Click to show/hide code
x <- c(1, 1, 1, 2, 2, 3, 3, 3, 3, 3)
quantile(
x,
probs = c(0.5, 0.75),
type = 1
)50% 75%
2 3 Quantile algorithms
1 Steps to calculate quantiles in R
The help of the
quantile()function in R describes the different algorithms used to calculate quantiles. However, I find the details very technical and not easy to understand for casual readers.Therefore, I tried to simplify the algorithms as discussed below.
R provides nine algorithms to calculate the quantiles, which can be specified using the
typeargument:Types \(1-3\) are used for discontinuous data, while types \(4-9\) are used for continuous data.
Type \(1\) and \(3\) are used for class “Date” and for ordered factors.
Type \(7\) is the default method for continuous data in R.
Type \(6\) is used to get results similar to
SPSS,Minitab, orGraphpad Prism.
Let \(n\) be the number of observations in the dataset.
Let \(p\) a number between \(0\) and \(1\), where \((p \times 100)\%\) is the quantile to be calculated, denoted by \(Q_p\).
The first step is to calculate the product \(np\).
Next, calculate the \(j^{th}\) rank and \(Q_p\) as follows:
Algorithm \(j\) Condition Quantile \((Q_p)\) \(1\) \(\lfloor np \rfloor\) \(np = j\)
\(np \ne j\)
\(x_j\)
\(x_{j+1}\)
\(2\) \(\lfloor np \rfloor\) \(np = j\)
\(np \ne j\)
\(\frac{1}{2}(x_j + x_{j+1})\)
\(x_{j+1}\)
\(3\) \(\lfloor np - \large \frac{1}{2}\rfloor\) \(np = j + \frac{1}{2}\) and \(j\) even
\(np \ne j+ \frac{1}{2}\)
\(x_j\)
\(x_{j+1}\)
\(4\) \(\lfloor np \rfloor\) \(x_j + (x_{j+1} - x_j)(np-j)\) \(5\) \(\lfloor np + \large \frac{1}{2}\rfloor\) \(x_j + (x_{j+1} - x_j)(np-j + \frac{1}{2})\) \(6\) \(\lfloor np + p \rfloor\) \(x_j + (x_{j+1} - x_j)(np-j +p)\) \(7\) \(\lfloor np -p +1 \rfloor\) \(x_j + (x_{j+1} - x_j)(np -j - p + 1)\) \(8\) \(\lfloor np + \large \frac{(p+1)}{3} \rfloor\) \(x_j + (x_{j+1} - x_j)(np - j + \frac{(p+1)}{3})\) \(9\) \(\lfloor np + \large \frac{(2p+3)}{8} \rfloor\) \(x_j + (x_{j+1} - x_j)(np - j + \frac{(2p+3)}{8})\) The symbol \(\lfloor x \rfloor\) reads as the floor of \(x\). This function returns the largest integer not greater than \(x\) (i.e., rounds down \(x\) to the nearest integer). For example, \(\lfloor 3.9 \rfloor = 3\).
\(x_j\) and \(x_{j+1}\) are the \(j^{th}\) and \((j+1)^{th}\) order statistics, respectively (i.e., the observations having the rank \(j\) and \(j+1\) in the ordered array of the observations).
Let’s illustrate the above algorithms with some examples.
Example \(1\) (ordered factor): consider the following set of observations pertaining to the pain severity (\(1\): mild, \(2\): moderate, \(3\): severe) of \(10\) patients: \(1, 1, 1, 2, 2, 3, 3, 3, 3, 3\). The \(50^{th}\ (Q_{0.5})\) and \(75^{th}\ (Q_{0.75})\) quantiles can be calculated using algorithms \(1-3\) as follows:
\(\small 50^{th}\) Quantile:
\(\small p = 0.5\), \(\small np = 10 \times 0.5 = 5\), and \(\small j = \lfloor np \rfloor = \lfloor 5 \rfloor = 5\)
Since \(\small np = j\), then \(\small Q_{0.5} = x_j = x_5\) (the observation having the rank \(\small 5\) in the ordered array), which has the value \(\small 2\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 10 \times 0.75 = 7.5\), and \(\small j = \lfloor np \rfloor = \lfloor 7.5 \rfloor = 7\)
Since \(\small np \ne j\), then \(\small Q_{0.75} = x_{j+1} = x_8\) (the observation having the rank \(\small 8\) in the ordered array), which has the value \(\small 3\)
Check the result using the
quantile()function in R:
\(\small 50^{th}\) Quantile:
\(\small p = 0.5\), \(\small np = 10 \times 0.5 = 5\), and \(\small j = \lfloor np \rfloor = \lfloor 5 \rfloor = 5\)
Since \(\small np = j\), then \(\small Q_{0.5} = \frac{1}{2}(x_j + x_{j+1}) =\) \(\small \frac{1}{2}(x_5 + x_6) = \frac{1}{2}(2 + 3) = 2.5\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 10 \times 0.75 = 7.5\), \(\small j = \lfloor np \rfloor = \lfloor 7.5 \rfloor = 7\)
Since \(\small np \ne \lfloor np \rfloor\), then \(\small Q_{0.75} = x_{j+1} = x_8 = 3\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( x, probs = c(0.5, 0.75), type = 2 )50% 75% 2.5 3.0
\(\small 50^{th}\) Quantile:
\(\small p = 0.5\), \(\small np = 10 \times 0.5 = 5\), and \(\small j = \lfloor np - 0.5 \rfloor = \lfloor 5 - 0.5 \rfloor = 4\)
Since \(\small np \ne j + 0.5\), then \(\small Q_{0.5} = x_{j+1} = x_5 = 2\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 10 \times 0.75 = 7.5\), and \(\small j = \lfloor np - 0.5 \rfloor = \lfloor 7.5 - 0.5 \rfloor = 7\)
