library(tidyverse)4 Data Wrangling I: Creating Variables
To keep it nice and simple, we’ll use yet another university dataset:
dat3 <- data.frame(
studs = c(14954, 47269, 23659, 9415, 38079),
profs = c(250, 553, 438, 150, 636),
prom_recht = c(FALSE, TRUE, TRUE, TRUE, FALSE),
gegr = c(1971, 1870, 1457, 1818, 1995),
uni = c("FH Aachen", "RWTH Aachen", "Uni Freiburg", "Uni Bonn", "FH Bonn-Rhein-Sieg")
)4.1 Creating Variables
Let’s take a closer look at creating variables in R. There are two basic ways to add variables to a data.frame:
Base R:
...$newvar <-{dplyr}:mutate(new_var= )
4.1.1 Base R: ...$newvar <-
dat3$studs_to_mean <- dat3$studs - mean(dat3$studs)
dat3 studs profs prom_recht gegr uni studs_to_mean
1 14954 250 FALSE 1971 FH Aachen -11721.2
2 47269 553 TRUE 1870 RWTH Aachen 20593.8
3 23659 438 TRUE 1457 Uni Freiburg -3016.2
4 9415 150 TRUE 1818 Uni Bonn -17260.2
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8You can also delete variables using <- NULL:
dat3$studs_to_mean <- NULL
dat3 studs profs prom_recht gegr uni
1 14954 250 FALSE 1971 FH Aachen
2 47269 553 TRUE 1870 RWTH Aachen
3 23659 438 TRUE 1457 Uni Freiburg
4 9415 150 TRUE 1818 Uni Bonn
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg4.1.2 {dplyr}: mutate(new_var= )
An alternative way to create variables is using mutate(new_variable = ) from {dplyr} ({tidyverse}):
dat3 %>% mutate(studs_to_mean = studs - mean(studs)) studs profs prom_recht gegr uni studs_to_mean
1 14954 250 FALSE 1971 FH Aachen -11721.2
2 47269 553 TRUE 1870 RWTH Aachen 20593.8
3 23659 438 TRUE 1457 Uni Freiburg -3016.2
4 9415 150 TRUE 1818 Uni Bonn -17260.2
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8You can also create multiple variables within a single mutate() command:
dat3 %>% mutate(
studs_to_mean = studs - mean(studs),
profs_to_mean = profs - mean(profs)
) studs profs prom_recht gegr uni studs_to_mean profs_to_mean
1 14954 250 FALSE 1971 FH Aachen -11721.2 -155.4
2 47269 553 TRUE 1870 RWTH Aachen 20593.8 147.6
3 23659 438 TRUE 1457 Uni Freiburg -3016.2 32.6
4 9415 150 TRUE 1818 Uni Bonn -17260.2 -255.4
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8 230.6Or variables can be reused within mutate():
dat3 %>% mutate(
rel_to_mean = studs - mean(studs),
above_mean = rel_to_mean > 0
) studs profs prom_recht gegr uni rel_to_mean above_mean
1 14954 250 FALSE 1971 FH Aachen -11721.2 FALSE
2 47269 553 TRUE 1870 RWTH Aachen 20593.8 TRUE
3 23659 438 TRUE 1457 Uni Freiburg -3016.2 FALSE
4 9415 150 TRUE 1818 Uni Bonn -17260.2 FALSE
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8 TRUEThe original dataset remains unchanged:
dat3 studs profs prom_recht gegr uni
1 14954 250 FALSE 1971 FH Aachen
2 47269 553 TRUE 1870 RWTH Aachen
3 23659 438 TRUE 1457 Uni Freiburg
4 9415 150 TRUE 1818 Uni Bonn
5 38079 636 FALSE 1995 FH Bonn-Rhein-SiegTo keep the results, store them in an object:
dat4 <- dat3 %>% mutate(
rel_to_mean = studs - mean(studs),
above_mean = rel_to_mean > 0
)
dat4 studs profs prom_recht gegr uni rel_to_mean above_mean
1 14954 250 FALSE 1971 FH Aachen -11721.2 FALSE
2 47269 553 TRUE 1870 RWTH Aachen 20593.8 TRUE
3 23659 438 TRUE 1457 Uni Freiburg -3016.2 FALSE
4 9415 150 TRUE 1818 Uni Bonn -17260.2 FALSE
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8 TRUE
Creating Dummy Variables with
as.numeric()
You can convert logical variables into numeric dummy variables (0/1) using as.numeric():
dat3 %>% mutate(
prom_dummy = as.numeric(prom_recht),
over10k = as.numeric(studs > 10000)
) studs profs prom_recht gegr uni prom_dummy over10k
1 14954 250 FALSE 1971 FH Aachen 0 1
2 47269 553 TRUE 1870 RWTH Aachen 1 1
3 23659 438 TRUE 1457 Uni Freiburg 1 1
4 9415 150 TRUE 1818 Uni Bonn 1 0
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 0 14.2 Grouping with .by=
The true power of mutate() becomes apparent when combined with other {dplyr} functions. A common task in data preparation involves grouped values.
