Primary exercises
- 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
- 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
- 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
- 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
- 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] -1.8424831 0.3771529 0.1408945 1.3716562 0.1535821 -0.5446351 -0.3262612 -1.2712089 -1.4035408 1.2726446
v <- rnorm( 10 ) # another vector of random numbers
v
[1] 0.94170936 -0.74960607 0.04293790 1.42695183 0.85984247 0.78012878 -1.27824130 -1.86219127 0.47255389 0.08519813
min( v )
[1] -1.862191
max( v )
[1] 1.426952
mean( v )
[1] 0.07192837
median( v )
[1] 0.278876
sd( v )
[1] 1.060146
- 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] FALSE TRUE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE
sum( v < 0 ) # calculates the number of negative elements in vector v
[1] 3
sum( !( v < 0 ) ) # calculates the number of non-negative (so, positive OR ZERO) elements in vector v
[1] 7
sum( v >= 0 ) # same as above
[1] 7
- 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] 0.658962328 0.405613240 -0.911405642 -1.726141428 0.205631753 1.361363410 0.485249157 -0.133469327 -0.119791503 1.207770961 1.313523435 0.397222703
[13] 0.806550618 -0.922736576 0.257475374 -0.680368244 -0.009816095 0.878533312 0.195450433 -0.019504327
head( v, 5 )
[1] 0.6589623 0.4056132 -0.9114056 -1.7261414 0.2056318
tail( v, 7 )
[1] -0.922736576 0.257475374 -0.680368244 -0.009816095 0.878533312 0.195450433 -0.019504327