1. Getting data buried in text

There are many cases in which your analysis pipeline needs to include steps to extract that data from the characters, rather than having to code it manually yourself.

  • This is the case with large datasets where important data is stored in column names or file names.

  • Dates in datasets are also often interpreted as characters in R.

  • In other cases, you might also conduct “text mining” analyses where you look for particular words and codes inside large amounts of documents, websites, etc.

R has different ways to deal with dates and texts, both in base R and in the tidyverse packages.

For this module, we will make use of two packages from the tidyverse family: stringr and lubridate:

library(tidyverse)
library(lubridate)

Get the stringr cheatsheet here

Get the lubridate cheatsheet here

2. Dealing with dates

2.1 Dates treated as characters when importing data:

Let’s take the presidential dataset that is included in the ggplot2 package:

presidential
## # A tibble: 12 × 4
##    name       start      end        party     
##    <chr>      <date>     <date>     <chr>     
##  1 Eisenhower 1953-01-20 1961-01-20 Republican
##  2 Kennedy    1961-01-20 1963-11-22 Democratic
##  3 Johnson    1963-11-22 1969-01-20 Democratic
##  4 Nixon      1969-01-20 1974-08-09 Republican
##  5 Ford       1974-08-09 1977-01-20 Republican
##  6 Carter     1977-01-20 1981-01-20 Democratic
##  7 Reagan     1981-01-20 1989-01-20 Republican
##  8 Bush       1989-01-20 1993-01-20 Republican
##  9 Clinton    1993-01-20 2001-01-20 Democratic
## 10 Bush       2001-01-20 2009-01-20 Republican
## 11 Obama      2009-01-20 2017-01-20 Democratic
## 12 Trump      2017-01-20 2021-01-20 Republican

As this is a pre-loaded package, you can see that the “start” and “end” dates of each president is already recognized as a “date” object:

class(presidential$start)
## [1] "Date"

This means that you can do things that you might expect to do with time data, like calculate the number of days between the start and end dates:

presidential$end - presidential$start
## Time differences in days
##  [1] 2922 1036 1886 2027  895 1461 2922 1461 2922 2922 2922 1461

But let’s what happens if we were to import this dataset from a .csv format!

First, write the data as a .csv file…

write.csv(presidential, "data/presidential.csv", row.names = F)

… then read it in, calling it something slightly different to make sure we can distinguish it from the original prepackaged dataset.

pres.dat=read.csv("data/presidential.csv")

Now see what this looks like.

pres.dat
##          name      start        end      party
## 1  Eisenhower 1953-01-20 1961-01-20 Republican
## 2     Kennedy 1961-01-20 1963-11-22 Democratic
## 3     Johnson 1963-11-22 1969-01-20 Democratic
## 4       Nixon 1969-01-20 1974-08-09 Republican
## 5        Ford 1974-08-09 1977-01-20 Republican
## 6      Carter 1977-01-20 1981-01-20 Democratic
## 7      Reagan 1981-01-20 1989-01-20 Republican
## 8        Bush 1989-01-20 1993-01-20 Republican
## 9     Clinton 1993-01-20 2001-01-20 Democratic
## 10       Bush 2001-01-20 2009-01-20 Republican
## 11      Obama 2009-01-20 2017-01-20 Democratic
## 12      Trump 2017-01-20 2021-01-20 Republican

You’ll notice that the “start” and “end” dates are now recognized as characters. This means that you can’t treat them like dates. For example, you wouldn’t be able to calculate the number of dates between start and end of the presidency:

pres.dat$end - pres.dat$start
## Error in pres.dat$end - pres.dat$start : 
##  non-numeric argument to binary operator

2.2. Turning dates-as-characters into “date” objects

You can take a date-as-character to turn it into a date object either in base R or using the lubridate package.

The base R way:

as.Date(pres.dat$start, format="%Y-%m-%d")
##  [1] "1953-01-20" "1961-01-20" "1963-11-22" "1969-01-20" "1974-08-09"
##  [6] "1977-01-20" "1981-01-20" "1989-01-20" "1993-01-20" "2001-01-20"
## [11] "2009-01-20" "2017-01-20"

In using the as.Date() function, you have to use the format= argument to specify the format the date is in.