We’ll make our example dataset a bit smaller:
dat5 <- dat3 %>%
select(-uni,-gegr) # to ensure everything is visibleSince {dplyr} version 1.1.1, we can specify a grouping directly in mutate() using the .by= argument. This .by= grouping is applied only to the immediate calculations within mutate():
dat5 %>%
mutate(m_studs = mean(studs),
m_profs = mean(profs)) %>%
mutate(m_studs2 = mean(studs),
.by = prom_recht) %>%
mutate(m_profs2 = mean(profs)) studs profs prom_recht m_studs m_profs m_studs2 m_profs2
1 14954 250 FALSE 26675.2 405.4 26516.5 405.4
2 47269 553 TRUE 26675.2 405.4 26781.0 405.4
3 23659 438 TRUE 26675.2 405.4 26781.0 405.4
4 9415 150 TRUE 26675.2 405.4 26781.0 405.4
5 38079 636 FALSE 26675.2 405.4 26516.5 405.4Using summarise() instead of mutate() provides an overview:
dat5 %>%
summarise(m_studs = mean(studs),.by = prom_recht) prom_recht m_studs
1 FALSE 26516.5
2 TRUE 26781.0group_by()
Before {dplyr} 1.1.1, grouping a dataset relied on group_by(). After setting group_by() along the values of a variable, all subsequent mutate() calculations are performed only within those groups:
dat5 %>%
mutate(m_studs = mean(studs),
m_profs = mean(profs)) %>%
group_by(prom_recht) %>%
mutate(m_studs2 = mean(studs),
m_profs2 = mean(profs))# A tibble: 5 × 7
# Groups: prom_recht [2]
studs profs prom_recht m_studs m_profs m_studs2 m_profs2
<dbl> <dbl> <lgl> <dbl> <dbl> <dbl> <dbl>
1 14954 250 FALSE 26675. 405. 26516. 443
2 47269 553 TRUE 26675. 405. 26781 380.
3 23659 438 TRUE 26675. 405. 26781 380.
4 9415 150 TRUE 26675. 405. 26781 380.
5 38079 636 FALSE 26675. 405. 26516. 443 After using group_by(), it’s good practice to remove the grouping with ungroup() once it’s no longer needed:
dat5 %>%
mutate(m_studs = mean(studs),
m_profs = mean(profs)) %>%
group_by(prom_recht) %>%
mutate(m_studs2 = mean(studs)) %>%
ungroup() %>%
mutate(m_profs2 = mean(profs))# A tibble: 5 × 7
studs profs prom_recht m_studs m_profs m_studs2 m_profs2
<dbl> <dbl> <lgl> <dbl> <dbl> <dbl> <dbl>
1 14954 250 FALSE 26675. 405. 26516. 405.
2 47269 553 TRUE 26675. 405. 26781 405.
3 23659 438 TRUE 26675. 405. 26781 405.
4 9415 150 TRUE 26675. 405. 26781 405.
