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R Quantile Calculations

• Define the sorted artificial dataset x, consisting of the first ten prime numbers:
  

  x <- c(2, 3, 5, 7, 11, 13, 17, 19, 23, 29) # Line 1
  

• In this document, we compute by hand the quantile 0.25 for each of the nine R methods of computing percentiles. See the Quantiles1 Example to see how to compute these quantiles using the R method quantile.
   
• The first three R methods to calculate quantiles are based on the empirical distribution function F:
   
  
  F(t) = (number of elements in sample ≤ t) / n
  
  where n is the sample size.
   
• R can be used to plot the empirical distribution function:
  

  > plot(ecdf(x))
  

  This command produces this graph of the cumulative distibution function.
   
• For each of the first two methods, we are computing the p quantile Q(p). In both cases, let np = j + g, where j is the integer part of np and g is the fractional part of np. We write
   
  
  j = ⌊np⌋ and g = np - ⌊np⌋
   
  
  where ⌊x⌋ is the floor of x (largest integer less than or equal to x).

R Method 1, SAS Percentile Definition 3

• Use the inverse of the empirical distribution function. Define Q(p) like this:
   
  
  Q₁(p) = x_j if g = 0,
  Q₁(p) = x_j+1 if g > 0
  
  To compute Q₁(0.25) for the sample x defined in Line 1, we see that pn = 0.25*10 = 2.5 = 2 + 0.5, so that j = 2 and g = 0.5. Since g > 0, Q₁(0.25) = x_j+1 = x₃ = 5. Check our answer with R:
  

  > quantile(x, 0.25, type=1)
  25% 
  5
  

R Method 2, SAS Percentile Definition 5

• Same as method 1, but use averaging at the discontinuities.
   
  
  Q₂(p) = (x_j + x_j+1) / 2 if g = 0,
  Q₂(p) = x_j+1 if g > 0
   
  
  The computation of Q₂(0.25) is the same as in Method 1, however, to compute Q₂(0.50), pn = 0.5*10 = 5.0, so j = ⌊5.0⌋ = 5 and g = 5.0 - j = 5.0 - 5 = 0.0. Therefore, Q₂(0.5) = (x_j + x_j+1) / 2 = (x₅ + x₆) / 2 = (11 + 13) / 2 = 12. This value differs from the value of 11 that Method 1 obtains. Check with R:
  

  > quantile(x, 0.50, type=2)
  50% 
  12 
  

R Method 3, SAS Percentile Definition 2

• Use the observation numbered closest to x_pn:
   
  
  Q₃(p) = x_⌊np⌋
  
  For the sample x defined in Line 1, Q₃ = ⌊np + 0.5⌋ = ⌊10*0.25 + 0.5⌋ = ⌊3.0⌋ = 3. Check with R:
  

  > quantile(x, 0.25, type=3)
  25% 
    3 
  

Methods using Interpolation

• Methods 4 through 9 use linear interpolation to obtain the percentile estimate.
   
• To remind us how to use linear interpolation to improve the accuracy of a table lookup, let's estimate the probability that P(x ≤ 1.234) using a standard normal table. Looking up values in the table gives us P(x ≤ 1.23) = 0.8907 and P(x ≤ 1.24) = 0.8925. We compute:
   
  
  xdiff_big = 1.24 - 1.23 = 0.01
   
  pdiff_big = 0.8925 - 0.8907 = 0.0018
  
  We also compute
   
  
  xdiff_small = 1.234 - 1.23 = 0.004
  
  Linear interpolation assumes that pdiff_small can be approximated from the proportion
   
  
  pdiff_small / xdiff_small = pdiff_big / xdiff_big
   
  pdiff_small / 0.004 = 0.0018 / 0.01
  
  Solve for pdiff_small:
   
  
  pdiff_small = 0.004 * 0.0018 / 0.01 = 0.00072
   
  
  Therefore the estimated probability P(x ≤ 1.234) = 0.8907 + pdiff_small = 0.8907 + 0.00072 = 0.8914. Check with R:
  

  > pnorm(1.234)
  [1] 0.8913985
  

• The points (1.23, 0.8907) and (1.24, 0.8925) are called knots.
   