Here are the options to include in the format argument:

Symbol Meaning Example
%d day as a number (0-31) 01-31
%a abbreviated weekday Mon
%A unabbreviated weekday Monday
%m month (00-12) 00-12
%b abbreviated month Jan
%B unabbreviated month January
%y 2-digit year 07
%Y 4-digit year 2007

Note that you have to be specific and correct. For example, if you use “%y” instead of “%Y” when you have a four-digit year, the function will not be able to parse it:

as.Date(pres.dat$start, format="%y-%m-%d")
##  [1] NA NA NA NA NA NA NA NA NA NA NA NA

But once you learn how to use this, it becomes easier.

Say you have your data in “month/date/year” format (common in the U.S.):

mdy_dates=c("5/17/1978", "3/23/2022")
as.Date(mdy_dates, format="%m/%d/%Y")
## [1] "1978-05-17" "2022-03-23"

Now, let’s try the same exercise using the lubridate package.

lubridate has a handy functions that simplify the processing of converting dates-as-characters into date objects:

For example, since the start and end dates are in “year-month-date” format, we can use a function called ymd():

ymd(pres.dat$start)
##  [1] "1953-01-20" "1961-01-20" "1963-11-22" "1969-01-20" "1974-08-09"
##  [6] "1977-01-20" "1981-01-20" "1989-01-20" "1993-01-20" "2001-01-20"
## [11] "2009-01-20" "2017-01-20"

To see that this actually converts these character strings into dates, let’s try calculating the number of days of each presidency:

ymd(pres.dat$end) - ymd(pres.dat$start)
## Time differences in days
##  [1] 2922 1036 1886 2027  895 1461 2922 1461 2922 2922 2922 1461

Now let’s try the “month/date/year” format. We’ll re-create the simple character string from before:

mdy_dates=c("5/17/1978", "3/23/2022")

Since the year, month and date are in a different order now, the ymd() function cannot process this:

ymd(mdy_dates)
## Warning: All formats failed to parse. No formats found.
## [1] NA NA

But we can simply use a friendly alternative function, mdy():

mdy(mdy_dates)
## [1] "1978-05-17" "2022-03-23"

3. Extracting, Replacing, and Splitting character strings

Let’s start with a hypothetical example. What if I had named my data files with the date:

filename="data_20220927.csv"

Now, let’s say I wanted to extract the dates from file names as part of my analyses. The basic way I might do this is through “text parsing”–i.e., extracting parts of text to get useful data. Here, I know that my file name has useful information (i.e., the year, month and date).

I’m going to show you three ways to do this, using base R and the stringr package for each.

3.1. Extracting substrings using fixed positions

One way to get the date from the file name is to just extract the part of this character string that contains the information I want.

If I just **count the start and stop position* of the information within this character string, I can just extract that part of the string (i.e., substring). In this case, the date (“20200927”) goes from position 6 to position 13 in this character.

I can use the substr() function in base R to do this:

substr(filename, start=6, stop=13)
## [1] "20220927"

Alternatively, I can use the str_sub() function in stringr in the same way.

str_sub(filename, start=6, end=13)
## [1] "20220927"

… and once I have this substring, I can convert this into a date object.

I can do this in base R, using the as.Date() function. To use this function, you need to specify the format in which the character string encodes the date information. Here, since it is a 4-number code for year, followed by month and date, you would put “%Y%m%d”.

as.Date(substr(filename, start=6, stop=13), format="%Y%m%d")
## [1] "2022-09-27"

Alternatively, you can use the ymd() function in lubridate. This function is a bit smarter in that it is more flexible–it can parse the text correctly as long as it’s in some form of year-month-date order. (If it was in month-date-year, you would use the mdy() function).

filename %>% str_sub(start=6, end=13) %>% ymd()
## [1] "2022-09-27"

Extracting substrings is probably the most straightforward way to parse text. HOWEVER, you can only do this if the text is formatted in a way that makes the position of your information of interest predictable

3.2. Replacing characters

You could also extract the date from the file name by substituting blanks for the parts that you don’t want (i.e., “data_” and “.csv”).