5 38079 636 FALSE 26675. 405. 26516. 405.4.3 across(): Processing Multiple Variables
A highly versatile addition to mutate() and summarise() is across(). This allows us to apply a function to multiple columns simultaneously, without repeating code:
dat3 %>%
summarise(studs = mean(studs),
profs = mean(profs)) studs profs
1 26675.2 405.4Here, across() offers a much shorter syntax for variable selection, and we can use ?select_helpers like matches():
dat3 %>%
summarise(across(.cols = matches("studs|profs"),.fns = ~mean(.x))) studs profs
1 26675.2 405.4This is also compatible with .by=:
dat3 %>%
summarise(across(matches("studs|profs"), ~mean(.x)), .by= prom_recht) prom_recht studs profs
1 FALSE 26516.5 443.0000
2 TRUE 26781.0 380.3333For more examples on how apply multiple functions, include renaming etc. see below
4.3.1 Exercise
4.4 Custom Functions
So far, we only used functions written by others. We can also use own functions to avoid repetition:
To do so, let’s examine three satisfaction variables for respondents in rows 12-16:
| variable | Important regarding job |
|---|---|
|
To earn a lot of money |
|
A job, that is fun |
|
Good career opportunities |
|
Job security |
|
Job where you can show off your abilities |
| -10 bis -1 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| t.n.z./k.A. | Very important | Rather important | Rather not important | Not important at all |
pend <- haven::read_dta("./orig/PENDDAT_cf_W13.dta")
sat_small <-
pend %>%
filter(welle == 1) %>%
select(matches("PEO0300(a|b|c)")) %>%
slice(12:16) %>%
haven::zap_labels() %>% haven::zap_label() # remove labels
sat_small# A tibble: 5 × 3
PEO0300a PEO0300b PEO0300c
<dbl> <dbl> <dbl>
1 2 3 2
2 1 1 3
3 1 1 3
4 2 1 1
5 1 1 2sat_small <- sat_small %>% mutate(across(everything(),~as.numeric(.x)))Sometimes we want to process multiple variables in the same way. Above, we saw how to handle this with across() for existing functions. But what if we want to perform a calculation that isn’t as simple as applying mean(), sd(), etc.?
sat_small %>%
mutate(dmean_PEO0300a = PEO0300a - mean(PEO0300a,na.rm = T),
dmean_PEO0300c = PEO0300c - mean(PEO0300c,na.rm = T))# A tibble: 5 × 5
PEO0300a PEO0300b PEO0300c dmean_PEO0300a dmean_PEO0300c
<dbl> <dbl> <dbl> <dbl> <dbl>
1 2 3 2 0.6 -0.200
2 1 1 3 -0.4 0.8
3 1 1 3 -0.4 0.8
4 2 1 1 0.6 -1.2
5 1 1 2 -0.4 -0.200…and now what about F1450_06? Typing this out three times would violate the “DRY” principle1, especially considering the PASS CampusFile contains 5 columns of similar satisfaction variables. Copying and pasting is not a practical option.
Custom functions allow us to adhere to the DRY principle in R. We’ll make our calculation steps part of a function() and apply it to the desired variables. A function takes an input, defined as a placeholder within the (). This placeholder is used within the function, and we return the result with return(). Only one output can be returned:
dtomean <- function(x){
d_x <- x - mean(x,na.rm = T)
return(d_x)
}How can we now apply our function dtomean() to the variables from our sat_small?
In principle, we saw at the beginning that a data.frame is simply a combined collection of vectors (the variables).
Accordingly, we can now apply our dtomean() to a variable (a vector) by calling it with data.frame$variablename:
dtomean(sat_small$PEO0300a)[1] 0.6 -0.4 -0.4 0.6 -0.4To apply our function to each variable of a data.frame, we can use lapply() - the output will then be a list, with elements named after the variable names:
lapply(sat_small,FUN = dtomean)$PEO0300a
[1] 0.6 -0.4 -0.4 0.6 -0.4
$PEO0300b
[1] 1.6 -0.4 -0.4 -0.4 -0.4
$PEO0300c
[1] -0.2 0.8 0.8 -1.2 -0.2res <- lapply(sat_small,FUN = dtomean)
class(res)[1] "list"map() from {purrr} is an alternative to lapply:
sat_small %>% map(~dtomean(.x))$PEO0300a
[1] 0.6 -0.4 -0.4 0.6 -0.4
$PEO0300b
[1] 1.6 -0.4 -0.4 -0.4 -0.4
$PEO0300c
[1] -0.2 0.8 0.8 -1.2 -0.2This formula syntax can also be found in across() - additionally, with .names = we have the option to modify the variable names for the results:
sat_small %>%
mutate(across(matches("PEO0300"),~dtomean(.x)) )# A tibble: 5 × 3
PEO0300a PEO0300b PEO0300c
<dbl> <dbl> <dbl>
1 0.6 1.6 -0.200
2 -0.4 -0.4 0.8
3 -0.4 -0.4 0.8
4 0.6 -0.4 -1.2
5 -0.4 -0.4 -0.2004.4.1 Exercise
4.5 Exercises
4.5.1 Exercise
- Use the
pend_smalldataset:
pend_small <- haven::read_dta("./orig/PENDDAT_cf_W13.dta",
col_select = c("welle","zpsex","PEO0400a","PEO0400b","PEO0400c","PEO0400d")
) %>%
haven::zap_labels() %>% # drop labels to have a clean data.frame
filter(welle == 2) %>%
slice(1:10)- Calculate the mean for the variables PEO0400a by gender (
zpsex): - Calculate the mean for the variables
PEO0400a,PEO0400b,PEO0400c, andPEO0400dby gender (zpsex): - Use
across()to calculate the means for all four variables.