• To compute the pth quantile with Methods 4 through 9 (Q₄(p), Q₅(p), ... , Q₉(p)), use linear interpolation. For each method we choose a constant m:
   
  
  Table 1:

  Method:	4	5	6	7	8	9
  Constant m:	0	1/2	p	1-p	(p+1)/3	p/4+3/8

  
  If j is chosen such that
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n,
  
  we can define the knots as (j, x_j) and (j+1, x_j+1).

• Method 4 uses linear interpolation based on the knots determined by the empirical pdf.
   
• Use x from Line 1. Use m = 0 for Method 4 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - 0) / 10 ≤ p ≤ (j - 0 + 1) / n
  j / 10 ≤ 0.25 ≤ (j + 1) / 10         (*4*)
  j ≤ 2.5 ≤ j + 1
  
  Therefore j = 2. The knots are obtained from (*4*)
   
  
  (2/10, x₂) and (3/10, x₃)
  
  which are (0.2, 3) and (0.3, 5). Use linear interpolation to obtain Q₄(0.5):
   
  
  pdiff_big = 0.3 - 0.2 = 1
  xdiff_big = 5 - 3 = 2
  pdiff_small = 0.25 - 0.2 = 0.05
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0.05 / 0.1
  
  This gives xdiff_small = 1, so Q₄(0.5) = x₂ + xdiff_small = 3 + 1 = 4. Check with R:
  

  > quantile(x, 0.25, type=4)
  25% 
  4 
  

R Method 5

• Method 5 uses linear interpolation where the knots are the values midway between the steps of the empirical cumulative distribution function. This method is popular among hydrologists.
   
• Use x from Line 1. Use m = 1/2 for Method 5 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - 0.5) / 10 ≤ 0.25 ≤ (j - 0.5 + 1) / n
  j - 0.5 ≤ 2.5 ≤ j + 0.5            (*5*)
  j ≤ 3 ≤ j + 1
  
  Therefore j = 3 and the knots are obtained from (*5*)
   
  
  ((3 - 0.5)/10, x₃) and ((4 - 0.5)/10, x₄)
  
  which are (0.25, 5) and (0.35, 7). Use linear interpolation to obtain Q₅(0.25):
   
  
  pdiff_big = 0.35 - 0.25 = 0.1
  xdiff_big = 7 - 5 = 2
  pdiff_small = 0.25 - 0.25 = 0
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0 / 0.1
  
  This gives xdiff_small = 0, so Q₅(0.5) = x₃ + xdiff_small = 5 + 0 = 5. Check with R:
  

  > quantile(x, 0.25, type=5)
  25% 
  5 
  

R Method 6, SAS Percentile Definition 4

• This method is also used by Minitab and SPSS.
   
• Use x from Line 1. Use m = p for Method 6 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - p) / 10 ≤ p ≤ (j - p + 1) / n
  (j - 0.25) / 10 ≤ 0.25 ≤ (j + 0.75) / 10          (*6*)
  j - 0.25 ≤ 2.5 ≤ j + 0.25
  j ≤ 2.75 ≤ j + 1
  
  
  Therefore j = 2 and the knots are obtained from (*6*)
   
  
  ((2 - 0.25)/10, x₂) and ((2 + 0.75)/10, x₃)
  
  which are (0.175, 3) and (0.275, 5). Use linear interpolation to obtain Q₆6(0.25):
   
  
  pdiff_big = 0.275 - 0.175 = 0.1
  xdiff_big = 5 - 3 = 2
  pdiff_small = 0.25 - 0.175 = 0.075
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0.075 / 0.1
  
  This gives xdiff_small = 1.5, so Q₆6(0.5) = x₂ + xdiff_small = 3 + 1.5 = 4.5. Check with R:
  

  > quantile(x, 0.25, type=6)
  25% 
  4.5
  

R Method 7

• Method 7 is the default method for R. It is also used by S and Excel.
   
