Primary exercises

  1. Dietary intakes.
    Four patients had daily dietary intakes of 2314, 2178, 1922, 2004 kcal.
    Make a vector intakesKCal of these four values.
    What is the class of this vector?
    Convert the values into in kJ using 1 kcal = 4.184 kJ.
intakesKCal <- c( 2314, 2178, 1922, 2004 )
intakesKCal
[1] 2314 2178 1922 2004
class( intakesKCal )
[1] "numeric"
intakesKCal * 4.184
[1] 9681.776 9112.752 8041.648 8384.736
  1. Combining (appending) vectors.
    Additional set of intakes is provided: 2122, 2616, NA, 1771 kcal.
    Use c() to append the new intakes after values in intakesKCal and store the result in allIntakesKCal.
    Print the combined vector and print its calculated length.
intakesKCal2 <- c( 2122, 2616, NA, 1771 )
allIntakesKCal <- c( intakesKCal, intakesKCal2 )
allIntakesKCal
[1] 2314 2178 1922 2004 2122 2616   NA 1771
length( allIntakesKCal )
[1] 8
  1. Mean and sum.
    Calculate mean intake for patients in vector intakesKCal.
    Next, calculate mean intake for patients in vector allIntakesKCal.
    Can you explain the result?
    Check help for ?mean, in particular the na.rm argument.
    Use the extra argument na.rm=TRUE to calculate the mean of non-NA elements of allIntakesKCal.
    Check help for ?sum how to omit NA elements in sum calculation.
    Now, calculate the total sum of allIntakesKCal intakes ignoring the NA element.
mean( intakesKCal )
[1] 2104.5
mean( allIntakesKCal )
[1] NA
# since one element is missing, the mean is unknown
# ?mean, adding argument na.rm=TRUE will omit NA elements
mean( allIntakesKCal, na.rm = TRUE )
[1] 2132.429
# ?sum also allows na.rm=TRUE argument to skip NA elements
sum( allIntakesKCal, na.rm = TRUE )
[1] 14927
  1. Selecting and counting (non)missing elements.
    Understand the result of is.na( allIntakesKCal ).
    Now, negate the above result with ! operator.
    Use above vectors as argument to sum to calculate the number of missing and non-missing elements in allIntakesKCal.
    Understand allIntakesKCal[ !is.na( allIntakesKCal ) ].
is.na( allIntakesKCal )         # TRUE marks positions with missing data
[1] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE
!is.na( allIntakesKCal )        # TRUE marks positions with available data
[1]  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE
sum( is.na( allIntakesKCal ) )                # number of missing elements
[1] 1
sum( !is.na( allIntakesKCal ) )               # number of non-missing elements
[1] 7
allIntakesKCal[ !is.na( allIntakesKCal ) ]    # keeps elements which are not NA
[1] 2314 2178 1922 2004 2122 2616 1771
sum( allIntakesKCal[ !is.na( allIntakesKCal ) ] )    # same as sum( allIntakesKCal, na.rm = TRUE )
[1] 14927
  1. Descriptive statistics of a vector; normally distributed random numbers.
    The code v <- rnorm( 10 ) would sample 10 numbers from the normal distribution and store them as a vector in v.
    Print v. Then repeat v <- rnorm( 10 ) and print v again. Has v changed? Print v and find by eye the smallest and the largest of these numbers.
    Try to use the functions min and max on v – have you found the same numbers?
    Calculate the mean, median and the standard deviation (sd) of v.
v <- rnorm( 10 ) # a vector of random numbers
v
 [1]  0.5935617 -0.5015409 -0.1647272 -0.9295109 -1.2055671 -0.7134088 -0.2267908 -2.3737123 -0.1872570  0.9924054
v <- rnorm( 10 ) # another vector of random numbers
v
 [1] -1.12371371 -0.67363352  1.85665769 -1.65825816 -0.01172825  1.32328767 -2.07902160 -1.33737725 -0.12177886 -0.40108252
min( v )
[1] -2.079022
max( v )
[1] 1.856658
mean( v )
[1] -0.4226648
median( v )
[1] -0.537358
sd( v )
[1] 1.254542
  1. Selecting and counting elements by a condition. Type v < 0 and understand the result.
    How to interpret the number produced by sum( v < 0 )? How to interpret the number produced by sum( !( v < 0 ) )?
v < 0             # TRUE corresponds to elements of vector v smaller than 0
 [1]  TRUE  TRUE FALSE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE  TRUE
sum( v < 0 )      # calculates the number of negative elements in vector v
[1] 8
sum( !( v < 0 ) ) # calculates the number of non-negative (so, positive OR ZERO) elements in vector v 
[1] 2
sum( v >= 0 )     # same as above
[1] 2
  1. Head and tail.
    Often there is a need to quickly look at the beginning (head) or at the end (tail) of a vector.
    Try these functions to show the first 5 and the last 7 elements of a randomly generated vector v <- rnorm( 20 ).
v <- rnorm( 20 )
v
 [1]  0.734170910  0.817425134  0.175779325  0.598587293  0.237837620 -0.015178547  1.085935906 -0.179678206 -0.741001462
[10] -0.429344990  0.002857104  0.637463848  0.491943373 -0.149613651  1.249617794 -0.643536065  0.789087896 -0.452782817
[19]  1.657816525  0.643141655
head( v, 5 )
[1] 0.7341709 0.8174251 0.1757793 0.5985873 0.2378376
tail( v, 7 )
[1] -0.1496137  1.2496178 -0.6435361  0.7890879 -0.4527828  1.6578165  0.6431417

