consists of a list of 14 matrices (lom), each corresponding to data from a country (with two-letter country codes).

Here is a full sample for Austria:

> lom["AT"]
$`AT`
   Year    AllKey    AllSub    SelKey    SelSub
1  2000  1.622279 0.5334964  1.892894 0.8057591
2  2001  1.903745 0.5827514  2.291335 0.8295899
3  2002  1.646538 0.4873866  2.006873 0.7360566
4  2003  1.405250 0.8692641  2.105648 1.2711968
5  2004  1.511154 1.5091751  1.970236 1.9407666
6  2005  1.459177 0.6781008  1.808982 1.1362805
7  2006  1.604652 0.5038658  1.942126 0.7992008
8  2007  2.107326 0.9260200  2.683072 1.3302627
9  2008  1.969735 0.6178362  2.994758 1.2051339
10 2009  1.955768 0.7365529  2.896198 1.2272024
11 2010  2.476157 0.7952590  3.715950 1.5686643
12 2011  2.092459 0.4970011  2.766169 0.6476707
13 2012  1.913122 0.5338756  2.450942 0.6022315
14 2013  2.086200 0.6739412  2.786736 0.9211941
15 2014  2.579428 0.8424793  3.152541 1.0225888
16 2015 10.662568 5.8472436  9.769320 3.8840780
17 2016 11.088286 4.6504581 10.567789 3.2383420
18 2017  7.225053 1.7528594  6.747515 1.2781224

I'd like to get all 14 countries plotted against x = Year and y = each of the other variables, i.e. four plots with 14 lines each. Hence the requirement in the question title.

I keep coming up with impossibilities involving some combination of a for loop and some apply function, for example:

for (i in colnames(lom$anyCountry)) {
    ggplot(lapply(lom, function(x) x[,1:14], aes(x=Year, y=i)   
}

which apart from many other problems I can now see throws:

Error: data must be a data frame, or other object coercible by fortify(), not a list

which led me to combine the list of matrices into a big matrix inspired by this:

bigDF <- do.call(rbind, lom)

I suppose I could restructure my data some other way, perhaps I'm missing some functionality that would help... probably both. I would appreciate any pointers as to how to achieve this as efficiently as possible.

Answer 1

Consider appending all matrix data into a master, single data frame with a country indicator that you can use for the color argument of line plots:

# CREATE LARGE DATAFRAME FROM MATRIX LIST
lom_df <- do.call(rbind, lapply(lom, data.frame))

# CREATE COLUMN NAMES FROM ROWNAMES
lom_df$country <- gsub("\\..*$", "", row.names(lom_df))
row.names(lom_df) <- NULL

# EXTRACT ALL FOUR Y COLUMN NAMES (MINUS Year AND country)
y_columns <- colnames(lom_df[2:(ncol(lom_df)-1)])

# PRODUCE LIST OF FOUR PLOTS EACH WITH COUNTRY LINES
plot_list <- lapply(y_columns, function(col)
  ggplot(lom_df, aes_string(x="Year", y=col, color="country")) +
     geom_line()
)

# OUTPUT EACH LIST 
plot_list

Answer 2

This solution uses package ggplot2.

It has two steps, data preparation and plotting.

First of all the list must be transformed into one large data frame, with a column as an id column. I have searched SO for a function that does this but couldn't find one so here it goes.

rbindWithID <- function(x, id.name = "ID", sep = "."){
    if(is.null(names(x))) names(x) <- paste(id.name, seq_along(x), sep = sep)
    res <- lapply(names(x), function(nm){
        DF <- x[[nm]]
        DF[[id.name]] <- nm
        x[[nm]] <- cbind(DF[ncol(DF)], DF[-ncol(DF)])
        x[[nm]]
    })
    do.call(rbind, res)
}

lom_df <- rbindWithID(lom, "Country")

Now reshape the data frame from wide to long.

molten <- reshape2::melt(lom_df, id.vars = c("Country", "Year"))

Finally, plot it.

library(ggplot2)

ggplot(molten, aes(Year, value, colour = Country)) +
    geom_line() +
    facet_wrap(~ variable)

DATA.

set.seed(1234)    # Make the results reproducible

lom <- lapply(1:4, function(i){
    data.frame(
        Year = 2000:2008,
        AllKey = runif(9, 1, 2),
        AllSub = runif(9, 0, 2),
        SelKey = runif(9, 1, 2),
        SelSub = runif(9, 0, 2)
    )
})

names(lom) <- c("AT", "DE", "FR", "PT")