• Use x from Line 1. Use m = 1- p for Method 7 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - (1 - p)) / 10 ≤ p ≤ (j - (1 - p) + 1) / n
  (j - (1 - 0.25)) / 10 ≤ 0.25 ≤ (j - (1 - 0.25) + 1) / 10
  (j - 0.75) / 10 ≤ 0.25 ≤ (j + 0.25) / 10          (*7*)
  j - 0.75 ≤ 2.5 ≤ j + 0.25
  j ≤ 3.25 ≤ j + 1
  
  
  Therefore j = 3 and the knots are obtained from (*7*)
   
  
  ((3 - 0.75)/10, x₃) and ((3 + 0.25)/10, x₄)
  
  which are (0.225, 5) and (0.325, 7). Use linear interpolation to obtain Q₇(0.25):
   
  
  pdiff_big = 0.325 - 0.225 = 0.1
  xdiff_big = 7 - 5 = 2
  pdiff_small = 0.25 - 0.225 = 0.025
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0.025 / 0.1
  
  This gives xdiff_small = 0.5, so Q₇(0.5) = x₂ + xdiff_small = 5 + 0.5 = 5.5. Check with R:
  

  > quantile(x, 0.25, type=7)
  25% 
  5.5 
  

R Method 8

• Q₈(p) is has the theoretical property of being median unbiased. See Reference 3. The authors of this reference recommend Method 8 as being the best of methods 1 though 9.
   
• Use x from Line 1. Use m = (p+1)/3 for Method 8 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - ((p+1)/3)) / 10 ≤ p ≤ (j - ((p+1)/3) + 1) / n
  (j - ((0.25+1)/3) / 10 ≤ 0.25 ≤ (j - ((0.25+1)/3) + 1) / 10
  (j - 0.4166667) / 10 ≤ 0.25 ≤ (j + 0.5833333) / 10          (*8*)
  j - 0.4166667 ≤ 2.5 ≤ j + 0.5833333
  j ≤ 2.916667 ≤ j + 1
  
  
  Therefore j = 2 and the knots are obtained from (*8*)
   
  
  ((2 - 0.4166667)/10, x₂) and ((2 + 0.5833333)/10, x₃)
  
  which are (0.15833333, 3) and (0.25833333, 5). Use linear interpolation to obtain Q₈(0.25):
   
  
  pdiff_big = 0.25833333 - 0.15833333 = 0.1
  xdiff_big = 5 - 3 = 2
  pdiff_small = 0.25 - 0.15833333 = 0.09166667
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0.09166667 / 0.1
  
  This gives xdiff_small = 1.833333, so Q₈(0.5) = x₂ + xdiff_small = 3 + 1.833333 = 4.833333. Check with R:
  

  > quantile(x, 0.25, type=8)
  25% 
  4.833333
  

Method 9

• Q₉(p) has the property that if the population is normal, Q₉(x) is asymptotically unbiased at the knots.
   
• Use x from Line 1. Use m = p/4+3/8 for Method 9 from Table 1.
   
• Find the locations of the knots:
   
  
  (j - m) / n ≤ p ≤ (j - m + 1) / n
  (j - (p/4 + 3/8) / 10 ≤ p ≤ (j - (p/4+3/8) + 1) / n
  (j - (0.25/4 + 3/8) / 10 ≤ 0.25 ≤ (j - (0.25/4 + 3/8) + 1) / 10
  (j - 0.4375) / 10 ≤ 0.25 ≤ (j + 0.5625) / 10          (*9*)
  j - 0.4375 ≤ 2.5 ≤ j + 0.5625
  j ≤ 2.9375 ≤ j + 1
  
  
  Therefore j = 2 and the knots are obtained from (*9*)
   
  
  ((2 - 0.4375)/10, x₂) and ((2 + 0.5625)/10, x₃)
  
  which are (0.15625, 3) and (0.25625, 5). Use linear interpolation to obtain Q₉(0.25):
   
  
  pdiff_big = 0.25625 - 0.15625 = 0.1
  xdiff_big = 5 - 3 = 2
  pdiff_small = 0.25 - 0.15625 = 0.09375
  
  Solve for xdiff_small in
   
  
  xdiff_small / xdiff_big = pdiff_small / pdiff_big
  xdiff_small / 2 = 0.09375 / 0.1
  
  This gives xdiff_small = 1.875, so Q₉(0.5) = x₂ + xdiff_small = 3 + 1.875 = 4.875. Check with R:
  

  > quantile(x, 0.25, type=9)
  25% 
  4.875