| var | lab |
|---|---|
|
Family/job: Woman should be willing to reduce her working hours for family |
|
Family/job: What women really want is a home and children |
|
Family/job: Working mother can have an equally warm relationship with her childr |
|
Family/job: Responsibility of husband: To earn money; responsibility of wife: Ho |
welle zpsex PEO0400a PEO0400b PEO0400c PEO0400d
1 2 2 1 1 4 1
2 2 1 2 1 3 2
3 2 2 1 3 1 4
4 2 2 1 1 4 1
5 2 2 1 1 1 1
6 2 1 1 1 1 1
7 2 1 1 2 1 4
8 2 1 3 2 3 2
9 2 2 3 3 3 3
10 2 1 2 3 2 2- Standardize the variables
PEO0400a-PEO0400dfrompend_smallusing the following pattern:
pend_small %>%
mutate(std_PEO0400b = (PEO0400b - mean(PEO0400b,na.rm = T))/sd(PEO0400b,na.rm = T))- Use a function so that you don’t have to repeatedly enter the same steps.
- Additionally, use
across()to apply the function to the desired variables. - Calculate the standardization separately by gender (
zpsex) using.by =.
4.5.2 Exercise
Continue using pend_small:
pend_small welle zpsex PEO0400a PEO0400b PEO0400c PEO0400d
1 2 2 1 1 4 1
2 2 1 2 1 3 2
3 2 2 1 3 1 4
4 2 2 1 1 4 1
5 2 2 1 1 1 1
6 2 1 1 1 1 1
7 2 1 1 2 1 4
8 2 1 3 2 3 2
9 2 2 3 3 3 3
10 2 1 2 3 2 2- Standardize the variables
PEO0400a-PEO0400dfrompend_smallusing the following pattern:
pend_small %>%
mutate(std_PEO0400b = (PEO0400b - mean(PEO0400b,na.rm = T))/sd(PEO0400b,na.rm = T))- Use a function so that you don’t have to repeatedly enter the same steps.
- Additionally, use
across()to apply the function to the desired variables.
4.6 Appendix
4.6.1 across(): Processing Multiple Variables
A highly versatile addition to mutate() and summarise() is across(). This allows us to apply a function to multiple columns simultaneously, without repeating code:
dat3 %>%
summarise(studs = mean(studs),
profs = mean(profs)) studs profs
1 26675.2 405.4Here, across() offers a much shorter syntax for variable selection, and we can use ?select_helpers like matches():
dat3 %>%
summarise(across(.cols = matches("studs|profs"),.fns = ~mean(.x))) studs profs
1 26675.2 405.4This is also compatible with .by=:
dat3 %>%
summarise(across(matches("studs|profs"), ~mean(.x)), .by= prom_recht) prom_recht studs profs
1 FALSE 26516.5 443.0000
2 TRUE 26781.0 380.3333We can apply multiple functions by placing them in a list():
dat3 %>%
summarise(across(matches("studs|profs"), list(mean = ~mean(.x), sd = ~sd(.x))), .by= prom_recht) prom_recht studs_mean studs_sd profs_mean profs_sd
1 FALSE 26516.5 16351.84 443.0000 272.9432
2 TRUE 26781.0 19119.14 380.3333 207.5966You can define this list() in advance and use it later:
wert_liste <- list(MEAN = ~mean(.x), SD = ~sd(.x))
dat3 %>%
summarise(across(matches("studs|profs"), wert_liste), .by= prom_recht) prom_recht studs_MEAN studs_SD profs_MEAN profs_SD
1 FALSE 26516.5 16351.84 443.0000 272.9432
2 TRUE 26781.0 19119.14 380.3333 207.5966The .names() argument allows us to control the naming of columns. {.fn} stands for the function being applied, and {.col} represents the name of the variable being processed.
