# Iterative Programming

Almost everything you do when dealing with data will need to be done again, and again, and again. If you are copy-pasting your way to repetitively do the same thing, you’re not only doing things inefficiently, you’re almost certainly setting yourself up for trouble if anything changes about the data or underlying process.

In order to avoid this, you need to be familiar with basic programming, and a starting point is to use an iterative approach to repetitive problems. Let’s look at the following. Let’s say we want to get the means of some columns in our data set. Do you do something like this?

means1 = mean(df$x) means2 = mean(df$y)
means3 = mean(df$z) means4 = mean(df$q)

Now consider what you have to change if you change a variable name, decide to do a median, or the data object name changes. Any minor change with the data will cause you to have to redo that code, and possibly every line of it.

## For Loops

A for loop will help us get around the problem. The idea is that we want to perform a particular action for every iteration of some sequence. That sequence may be over columns, rows, lines in a text, whatever. Here is a loop.

for (column in c('x','y','z','q')) {
mean(df[[column]])
}

What’s going on here? We’ve created an iterative process in which, for every element in c('x','y','z','q'), we are going to do something. We use the completely arbitrary word column as a placeholder to index which of the four columns we’re dealing with at a given point in the process. On the first iteration, column will equal x, on the second y, and so on. We then take the mean of df[[column]], which will be df[['x']], then df[['y']], etc.

Here is an example with the nycflights data, which regards flights that departed New York City in 2013. The weather data set has columns for things like temperature, humidity, and so forth.

weather = nycflights13::weather

for (column in c('temp', 'humid', 'wind_speed', 'precip')) {
print(mean(weather[[column]], na.rm = TRUE))
}
[1] 55.26039
[1] 62.53006
[1] 10.51749
[1] 0.004469079