In base R, you can do this with gsub(). For example:

gsub("data_", "", filename) #this removes "data_" from the character string
## [1] "20220927.csv"

So you can string this together, then convert to date:

a=gsub("data_", "", filename) #remove the prefix
b=gsub(".csv", "", a) #remove the suffix
as.Date(b, format="%Y%m%d") #convert to date object
## [1] "2022-09-27"

Using the stringr function str_replace() in the same way:

Let’s try using the pipe and str_replace() to get rid of the first part of the file name:

filename %>% str_replace("data_", "")
## [1] "20220927.csv"

Using the same principle, we can string together the str_replace() functions and convert the result to a date object using ymd():

filename %>% 
  str_replace("data_", "") %>% 
  str_replace(".csv", "") %>% 
  ymd()
## [1] "2022-09-27"

3.3. Splitting characters

Another option is to split the character string into multiple elements. You can specify the character you use as the marker for where you want to split the character string. This is a very powerful and flexible tool, though it requires you to have a good understanding of how to deal with lists… So it is probably best reserved for complex tasks that you can’t do with substrings or text replacement.

For example, if we split the file name at the underscore (“_“):

strsplit(filename, split=c("_"))
## [[1]]
## [1] "data"         "20220927.csv"
strsplit(filename, split=c("."), fixed=T)
## [[1]]
## [1] "data_20220927" "csv"
a=strsplit(filename, split=c("_"))
a
## [[1]]
## [1] "data"         "20220927.csv"

Now, the next step is to split the remaining string “20220927.csv” by the period (“.”). However, this is a bit challenging because R does not handle the character “.” simply as a period. To do this, you have to use the Regular Expression for a period, which is “\.”

strsplit(a[[1]], split=c("\\."))
## [[1]]
## [1] "data"
## 
## [[2]]
## [1] "20220927" "csv"
b=strsplit(a[[1]], split=c("\\."))
b
## [[1]]
## [1] "data"
## 
## [[2]]
## [1] "20220927" "csv"
b[[2]][1]
## [1] "20220927"
s=b[[2]][1]
as.Date(s, format="%Y%m%d") #convert to date object
## [1] "2022-09-27"
str_split_1(filename, pattern=c("_")) %>%
  str_split(.,pattern=c("\\."))
## [[1]]
## [1] "data"
## 
## [[2]]
## [1] "20220927" "csv"

From here, you can use the same process as the base R case to extract the number.

3.4. Pattern Matching

Finally, there may be cases where you don’t need to extract the whole information, but you simply need to know if the character string contains some pattern.

filenames_vector=c("data_20220927.csv", "data_20221121.csv", "data_20230103.csv", "data_20230308.csv")

For example, let’s say you are scouring the file names for all files pertaining to the year 2022. Then, perhaps all you need to know is whether the file name contains “2022”.

You can find this out one of two ways.

The grepl() function will tell you TRUE or FALSE if the character string does or does not contain the string you are looking for:

grepl("2022", filenames_vector)
## [1]  TRUE  TRUE FALSE FALSE

In contrast, grep() will return the position of the elements that matched your pattern:

grep("2022", filenames_vector)
## [1] 1 2

Then, you can extract the file names that contain “2022”.

filenames_vector[grep("2022", filenames_vector)]
## [1] "data_20220927.csv" "data_20221121.csv"

4. Example using WHO tuberculosis data (UNDER CONSTRUCTION)

Let’s do an example of how one might use text parsing in the process of data analysis.

For this example, we will use the who dataset, included in tidyverse, which has data from the World Health Organization Global Tuberculosis Report.