Extra exercises

  1. More complex sequences of numbers.
    Read help (see Help pane) about seq function.
    Use it to generate sequence: 10, 7, 4, 1, -2, -5.
    Understand the error message of seq( 10, -5, 3 ).
seq( 10, -5, -3 )
[1] 10  7  4  1 -2 -5
seq( from = 10, to = -5, by = -3 )
[1] 10  7  4  1 -2 -5
  1. Repetitions.
    Read help (see Help pane) about rep function.
    Use it to generate sequence: 0, 1, 0, 1, 0, 1.
rep( c( 0, 1 ), 3 )
[1] 0 1 0 1 0 1
  1. Type conversion to a character vector.
    Sometimes it is necessary to convert a numerical vector to a character vector.
    Understand what the function as.character does for argument 1:5.
1:5
[1] 1 2 3 4 5
as.character( 1:5 )
[1] "1" "2" "3" "4" "5"
class( 1:5 )
[1] "integer"
class( as.character( 1:5 ) )
[1] "character"
  1. Type conversion to a numerical vector.
    Sometimes it is necessary to convert a character vector to a numerical vector.
    Understand what the function as.numeric does for argument c( "1", "-1", "x" ).
    Note the warning message. Why is there NA?
as.numeric( c( "1", "-1", "x" ) )
Warning: NAs introduced by coercion
[1]  1 -1 NA
  1. Naming vector elements.
    It is possible to give names to vector elements.
    Define ages <- c( Amy = 10, 'Dan' = 6, "Eve" = 11, "Eve 2" = 3, Grandma = NA ).
    Print ages and understand names( ages ).
    Use square brackets to access age of Dan. Try also for Eve 2.
ages <- c( Amy = 10, 'Dan' = 6, "Eve" = 11, "Eve 2" = 3, Grandma = NA )
ages
    Amy     Dan     Eve   Eve 2 Grandma 
     10       6      11       3      NA 
names( ages )
[1] "Amy"     "Dan"     "Eve"     "Eve 2"   "Grandma"
ages[ 'Dan' ]
Dan 
  6 
ages[ 'Eve 2' ]
Eve 2 
    3 
# Another way to create a vector with named elements
ages2 <- c( 10, 6, 11, 3, NA )
names( ages2 ) <- c( "Amy", "Dan", "Eve", "Eve 2", "Grandma" )
ages2
    Amy     Dan     Eve   Eve 2 Grandma 
     10       6      11       3      NA
  1. (ADV) Write a text vector to a file and read it back.
    This exercise demonstrates writing a single-column vector (later multicolumn tables will be discussed).
    First choose a name for the file (e.g. test.txt) and store it in the variable fileName.
    Next, create a character/text vector v with several text elements.
    Check manual for writeLines and try writeLines(v) to see in the console what will be written to a file.
    Now, set the argument con = fileName and write to the file.
    Use readLines( con = fileName ) to read the file and put it back to variable w.
    Understand identical( v, w ).
fileName <- "test.txt"
v <- c( "First line", "Second", "Third", "4th", "5th", "6th" )
v
[1] "First line" "Second"     "Third"      "4th"        "5th"        "6th"       
writeLines( v )                     # writes to the console
First line
Second
Third
4th
5th
6th
writeLines( v, con = fileName )     # writes to a file
w <- readLines( con = fileName )
identical( v, w )   # checks whether v and w are exactly equal
[1] TRUE
unlink( fileName )  # removes the file
  1. (ADV) Write/read a numerical vector; problems.
    In the previous exercise change v to be a vector of some numbers.
    Use as.character to make writeLines work (do not change v).
    Why identical( v, w ) fails? Check class(v) and class(w).
    What conversions of w would be needed to make identical work?
fileName <- "test.txt"
v <- sample( 1:100, 10 )
v
 [1] 70 17 50 44 20 96 31 76 84  3
writeLines( as.character( v ) )                 # conversion to character needed
70
17
50
44
20
96
31
76
84
3
writeLines( as.character( v ), con = fileName )
w <- readLines( con = fileName )
identical( v, w )     # numbers are not the same as their text representation
[1] FALSE
w <- as.numeric( w )
identical( v, w )     # still not identical; class(v) is different than class(w)
[1] FALSE
w <- as.integer( w )
identical( v, w )     # now identical
[1] TRUE
unlink( fileName )    # removes the file

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