You can check this for yourself by testing a column or two directly with just mean(df$x). Now if the data name changes, the columns we want change, or we want to calculate something else, we usually end up only changing one thing, rather than at least changing one at a minimum, and probably many more things. In addition, the amount of code is the same whether the loop goes over 100 columns or 4. Let’s do things a little differently. columns = c('temp', 'humid', 'wind_speed', 'precip') nyc_means = rep(NA, length(columns)) for (i in seq_along(columns)) { column = columns[i] nyc_means[i] = mean(weather[[column]], na.rm = TRUE) # alternative without the initial first step # nyc_means[i] = mean(weather[[columns[i]]], na.rm = TRUE) } nyc_means [1] 55.260392127 62.530058972 10.517488384 0.004469079 By creating a columns object, if anything changes about the columns we want, that’s the only line in the code that would need to be changed. The i is now a place holder for a number that goes from 1 to the length of columns (i.e. 4). We make an empty nyc_means object that’s the length of the columns, so that each element will eventually be the mean of the corresponding column. In the following I remove precipitation and add visibility and air pressure. columns = c('temp', 'humid', 'wind_speed', 'visib', 'pressure') nyc_means = rep(NA, length(columns)) for (i in seq_along(columns)) { nyc_means[i] = mean(weather[[columns[i]]], na.rm = TRUE) } nyc_means %>% round(2) [1] 55.26 62.53 10.52 9.26 1017.90 Had we been copy-pasting, this would require deleting or commenting out a line in our code, pasting two more, and changing each one after pasting to represent the new columns. That’s tedious, and not a fun way to code. ### A slight speed gain Note that you do not have to create an empty object like we did. The following works also. columns = c('temp', 'humid', 'wind_speed', 'visib', 'pressure') nyc_means = numeric() for (i in seq_along(columns)) { nyc_means[i] = mean(weather[[columns[i]]], na.rm = TRUE) } nyc_means %>% round(2) [1] 55.26 62.53 10.52 9.26 1017.90 However, the other approach is slightly faster, because memory is already allocated for all elements of nyc_means, rather than updating it every iteration of the loop. This speed gain can become noticeable when dealing with thousands of columns and complex operations. ### While alternative When you look at some people’s R code, you may see a loop of a different sort. columns = c('temp','humid','wind_speed', 'visib', 'pressure') nyc_means = c() i = 1 while (i <= length(columns)) { nyc_means[i] = mean(weather[[columns[i]]], na.rm = TRUE) i = i + 1 } nyc_means %>% round(2) [1] 55.26 62.53 10.52 9.26 1017.90 This involves a while statement. It states, while i is less than or equal to the length (number) of columns, compute the value of the ith element of nyc_means as the mean of ith column of weather. After that, increase the value of i. So, we start with i = 1, compute that subsequent mean, i now equals 2, do the process again, and so on. The process will stop as soon as the value of i is greater than the length of columns. There is zero difference to using the while approach vs. the for loop. While is often used when there is a check to be made, e.g. in modeling functions that have to stop the estimation process at some point, or else they’d go on indefinitely. In that case the while syntax is probably more natural. Either is fine. ### Loops summary Understanding loops is fundamental toward spending less time processing data and more time toward exploring it. Your code will be more succinct and more able to handle the usual changes that come with dealing with data. Now that you have a sense of it, know that once you are armed with the sorts of things we’ll be talking about next- apply functions, writing functions, and vectorization - you’ll likely have little need to write explicit loops. While there is always a need for iterative processing of data, R provides even more efficient means to do so. ## Implicit Loops Writing loops is straightforward once you get the initial hang of it. However, R offers alternative ways to do loops that can simplify code without losing readability. As such, even when you loop in R, you don’t have to do so explicitly. ### apply family A family of functions comes with R that allows for a succinct way of looping when it is appropriate. Common functions in this family include: • apply • arrays, matrices, data.frames • lapply, sapply, vapply • lists, data.frames, vectors • tapply • grouped operations (table apply) • mapply • multivariate version of sapply • replicate • performs an operation N times As an example we’ll consider standardizing variables, i.e. taking a set of numbers, subtracting the mean, and dividing by the standard deviation. This results in a variable with mean of 0 and standard deviation of 1. Let’s start with a loop approach. for (i in 1:ncol(mydf)) { x = mydf[, i] for (j in 1:length(x)) { x[j] = (x[j] - mean(x)) / sd(x) } } The above would be a really bad way to use R. It goes over each column individually, then over each value of the column. Conversely, apply will take a matrix or data frame, and apply a function over the margin, row or column, you want to loop over. The first argument is the data you’re considering, the margin is the second argument (1 for rows, 2 for columns12), and