who
## # A tibble: 7,240 × 60
##    country  iso2  iso3   year new_sp_m014 new_sp_m1524 new_sp_m2534 new_sp_m3544
##    <chr>    <chr> <chr> <dbl>       <dbl>        <dbl>        <dbl>        <dbl>
##  1 Afghani… AF    AFG    1980          NA           NA           NA           NA
##  2 Afghani… AF    AFG    1981          NA           NA           NA           NA
##  3 Afghani… AF    AFG    1982          NA           NA           NA           NA
##  4 Afghani… AF    AFG    1983          NA           NA           NA           NA
##  5 Afghani… AF    AFG    1984          NA           NA           NA           NA
##  6 Afghani… AF    AFG    1985          NA           NA           NA           NA
##  7 Afghani… AF    AFG    1986          NA           NA           NA           NA
##  8 Afghani… AF    AFG    1987          NA           NA           NA           NA
##  9 Afghani… AF    AFG    1988          NA           NA           NA           NA
## 10 Afghani… AF    AFG    1989          NA           NA           NA           NA
## # ℹ 7,230 more rows
## # ℹ 52 more variables: new_sp_m4554 <dbl>, new_sp_m5564 <dbl>,
## #   new_sp_m65 <dbl>, new_sp_f014 <dbl>, new_sp_f1524 <dbl>,
## #   new_sp_f2534 <dbl>, new_sp_f3544 <dbl>, new_sp_f4554 <dbl>,
## #   new_sp_f5564 <dbl>, new_sp_f65 <dbl>, new_sn_m014 <dbl>,
## #   new_sn_m1524 <dbl>, new_sn_m2534 <dbl>, new_sn_m3544 <dbl>,
## #   new_sn_m4554 <dbl>, new_sn_m5564 <dbl>, new_sn_m65 <dbl>, …

Also make sure to look at the help file for the dataset:

?who

First thing we see is that this data set is terribly wide. There are 60 columns, 56 of which contain data on counts of TB cases, and each of those 56 columns represents different combinations of method of diagnosis, gender and age group.

These column names are where a lot of the important information are, and what we might want to use as variables of interest.

So, let’s refresh our memory on how we can use the pivot_longer() function to make this data into a “long format”.

Start by using pivot_longer(), specifying that we want to keep the first four columns as is:

who_long=who %>% pivot_longer(c(-country, -iso2, -iso3, -year))
who_long
## # A tibble: 405,440 × 6
##    country     iso2  iso3   year name         value
##    <chr>       <chr> <chr> <dbl> <chr>        <dbl>
##  1 Afghanistan AF    AFG    1980 new_sp_m014     NA
##  2 Afghanistan AF    AFG    1980 new_sp_m1524    NA
##  3 Afghanistan AF    AFG    1980 new_sp_m2534    NA
##  4 Afghanistan AF    AFG    1980 new_sp_m3544    NA
##  5 Afghanistan AF    AFG    1980 new_sp_m4554    NA
##  6 Afghanistan AF    AFG    1980 new_sp_m5564    NA
##  7 Afghanistan AF    AFG    1980 new_sp_m65      NA
##  8 Afghanistan AF    AFG    1980 new_sp_f014     NA
##  9 Afghanistan AF    AFG    1980 new_sp_f1524    NA
## 10 Afghanistan AF    AFG    1980 new_sp_f2534    NA
## # ℹ 405,430 more rows

We can see that this creates a “name” column, which has the column names we have pivoted, and a “value” column, which is where the count data goes (the first 10 country x year combo happens to not have data).

Ok, so using this, we are going to first ‘pull’ the variable called “name” and see all of the unique values:

#get the codes: pull() is a function to get one variable out of a dataframe
who_long %>% pull(name) %>% unique()
##  [1] "new_sp_m014"  "new_sp_m1524" "new_sp_m2534" "new_sp_m3544" "new_sp_m4554"
##  [6] "new_sp_m5564" "new_sp_m65"   "new_sp_f014"  "new_sp_f1524" "new_sp_f2534"
## [11] "new_sp_f3544" "new_sp_f4554" "new_sp_f5564" "new_sp_f65"   "new_sn_m014" 
## [16] "new_sn_m1524" "new_sn_m2534" "new_sn_m3544" "new_sn_m4554" "new_sn_m5564"
## [21] "new_sn_m65"   "new_sn_f014"  "new_sn_f1524" "new_sn_f2534" "new_sn_f3544"
## [26] "new_sn_f4554" "new_sn_f5564" "new_sn_f65"   "new_ep_m014"  "new_ep_m1524"
## [31] "new_ep_m2534" "new_ep_m3544" "new_ep_m4554" "new_ep_m5564" "new_ep_m65"  
## [36] "new_ep_f014"  "new_ep_f1524" "new_ep_f2534" "new_ep_f3544" "new_ep_f4554"
## [41] "new_ep_f5564" "new_ep_f65"   "newrel_m014"  "newrel_m1524" "newrel_m2534"
## [46] "newrel_m3544" "newrel_m4554" "newrel_m5564" "newrel_m65"   "newrel_f014" 
## [51] "newrel_f1524" "newrel_f2534" "newrel_f3544" "newrel_f4554" "newrel_f5564"
## [56] "newrel_f65"

