# Statistical Methods in Medical Research - 
# Laaketieteellisen tutkimuksen tilastolliset menetelmat
# At University of Helsinki
# 5.8.2025
# Matti Pirinen



###
### 3. Learn R: Logical comparisons and 'if'-statement
###


# We can assess whether a statement is true and R returns one of
# two possible logical values: TRUE or FALSE
1 > 2  # is 1 greater than 2?
1 < 2  # is 1 smaller than 2?
1 > 1  # is 1 greater than 1?
1 >= 1 # is 1 greater or equal to 1?
1 <= 1 # is 1 less or equal to 1?
x = 1  # this is an assignment not a comparison! -- from now on x has value 1
x == 2 # this is a comparison: Is x equal to 2? Note the '=' symbol is twice in the comparison operator!
x != 2 # is x NOT equal to 2?
x == 1 # is x equal to 1?
x != 1 # is x NOT equal to 1?

# To combine several comparisons we can use "and" or "or".
# These are denoted by symbols '&' for AND and '|' for OR

# To check whether 'x' is in interval (0,1) we can ask whether 
#  'x' is both >0 and <1 
(x > 0) & (x < 1)
# If we allow the upper limit to be exactly 1 then the statement becomes TRUE since currently x equals 1:
(x > 0) & (x <= 1)

# To check whether at least one of these is true:
# 'x' is > 0 or -2 < x < -1, 
(x > 0) | (x > -2 & x < -1)
# If we increase the first comparison point from 0 to 1, 
# then the statement becomes FALSE because currently x equals 1
(x > 1) | (x > -2 & x < -1)



# We can use such logical TRUE/FALSE values to control what R does.
if(x > 0) {
    print("First IF was true.")
    x = x + 1
}

if(x <= 0) {
    print("Second IF was true.")
    x = x - 1
}

# If the comparison within the 'if()' statement is TRUE, 
# then the code within the following block '{ }' is run.
# If it is FALSE, then the code block within '{ }' will be skipped.
# Above, only the first IF is true, so R changes x from 1 to 2 = 1 + 1, but
# since the second IF is not true, R does not subtract 1 from x at the second step.
# (Note also how the print("") command has only been run for the first if-statement that was true.)
# So currently x should be 2.
x

# Note that the curly brackets don't need to be on their own lines so this works as well
if(x <= 0) {x = x - 1}
# But if you have more than one expression within the brackets, 
# then it is good to use a separate line for each command.

# Define a vector of 3 positive and 2 negative values
x = c(1, -2, 4, -10, 1)
# If we apply a logical comparison to the vector, 
# the result is an element-wise comparison
x < 0
# We can ask which indexes of the vector correspond to value TRUE in comparison
which(x < 0) #returns 2 and 4

# We can apply mathematical operations, like sum() or mean(), to logical values
# and then R will replace TRUE with 1  and FALSE with 0.
# So sum(x < 0) will compute how many elements of x are < 0
sum(x < 0)
# and mean(x < 0) is the same as sum(x < 0)/length(x), that is, the proportion of TRUEs.
mean(x < 0)


#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
# Test Yourself 3.1. (Answers are at the end of this file.)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#

#    BMI of 6 individuals are
#    24.5, 38, 27.4, 30.4, 26.3, 17.5
#(a) Make a vector of these 6 BMI values
#(b) Use logical comparisons to determine which individuals
#    have BMI value within interval (20,25), i.e., 20 < BMI < 25.
#(c) Apply sum() to the comparison in (b) to determine how many are in the interval (20,25).
#(d) Apply mean() to the comparison in (b) to determine which proportion is in the interval (20,25).
#(e) Use logical comparisons to determine which individuals
#    have either (BMI > 30) or (BMI < 20).
#(f) Set value of variable x to 3. Write an if-sentence that increases
#    value of variable x by +1 in case the value of x is < 4.
#    Show the value of 'x' after running the if-sentence.
#    Rerun the if-sentence and show again the value of 'x'.
#    Note that it should change only in the first round.





#
##
###  ANSWERS
##
#





#    BMI of 6 individuals are
#    24.5, 38, 27.4, 30.4, 26.3, 17.5
#(a) Make a vector of these 6 BMI values

bmi = c(24.5, 38, 27.4, 30.4, 26.3, 17.5)

#(b) Use logical comparisons to determine which individuals
#    have BMI value within interval (20,25), i.e., 20 < BMI < 25.
(bmi > 20) & (bmi < 25)
which((bmi > 20) & (bmi < 25)) #show indexes


#(c) Apply sum() to the comparison in (b) to determine how many are in the interval (20,25).

sum( (bmi > 20) & (bmi < 25) )

#(d) Apply mean() to the comparison in (b) to determine which proportion is in the interval (20,25).

mean( (bmi > 20) & (bmi < 25) )


#(e) Use logical comparisons to determine which individuals
#    have either (BMI > 30) or (BMI < 20).

(bmi > 30) | (bmi < 20)
which((bmi > 30) | (bmi < 20))


#(f) Set value of variable x to 3. Write an if-sentence that increases
#    value of variable x by +1 in case value of x is < 4.
#    Show the value of 'x' after running the if-sentence.
#    Rerun the if-sentence and show again the value of 'x'.
#    Note that it should change only in the first round.

x = 3
x #x is 3
if( x < 4) { x = x + 1}
x #x increased to 4
if( x < 4) { x = x + 1}
x #x is still 4