the function you want to apply to those rows is the third argument. The following example is much cleaner compared to the loop, and now you’d have a function you can use elsewhere if needed. stdize <- function(x) { (x - mean(x)) / sd(x) } apply(mydf, 2, stdize) # 1 for rows, 2 for columnwise application Many of the other apply functions work similarly, taking an object and a function to do the work on the object (possibly implicit), possibly with other arguments specified if necessary. #### lapply Let’s say we have a list object, or even just a vector of values. There are no rows or columns to iterate over, so what do we do here? x = list('aba', 'abb', 'abc', 'abd', 'abe') lapply(x, str_remove, pattern = 'ab') [[1]] [1] "a" [[2]] [1] "b" [[3]] [1] "c" [[4]] [1] "d" [[5]] [1] "e" The lapply operation iterates over each element of the list and applies a function to them. In this case, the function is str_remove. It has an argument for the string pattern we want to take out of the character string that is fed to it (‘ab’). For example, for ‘aba’ we will be left with just the ‘a’. As can be seen, lapply starts with a list and returns a list. The only difference with sapply is that sapply will return a simplified form if possible13. sapply(x, str_remove, pattern = 'ab') [1] "a" "b" "c" "d" "e" In this case we just get a vector back. ### Apply functions It is important to be familiar with the apply family for efficient data processing, if only because you’ll regularly come code employing these functions. A summary of benefits includes: • Cleaner/simpler code • Environment kept clear of unnecessary objects • Potentially more reproducible • more likely to use generalizable functions • Parallelizable Note that apply functions are NOT necessarily faster than explicit loops, and if you create an empty object for the loop as discussed previously, the explicit loop will likely be faster. On top of that, functions like replicate and mapply are especially slow. However, the apply family can ALWAYS potentially be faster than standard R loops do to parallelization. With base R’s parallel package, there are parallel versions of the apply family, e.g.parApply, parLapply etc. As every modern computer has at least four cores to play with, you’ll always potentially have nearly a 4x speedup by using the parallel apply functions. Apply functions and similar approaches should be a part of your regular R experience. We’ll talk about other options that may have even more benefits, but you need to know the basics of how apply functions work in order to use those. I use R every day, and very rarely use explicit loops. Note that there is no speed difference for a for loop vs. using while. And if you must use an explicit loop, create an empty object of the dimension/form you need, and then fill it in via the loop. This will be notably faster. I pretty much never use an explicit double loop, as a little more thinking about the problem will usually provide a more efficient path to solving the problem. ### purrr The purrr package allows you to take the apply family approach to the tidyverse. And with packages future + furrr, they too are parallelizable. Consider the following. We’ll use the map function to map the sum function to each element in the list, the same way we would with lapply. x = list(1:3, 4:6, 7:9) map(x, sum) [[1]] [1] 6 [[2]] [1] 15 [[3]] [1] 24 The map functions take some getting used to, and in my experience they are typically slower than the apply functions, sometimes notably so. However they allow you stay within the tidy realm, which has its own benefits, and have more control over the nature of the output14, which is especially important in reproducibility, package development, producing production-level code, etc. The key idea is that the map functions will always return something the same length as the input given to it. The purrr functions want a list or vector, i.e. they don’t work with data.frame objects in the same way we’ve done with mutate and summarize except in the sense that data.frames are lists. ## mtcars %>% ## map(scale) # returns a list, not shown mtcars %>% map_df(scale) # returns a df # A tibble: 32 x 11 mpg[,1] cyl[,1] disp[,1] hp[,1] drat[,1] wt[,1] qsec[,1] vs[,1] am[,1] gear[,1] carb[,1] <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> 1 0.151 -0.105 -0.571 -0.535 0.568 -0.610 -0.777 -0.868 1.19 0.424 0.735 2 0.151 -0.105 -0.571 -0.535 0.568 -0.350 -0.464 -0.868 1.19 0.424 0.735 3 0.450 -1.22 -0.990 -0.783 0.474 -0.917 0.426 1.12 1.19 0.424 -1.12 4 0.217 -0.105 0.220 -0.535 -0.966 -0.00230 0.890 1.12 -0.814 -0.932 -1.12 5 -0.231 1.01 1.04 0.413 -0.835 0.228 -0.464 -0.868 -0.814 -0.932 -0.503 6 -0.330 -0.105 -0.0462 -0.608 -1.56 0.248 1.33 1.12 -0.814 -0.932 -1.12 7 -0.961 1.01 1.04 1.43 -0.723 0.361 -1.12 -0.868 -0.814 -0.932 0.735 8 0.715 -1.22 -0.678 -1.24 0.175 -0.0278 1.20 1.12 -0.814 0.424 -0.503 9 0.450 -1.22 -0.726 -0.754 0.605 -0.0687 2.83 1.12 -0.814 0.424 -0.503 10 -0.148 -0.105 -0.509 -0.345 0.605 0.228 0.253 1.12 -0.814 0.424 0.735 # … with 22 more rows mtcars %>% map_dbl(sum) # returns a numeric (double) vector of column sums  mpg cyl disp hp drat wt qsec vs am gear carb 642.900 198.000 7383.100 4694.000 115.090 102.952 571.160 14.000 13.000 118.000 90.000  diamonds %>% map_at( vars(carat, depth, price), function(x) as.integer(x > median(x)) ) %>% as_tibble() # A tibble: 53,940 x 10 carat cut color clarity depth table price x y z <int> <ord> <ord> <ord> <int> <dbl> <int> <dbl> <dbl> <dbl> 1 0 Ideal E SI2 0 55 0 3.95 3.98 2.43 2 0 Premium E SI1 0 61 0 3.89 3.84 2.31 3 0 Good E VS1 0 65 0 4.05 4.07 2.31 4 0 Premium I VS2 1 58 0 4.2 4.23 2.63 5 0 Good J SI2 1 58 0 4.34 4.35 2.75 6 0 Very Good J VVS2 1 57 0 3.94 3.96 2.48 7 0 Very Good I VVS1 1 57 0 3.95 3.98 2.47 8 0 Very Good H SI1 1 55 0 4.07 4.11 2.53 9 0 Fair E VS2 1 61 0 3.87 3.78 2.49 10 0 Very Good H VS1 0 61 0 4 4.05 2.39 # … with 53,930 more rows However, working with lists is very useful, so let’s turn to that. ## Looping with Lists Aside from data frames, you may think you don’t have much need for list objects. However, list objects make it very easy to iterate some form of data processing. Let’s say you have models of increasing complexity, and you want to easily summarise and/or compare them. We create a list for which each element is a model object. We then apply a function, e.g. to get the AIC value for each, or adjusted R square (this requires a custom function). library(mgcv) # for gam mtcars$cyl = factor(mtcars$cyl) mod_lm = lm(mpg ~ wt, data = mtcars) mod_poly = lm(mpg ~ poly(wt, 2), data = mtcars) mod_inter = lm(mpg ~ wt * cyl, data = mtcars) mod_gam = gam(mpg ~ s(wt), data = mtcars) mod_gam_inter = gam(mpg ~ cyl + s(wt, by = cyl), data = mtcars) model_list = list( mod_lm = mod_lm, mod_poly = mod_poly, mod_inter = mod_inter, mod_gam = mod_gam, mod_gam_inter = mod_gam_inter ) # lowest wins model_list %>% map_dbl(AIC) %>% sort() mod_gam_inter mod_inter mod_poly mod_gam mod_lm 150.6324 155.4811 158.0484 158.5717 166.0294  # highest wins model_list %>% map_dbl( function(x) if_else(inherits(x, 'gam'), summary(x)$r.sq,
summary(x)$adj) ) %>% sort(decreasing = TRUE) mod_gam_inter mod_inter mod_poly mod_gam mod_lm 0.8643020 0.8349382 0.8065828 0.8041651 0.7445939  Let’s go further and create a plot of these results. We’ll map to a data frame, use pivot_longer to melt it to two columns of model and value, then use ggplot2 to plot the results15. model_list %>% map_df( function(x) if_else(inherits(x, 'gam'), summary(x)$r.sq,
summary(x)$adj) ) %>% pivot_longer(cols = starts_with('mod'), names_to = 'model', values_to = "Adj. Rsq") %>% arrange(desc(Adj. Rsq)) %>% mutate(model = factor(model, levels = model)) %>% # sigh ggplot(aes(x = model, y = Adj. Rsq)) + geom_point(aes(color = model), size = 10, show.legend = F) Why not throw in AIC also? mod_rsq = model_list %>% map_df( function(x) if_else( inherits(x, 'gam'), summary(x)$r.sq,
summary(x)$adj ) ) %>% pivot_longer(cols = starts_with('mod'), names_to = 'model', values_to = 'Rsq') mod_aic = model_list %>% map_df(AIC) %>% pivot_longer(cols = starts_with('mod'), names_to = 'model', values_to = 'AIC') left_join(mod_rsq, mod_aic) %>% arrange(AIC) %>% mutate(model = factor(model, levels = model)) %>% pivot_longer(cols = -model, names_to = 'measure', values_to = 'value') %>% ggplot(aes(x = model, y = value)) + geom_point(aes(color = model), size = 10, show.legend = F) + facet_wrap(~ measure, scales = 'free') #### List columns As data.frames are lists, anything can be put into a column just as you would a list element. We’ll use pmap here, as it can take more than one argument, and we’re feeding all columns of the data.frame. You don’t need to worry about the details here, we just want to create a column that is actually a list. In this case the column will contain a data frame in each entry. mtcars2 = as.matrix(mtcars) mtcars2[sample(1:length(mtcars2), 50)] = NA # add some missing data mtcars2 = data.frame(mtcars2) %>% rownames_to_column(var = 'observation') %>% as_tibble() head(mtcars2) # A tibble: 6 x 12 observation mpg cyl disp hp drat wt qsec vs am gear carb <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> 1 Mazda RX4 21.0 6 160.0 "110" 3.90 2.620 <NA> 0 1 4 4 2 Mazda RX4 Wag 21.0 6 160.0 "110" 3.90 2.875 17.02 0 1 4 4 3 Datsun 710 22.8 4 108.0 " 93" 3.85 2.320 18.61 1 1 4 1 4 Hornet 4 Drive 21.4 6 258.0 "110" 3.08 3.215 19.44 1 <NA> 3 1 5 Hornet Sportabout 18.7 <NA> 360.0 "175" 3.15 3.440 17.02 0 0 3 2 6 Valiant <NA> 6 225.0 "105" <NA> 3.460 20.22 <NA> 0 3 1  mtcars2 = mtcars2 %>% mutate( newvar = pmap(., ~ data.frame( N = sum(!is.na(c(...))), Missing = sum(is.na(c(...))) ) ) ) Now check out the list column. mtcars2 # A tibble: 32 x 13 observation mpg cyl disp hp drat wt qsec vs am gear carb newvar <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <list> 1 Mazda RX4 21.0 6 160.0 "110" 3.90 2.620 <NA> 0 1 4 4 <df[,2] [1 × 2]> 2 Mazda RX4 Wag 21.0 6 160.0 "110" 3.90 2.875 17.02 0 1 4 4 <df[,2] [1 × 2]> 3 Datsun 710 22.8 4 108.0 " 93" 3.85 2.320 18.61 1 1 4 1 <df[,2] [1 × 2]> 4 Hornet 4 Drive 21.4 6 258.0 "110" 3.08 3.215 19.44 1 <NA> 3 1 <df[,2] [1 × 2]> 5 Hornet Sportabout 18.7 <NA> 360.0 "175" 3.15 3.440 17.02 0 0 3 2 <df[,2] [1 × 2]> 6 Valiant <NA> 6 225.0 "105" <NA> 3.460 20.22 <NA> 0 3 1 <df[,2] [1 × 2]> 7 Duster 360 <NA> 8 360.0 "245" 3.21 3.570 15.84 0 0 3 4 <df[,2] [1 × 2]> 8 Merc 240D 24.4 4 <NA> " 62" 3.69 3.190 20.00 1 0 4 2 <df[,2] [1 × 2]> 9 Merc 230 22.8 4 140.8 <NA> 3.92 3.150 22.90 1 0 4 <NA> <df[,2] [1 × 2]> 10 Merc 280 19.2 6 <NA> "123" 3.92 <NA> 18.30 1 <NA> 4 4 <df[,2] [1 × 2]> # … with 22 more rows mtcars2$newvar %>% head(3)
[[1]]
N Missing
1 11       1