The help file provides the codes:

  • all codes start with “new”
  • the next part codes the method of diagnosis. It is either “rel”, “_sn”, “_sp”, or “_ep”… notice that the under score ensures that they are all 3 characters long.
  • next part is “_m” for male and “_f” for female. These are all 2 characters long
  • final part has different number of characters: 014 = 0-14 yrs of age, 1524 = 15-24 years of age, 2534 = 25 to 34 years of age, 3544 = 35 to 44 years of age, 4554 = 45 to 54 years of age, 5564 = 55 to 64 years of age, 65 = 65 years of age or older

So, let’s get the gender from those columns.

We can see that the code for gender (“f” for female, “m” for male) is ALWAYS the 8th character of the string. This makes our life easy–we just need to get the 8th character. We can do that using str_sub() to extract the substring:

who_long %>% pull(name) %>% str_sub(start=8, end=8) %>% unique()
## [1] "m" "f"

So we can use this to create a new column called “gender” in the who_long dataset.

To do this, we will use the mutate() function, and use the str_sub() within it. In that function, we specify that we are taking the variable “name” and taking the 8th character.

who_long %>%
  mutate(gender=str_sub(name, start=8, end=8))
## # A tibble: 405,440 × 7
##    country     iso2  iso3   year name         value gender
##    <chr>       <chr> <chr> <dbl> <chr>        <dbl> <chr> 
##  1 Afghanistan AF    AFG    1980 new_sp_m014     NA m     
##  2 Afghanistan AF    AFG    1980 new_sp_m1524    NA m     
##  3 Afghanistan AF    AFG    1980 new_sp_m2534    NA m     
##  4 Afghanistan AF    AFG    1980 new_sp_m3544    NA m     
##  5 Afghanistan AF    AFG    1980 new_sp_m4554    NA m     
##  6 Afghanistan AF    AFG    1980 new_sp_m5564    NA m     
##  7 Afghanistan AF    AFG    1980 new_sp_m65      NA m     
##  8 Afghanistan AF    AFG    1980 new_sp_f014     NA f     
##  9 Afghanistan AF    AFG    1980 new_sp_f1524    NA f     
## 10 Afghanistan AF    AFG    1980 new_sp_f2534    NA f     
## # ℹ 405,430 more rows

Next, let’s try to extract the method of diagnosis. This information is always contained between the 4th and 6th characters of the “name” variable:

Let’s pull the “name” variable, get the 4th through 6th character, and see all unique values:

who_long %>% pull(name) %>% str_sub(start=4, end=6) %>% unique()
## [1] "_sp" "_sn" "_ep" "rel"

You can see that this value can be either “rel” for relapse, or “sp”, “sn”, “ep”. For the latter three, there is a preceding “_” to make it the same number of character (I see why, but not ideal… they could have just use the same number of letters throughout…).

So what we can do is to go ahead and extract the 4th to 6th character and then remove the underscore if it exists.

Remove the underscore “_” if it is exists.

who_long %>%
  mutate(gender=str_sub(name, start=8, end=8)) %>%
  mutate(method=str_sub(name, 4, 6)) %>%
  mutate(method=str_replace(method, "_", ""))
## # A tibble: 405,440 × 8
##    country     iso2  iso3   year name         value gender method
##    <chr>       <chr> <chr> <dbl> <chr>        <dbl> <chr>  <chr> 
##  1 Afghanistan AF    AFG    1980 new_sp_m014     NA m      sp    
##  2 Afghanistan AF    AFG    1980 new_sp_m1524    NA m      sp    
##  3 Afghanistan AF    AFG    1980 new_sp_m2534    NA m      sp    
##  4 Afghanistan AF    AFG    1980 new_sp_m3544    NA m      sp    
##  5 Afghanistan AF    AFG    1980 new_sp_m4554    NA m      sp    
##  6 Afghanistan AF    AFG    1980 new_sp_m5564    NA m      sp    
##  7 Afghanistan AF    AFG    1980 new_sp_m65      NA m      sp    
##  8 Afghanistan AF    AFG    1980 new_sp_f014     NA f      sp    
##  9 Afghanistan AF    AFG    1980 new_sp_f1524    NA f      sp    
## 10 Afghanistan AF    AFG    1980 new_sp_f2534    NA f      sp    
## # ℹ 405,430 more rows