Since \(\small np = j + 0.5\) but \(j\) is not even, then \(\small Q_{0.75} = x_{j+1} = x_8 = 3\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( x, probs = c(0.5, 0.75), type = 3 )50% 75% 2 3
- Example \(2\) (continuous data): consider the following ordered set of observations: \((10.2,\ 10.4,\ 11.6,\ 12.3,\ 13.2,\ 14.7,\ 15.4,\ 16.1)\). The \(25^{th}\ (Q_{0.25})\) and \(75^{th}\ (Q_{0.75})\) quantiles can be calculated using algorithms \(4-9\) as follows:
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np \rfloor = \lfloor 2 \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j) = x_2 + (x_3 - x_2)(2 - 2) = 10.4\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np \rfloor = \lfloor 6 \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j) = x_6 + (x_7 - x_6)(6 - 6) = 14.7\)
Check the result using the
quantile()function in R:Click to show/hide code
y <- c(10.2, 10.4, 11.6, 12.3, 13.2, 14.7, 15.4, 16.1) quantile( y, probs = c(0.25, 0.75), type = 4 )25% 75% 10.4 14.7
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np + 0.5 \rfloor = \lfloor 2.5 \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j + 0.5) = x_2 + (x_3 - x_2)(2 - 2 + 0.5) =\)
\(\small 10.4 + (11.6 - 10.4) \times 0.5 = 11\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np + 0.5 \rfloor = \lfloor 6.5 \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j + 0.5) = x_6 + (x_7 - x_6)(6 - 6 + 0.5) =\) \(\small 14.7 + (15.4 - 14.7) \times 0.5 = 15.05\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( y, probs = c(0.25, 0.75), type = 5 )25% 75% 11.00 15.05
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np + p \rfloor = \lfloor 2 + 0.25 \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j + p) = x_2 + (x_3 - x_2)(2 - 2 + 0.25) =\)
\(\small 10.4 + (11.6 - 10.4) \times 0.25 = 10.7\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np + p \rfloor = \lfloor 6 + 0.75 \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j + 0.75) = x_6 + (x_7 - x_6)(6 - 6 + 0.75) =\) \(\small 14.7 + (15.4 - 14.7) \times 0.75 = 15.225\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( y, probs = c(0.25, 0.75), type = 6 )25% 75% 10.700 15.225
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np - p + 1 \rfloor = \lfloor 2 - 0.25 + 1 \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j - p + 1) = x_2 + (x_3 - x_2)(2 - 2 - 0.25 + 1) =\)
\(\small 10.4 + (11.6 - 10.4) \times 0.75 = 11.3\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np - p + 1 \rfloor = \lfloor 6 - 0.75 + 1 \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j - 0.75 + 1) = x_6 + (x_7 - x_6)(6 - 6 - 0.75 + 1) =\) \(\small 14.7 + (15.4 - 14.7) \times 0.25 = 14.875\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( y, probs = c(0.25, 0.75), type = 7 )25% 75% 11.300 14.875
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np + \frac{p + 1}{3} \rfloor = \lfloor 2 + \frac{0.25 + 1}{3} \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j + \frac{p + 1}{3}) = x_2 + (x_3 - x_2)(2 - 2 + \frac{0.25 + 1}{3}) =\)
\(\small 10.4 + (11.6 - 10.4) \times 0.41667 = 10.9\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np + \frac{p + 1}{3} \rfloor = \lfloor 6 + \frac{0.75 + 1}{3} \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j + \frac{p + 1}{3}) = x_6 + (x_7 - x_6)(6 - 6 + \frac{0.75 + 1}{3}) =\) \(\small 14.7 + (15.4 - 14.7) \times 0.58333 = 15.10833\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( y, probs = c(0.25, 0.75), type = 8 )25% 75% 10.90000 15.10833
\(\small 25^{th}\) Quantile:
\(\small p = 0.25\), \(\small np = 8 \times 0.25 = 2\), and \(\small j = \lfloor np + \frac{2p + 3}{8} \rfloor = \lfloor 2 + \frac{2 \times 0.25 + 3}{8} \rfloor = 2\)
\(\small Q_{0.25} = x_j + (x_{j+1} - x_j)(np - j + \frac{2p + 3}{8}) = x_2 + (x_3 - x_2)(2 - 2 + \frac{2 \times 0.25 + 3}{8}) =\)
\(\small 10.4 + (11.6 - 10.4) \times 0.4375 = 10.925\)
\(\small 75^{th}\) Quantile:
\(\small p = 0.75\), \(\small np = 8 \times 0.75 = 6\), and \(\small j = \lfloor np + \frac{2p + 3}{8} \rfloor = \lfloor 6 + \frac{2 \times0.75 + 3}{8} \rfloor = 6\)
\(\small Q_{0.75} = x_j + (x_{j+1} - x_j)(np - j + \frac{2p + 3}{8}) = x_6 + (x_7 - x_6)(6 - 6 + \frac{2 \times 0.75 + 3}{8}) =\) \(\small 14.7 + (15.4 - 14.7) \times 0.5625 = 15.09375\)
Check the result using the
quantile()function in R:Click to show/hide code
quantile( y, probs = c(0.25, 0.75), type = 9 )25% 75% 10.92500 15.09375
2 References
quantile: Sample Quantiles. Retrieved October 20, 2024, from https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/quantile