[[2]]
N Missing
1 12       0

[[3]]
N Missing
1 12       0

Unnest it with the tidyr function.

mtcars2 %>%
unnest(newvar)
# A tibble: 32 x 14
observation       mpg   cyl   disp  hp    drat  wt    qsec  vs    am    gear  carb      N Missing
<chr>             <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <int>   <int>
1 Mazda RX4         21.0  6     160.0 "110" 3.90  2.620 <NA>  0     1     4     4        11       1
2 Mazda RX4 Wag     21.0  6     160.0 "110" 3.90  2.875 17.02 0     1     4     4        12       0
3 Datsun 710        22.8  4     108.0 " 93" 3.85  2.320 18.61 1     1     4     1        12       0
4 Hornet 4 Drive    21.4  6     258.0 "110" 3.08  3.215 19.44 1     <NA>  3     1        11       1
5 Hornet Sportabout 18.7  <NA>  360.0 "175" 3.15  3.440 17.02 0     0     3     2        11       1
6 Valiant           <NA>  6     225.0 "105" <NA>  3.460 20.22 <NA>  0     3     1         9       3
7 Duster 360        <NA>  8     360.0 "245" 3.21  3.570 15.84 0     0     3     4        11       1
8 Merc 240D         24.4  4     <NA>  " 62" 3.69  3.190 20.00 1     0     4     2        11       1
9 Merc 230          22.8  4     140.8  <NA> 3.92  3.150 22.90 1     0     4     <NA>     10       2
10 Merc 280          19.2  6     <NA>  "123" 3.92  <NA>  18.30 1     <NA>  4     4         9       3
# … with 22 more rows

This is a pretty esoteric demonstration, and not something you’d normally want to do, as mutate or other approaches would be far more efficient and sensical. However, the idea is that you might want to retain the information you might otherwise store in a list with the data that was used to create it. As an example, you could potentially attach models as a list column to a dataframe that contains meta-information about each model. Once you have a list column, you can use that column as you would any list for iterative programming.

## Iterative Programming Exercises

### Exercise 1

With the following matrix, use apply and the sum function to get row or column sums of the matrix x.

x = matrix(1:9, 3, 3)

### Exercise 2

With the following list object x, use lapply and sapply and the sum function to get sums for the elements. There is no margin to specify for a list, so just supply the list and the sum function.

x = list(1:3, 4:10, 11:100)

### Exercise 3

As in the previous example, use a map function to create a data frame of the column means. See ?map to see all your options.

d = tibble(
x = rnorm(100),
y = rnorm(100, 10, 2),
z = rnorm(100, 50, 10),
)

1. You can have 3 or more dimensional arrays, and apply will work over those dimensions (or any combination of them), but this doesn’t come up too often.↩︎

2. sapply is actually just a wrapper for lapply. If you supply the argument simplified=F, it is identical. Otherwise, it attempts to return a vector or matrix.↩︎

3. The main thing to note is that both the benefit and difficulty of using the map_* functions is they are more strict with inputs, which typically should be lists or list-like objects like data frames, and outputs.↩︎

4. Note that ggplot2 will change the ordering of the variable unless you coerce it by creating a factor. This pretty much defeats the entire purpose of retaining characters over factors everywhere else in the tidyverse.↩︎