dat3 %>%
summarise(across(matches("studs|profs"),
wert_liste,
.names = "{.fn}_{.col}"),
.by= prom_recht) prom_recht MEAN_studs SD_studs MEAN_profs SD_profs
1 FALSE 26516.5 16351.84 443.0000 272.9432
2 TRUE 26781.0 19119.14 380.3333 207.5966All these functions also work with mutate():
dat3 %>%
mutate(across(matches("studs|profs"),
wert_liste,
.names = "{.col}XX{.fn}")) studs profs prom_recht gegr uni studsXXMEAN studsXXSD
1 14954 250 FALSE 1971 FH Aachen 26675.2 15799.92
2 47269 553 TRUE 1870 RWTH Aachen 26675.2 15799.92
3 23659 438 TRUE 1457 Uni Freiburg 26675.2 15799.92
4 9415 150 TRUE 1818 Uni Bonn 26675.2 15799.92
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 26675.2 15799.92
profsXXMEAN profsXXSD
1 405.4 203.349
2 405.4 203.349
3 405.4 203.349
4 405.4 203.349
5 405.4 203.349More examples in the across() documentation
4.6.1.1 Exercise
4.6.2 Helper functions
4.6.2.1 ifelse()
ifelse() is a great help for all recoding tasks: we formulate a condition and if it is met, the first value is used; if not, the second value is used. Here we check whether studs-mean(studs) is greater than 0 - if so, above is used, otherwise below:
dat3 %>% mutate(rel_to_mean = studs-mean(studs),
ab_mean_lab = ifelse(rel_to_mean > 0,"above","below")) studs profs prom_recht gegr uni rel_to_mean ab_mean_lab
1 14954 250 FALSE 1971 FH Aachen -11721.2 below
2 47269 553 TRUE 1870 RWTH Aachen 20593.8 above
3 23659 438 TRUE 1457 Uni Freiburg -3016.2 below
4 9415 150 TRUE 1818 Uni Bonn -17260.2 below
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg 11403.8 aboveThis can be helpful to replace negative values with NA, for example in the PASS data:
pend_small2 <- haven::read_dta("./orig/PENDDAT_cf_W13.dta",
col_select = c("palter","PEO0400a","PEO0400b","PEO0400c","statakt")) %>%
slice(5624:5640)The basic idea is to use ifelse() to replace negative values in a variable with NA:
pend_small2 %>% mutate(PEO0400a = ifelse(PEO0400a<0,NA,PEO0400a))across() allows us to apply this ifelse()-function to replace NA in PEO0400a,PEO0400b, PEO0400c and statakt:
pend_small2 %>% mutate(across(c("PEO0400a","PEO0400b","PEO0400c","statakt"), ~ifelse(.x<0,NA,.x))) # A tibble: 17 × 5
palter PEO0400a PEO0400b PEO0400c statakt
<dbl+lbl> <dbl> <dbl> <dbl> <dbl>
1 77 1 3 3 NA
2 78 NA NA NA NA
3 51 2 4 1 NA
4 23 3 3 2 NA
5 17 3 2 1 NA
6 47 3 2 2 NA
7 24 3 4 1 1
8 52 2 3 1 1
9 19 2 3 2 3
10 48 2 3 1 1
11 49 NA NA NA NA
12 47 2 3 1 NA
13 48 2 3 1 1
14 49 NA NA NA 1
15 39 4 3 1 NA
16 37 3 4 1 NA
17 38 3 3 1 1pend_small2 %>% mutate(across(matches("PEO0400|statakt"), ~ifelse(.x<0,NA,.x))) # even shorter: matches()# A tibble: 17 × 5
palter PEO0400a PEO0400b PEO0400c statakt
<dbl+lbl> <dbl> <dbl> <dbl> <dbl>
1 77 1 3 3 NA
2 78 NA NA NA NA
3 51 2 4 1 NA
4 23 3 3 2 NA
5 17 3 2 1 NA
6 47 3 2 2 NA
7 24 3 4 1 1
8 52 2 3 1 1
9 19 2 3 2 3
10 48 2 3 1 1
11 49 NA NA NA NA
12 47 2 3 1 NA
13 48 2 3 1 1
14 49 NA NA NA 1
15 39 4 3 1 NA
16 37 3 4 1 NA
17 38 3 3 1 14.6.2.2 case_when()
case_when() ({dplyr}) extends the principle ifelse(), allowing us to specify more than two options.