Finally, let’s get the age class:

This code part of the code starts on the 9th character, but can contain two characters (“65”), three characters (“014”) or four characters (“1524”, “2534”, “3544”, “4554”, or “5564”). What we can do is to use the str_sub() function but just give it the starting character number:

who_long %>% pull(name) %>% str_sub(start=9) %>% unique()
## [1] "014"  "1524" "2534" "3544" "4554" "5564" "65"

Ok, now we want to convert this into something that makes more sense when we are plotting it…

What we can do is to replace each of the age codes to readable age categories and then convert it into a factor using as_factor()

who_long %>% pull(name) %>% str_sub(start=9) %>%
  str_replace("014", "0-14") %>%
  str_replace("1524", "15-24") %>%
  str_replace("2534", "25-34") %>%
  str_replace("3544", "35-44") %>%
  str_replace("4554", "45-54") %>%
  str_replace("5564", "55-64") %>%
  str_replace("65", "65+") %>%
  as_factor() %>%
  head()
## [1] 0-14  15-24 25-34 35-44 45-54 55-64
## Levels: 0-14 15-24 25-34 35-44 45-54 55-64 65+

We can use this within a mutate function: Let’s do that and save it as a new dataframe:

who_use=who_long %>%
  mutate(gender=str_sub(name, start=8, end=8)) %>%
  mutate(method=str_sub(name, 4, 6)) %>%
  mutate(method=str_replace(method, "_", "")) %>%
  mutate(age=str_sub(name, start=9) %>%
           str_replace("014", "0-14") %>%
           str_replace("1524", "15-24") %>%
           str_replace("2534", "25-34") %>%
           str_replace("3544", "35-44") %>%
           str_replace("4544", "45-54") %>%
           str_replace("5564", "55-64") %>%
           str_replace("65", "65+") %>%
           as_factor())

who_use
## # A tibble: 405,440 × 9
##    country     iso2  iso3   year name         value gender method age  
##    <chr>       <chr> <chr> <dbl> <chr>        <dbl> <chr>  <chr>  <fct>
##  1 Afghanistan AF    AFG    1980 new_sp_m014     NA m      sp     0-14 
##  2 Afghanistan AF    AFG    1980 new_sp_m1524    NA m      sp     15-24
##  3 Afghanistan AF    AFG    1980 new_sp_m2534    NA m      sp     25-34
##  4 Afghanistan AF    AFG    1980 new_sp_m3544    NA m      sp     35-44
##  5 Afghanistan AF    AFG    1980 new_sp_m4554    NA m      sp     4554 
##  6 Afghanistan AF    AFG    1980 new_sp_m5564    NA m      sp     55-64
##  7 Afghanistan AF    AFG    1980 new_sp_m65      NA m      sp     65+  
##  8 Afghanistan AF    AFG    1980 new_sp_f014     NA f      sp     0-14 
##  9 Afghanistan AF    AFG    1980 new_sp_f1524    NA f      sp     15-24
## 10 Afghanistan AF    AFG    1980 new_sp_f2534    NA f      sp     25-34
## # ℹ 405,430 more rows

Now, we can plot some trends, making use of out new variables for filtering data and using them as variables of interest:

For example, how has the prevalence of TB changed across years in the U.S. for those over 65, detected using positive pulmonary smears (method==“sp”), and breaking down the trend by gender?

ggplot(who_use %>% filter(iso3=="USA") %>% filter(method=="sp") %>% filter(age=="65+"), aes(x=year, y=value, color=gender)) +
  geom_point() +
  geom_line()
## Warning: Removed 32 rows containing missing values or values outside the scale range
## (`geom_point()`).
## Warning: Removed 32 rows containing missing values or values outside the scale range
## (`geom_line()`).