The syntax is slightly different: first, we specify the condition, then after a ~ the values to be used:
dat3 %>% mutate(age = case_when(gegr < 1500 ~ "very old",
gegr < 1900 ~ "old")) studs profs prom_recht gegr uni age
1 14954 250 FALSE 1971 FH Aachen <NA>
2 47269 553 TRUE 1870 RWTH Aachen old
3 23659 438 TRUE 1457 Uni Freiburg very old
4 9415 150 TRUE 1818 Uni Bonn old
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg <NA>With TRUE, we can address all cases that have not met any conditions so far:
dat3 %>% mutate(age = case_when(gegr < 1500 ~ "very old",
gegr < 1900 ~ "old",
TRUE ~ "relatively new")) studs profs prom_recht gegr uni age
1 14954 250 FALSE 1971 FH Aachen relatively new
2 47269 553 TRUE 1870 RWTH Aachen old
3 23659 438 TRUE 1457 Uni Freiburg very old
4 9415 150 TRUE 1818 Uni Bonn old
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg relatively newThis doesn’t have to be limited to one variable:
dat3 %>% mutate(age = case_when(gegr < 1500 & prom_recht == T ~ "very old university",
gegr < 1900 & prom_recht == T ~ "old university",
gegr > 1900 & prom_recht == T ~ "young university",
gegr < 1900 & prom_recht == F ~ "old college",
gegr > 1900 & prom_recht == F ~ "young college")) studs profs prom_recht gegr uni age
1 14954 250 FALSE 1971 FH Aachen young college
2 47269 553 TRUE 1870 RWTH Aachen old university
3 23659 438 TRUE 1457 Uni Freiburg very old university
4 9415 150 TRUE 1818 Uni Bonn old university
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg young college4.6.2.3 cut(): creating classes
dat3 studs profs prom_recht gegr uni
1 14954 250 FALSE 1971 FH Aachen
2 47269 553 TRUE 1870 RWTH Aachen
3 23659 438 TRUE 1457 Uni Freiburg
4 9415 150 TRUE 1818 Uni Bonn
5 38079 636 FALSE 1995 FH Bonn-Rhein-SiegA common task in data preparation is classifying a continuous variable, such as the number of professors. We want to group profs in steps of 150. To create these classes, we use cut() and specify the class boundaries with breaks. We can use seq() to generate the breakpoints. In seq(), we specify the lower and upper limits along with the step size.
cut(dat3$profs,breaks = c(50, 200, 350, 500, 650))[1] (200,350] (500,650] (350,500] (50,200] (500,650]
Levels: (50,200] (200,350] (350,500] (500,650]cut(dat3$profs,breaks = seq(50,650,150))[1] (200,350] (500,650] (350,500] (50,200] (500,650]
Levels: (50,200] (200,350] (350,500] (500,650]We store these values in a new variable in the dat3 dataset:
dat3$prof_class <- cut(dat3$profs,breaks = seq(50,650,150))
dat3 studs profs prom_recht gegr uni prof_class
1 14954 250 FALSE 1971 FH Aachen (200,350]
2 47269 553 TRUE 1870 RWTH Aachen (500,650]
3 23659 438 TRUE 1457 Uni Freiburg (350,500]
4 9415 150 TRUE 1818 Uni Bonn (50,200]
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg (500,650]For this new variable, we can request a frequency table using count():
dat3 %>% count(prof_class) prof_class n
1 (50,200] 1
2 (200,350] 1
3 (350,500] 1
4 (500,650] 2The parentheses ( indicate exclusion, while the brackets ] indicate inclusion. There are 1 universities in the dataset that have more than 200 and up to 350 professors.
For the following examples, we delete the prof_class variable again:
dat3$prof_class <- NULLSome useful options for cut() in the appendix
bsp <- c(1990,1998,2001,2009)
bsp[1] 1990 1998 2001 2009cut(bsp,breaks = c(1990,2000,2010)) [1] <NA> (1.99e+03,2e+03] (2e+03,2.01e+03] (2e+03,2.01e+03]
Levels: (1.99e+03,2e+03] (2e+03,2.01e+03]# Specify the number of digits in the labels
cut(bsp,breaks = c(1990,2000,2010),dig.lab = 4) [1] <NA> (1990,2000] (2000,2010] (2000,2010]
Levels: (1990,2000] (2000,2010]# Include the lower boundary
cut(bsp,breaks = c(1990,2000,2010),dig.lab = 4,include.lowest = T) [1] [1990,2000] [1990,2000] (2000,2010] (2000,2010]
Levels: [1990,2000] (2000,2010]# Number the categories instead of labels:
cut(bsp,breaks = c(1990,2000,2010),labels = FALSE)[1] NA 1 2 2# Specify your own labels:
cut(bsp,breaks = c(1990,2000,2010),labels = c("90s","00s"))[1] <NA> 90s 00s 00s
Levels: 90s 00s4.6.3 Renaming variables
To rename variables, use rename(new_name = old_name)
sat_small %>% rename(newname = PEO0300a)# A tibble: 5 × 3
newname PEO0300b PEO0300c
<dbl> <dbl> <dbl>
1 2 3 2
2 1 1 3
3 1 1 3
4 2 1 1
5 1 1 2For advanced transformations, it’s worth looking into rename_with(). This allows us to use Regular Expressions, for example from {stringr}. Here’s just an example:
sat_small %>% rename_with(~tolower(.))# A tibble: 5 × 3
peo0300a peo0300b peo0300c
<dbl> <dbl> <dbl>
1 2 3 2
2 1 1 3
3 1 1 3
4 2 1 1
5 1 1 2sat_small %>% rename_with(~str_remove(.x,"PEO0300"))# A tibble: 5 × 3
a b c
<dbl> <dbl> <dbl>
1 2 3 2
2 1 1 3
3 1 1 3
4 2 1 1
5 1 1 2sat_small %>% rename_with(~str_replace(.x,"PEO0300","Occupation_"))# A tibble: 5 × 3
Occupation_a Occupation_b Occupation_c
<dbl> <dbl> <dbl>
1 2 3 2
2 1 1 3
3 1 1 3
4 2 1 1
5 1 1 24.6.4 String Functions for regex
{stringr} provides a series of very useful string functions with regular expressions. You can get an overview from this cheatsheet.
dat3 %>% mutate(uni_fh = str_detect(uni,"Uni")) studs profs prom_recht gegr uni uni_fh
1 14954 250 FALSE 1971 FH Aachen FALSE
2 47269 553 TRUE 1870 RWTH Aachen FALSE
3 23659 438 TRUE 1457 Uni Freiburg TRUE
4 9415 150 TRUE 1818 Uni Bonn TRUE
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg FALSEdat3 %>% mutate(bula = case_when(str_detect(uni,"Bremen")~ "HB",
str_detect(uni,"Oldenb|Vechta")~ "NDS",
str_detect(uni,"Bonn|Aachen")~ "NRW",
str_detect(uni,"Freiburg")~ "BW"
)) studs profs prom_recht gegr uni bula
1 14954 250 FALSE 1971 FH Aachen NRW
2 47269 553 TRUE 1870 RWTH Aachen NRW
3 23659 438 TRUE 1457 Uni Freiburg BW
4 9415 150 TRUE 1818 Uni Bonn NRW
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg NRWdat3 %>% mutate(ort = str_remove(uni,"Uni |FH |RWTH ")) studs profs prom_recht gegr uni ort
1 14954 250 FALSE 1971 FH Aachen Aachen
2 47269 553 TRUE 1870 RWTH Aachen Aachen
3 23659 438 TRUE 1457 Uni Freiburg Freiburg
4 9415 150 TRUE 1818 Uni Bonn Bonn
5 38079 636 FALSE 1995 FH Bonn-Rhein-Sieg Bonn-Rhein-SiegDo not repeat yourself, see Wickham et al: “You should consider writing a function whenever you’ve copied and pasted a block of code more than twice (i.e. you now have three copies of the same code).”↩︎