Differences

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--- quartile [2019/09/19 10:42] – [e.g. 1] hkimscil
+++ quartile [2023/09/11 08:42] (current) – [r method] hkimscil
@@ Line 12: / Line 12: @@
 ====== Finding lower and upper quartile ======
-===== e.g. 1 =====
+===== e.g. 1, Head First method =====
 <code>> k <- c(1:8)
 > k
@@ Line 34: / Line 33: @@
 위의 방법으로는
-lower quartile: 2
+lower quartile: 2.5
-upper quartile: 6
+upper quartile: 6.5
 Ordered Data Set: 6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49
@@ Line 45: / Line 44: @@
-===== e.g. 2 =====
+===== r method =====
 in r
-<code>> j <- c(6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49)
+<code>
+j <- c(1,2,3,4,5)
+j <- sort(j)
+quantile(j)
+</code>
+<code>
+> j <- c(1,2,3,4,5)
+> j <- sort(j)
 > quantile(j)
 %  25%  50%  75% 100%
-.0 25.5 40.0 42.5 49.0 </code>
+    2    3    4    5
+>
-Method r
+</code>
+Odd number of elements
   * Use the median to divide the ordered data set into two halves.
-  * If there are an odd number of data points in the original ordered data set, include the median (the central value in the ordered list) in both halves.
+  * If there are an odd number of data points in the original ordered data set, include the median (the central value in the ordered list) in both halves. (가운데 숫자)
-  * If there are an even number of data points in the original ordered data set, split this data set exactly in half.
-  * The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.
-  * The values found by this method are also known as "Tukey's hinges."
+<code>
+j2 <- c(1,2,3,4,5,6)
+j2 <- sort(j2)
+quantile(j2)
+</code>
+<code>
+> j2 <- c(1,2,3,4,5,6)
+> j2 <- sort(j2)
+> quantile(j2)
+%  25%  50%  75% 100%
+.00 2.25 3.50 4.75 6.00
+>
+>
+</code>
+Even number of elements
+  * If there are an even number of data points in the original ordered data set, split this data set exactly in half. 즉, 3과 4의 가운데 값 (50%) = 3.5
+  * lower bound (lower quartile) 앞부분을 반으로 쪼갯을 때의 숫자 (여기서는 2) 더하기, 그 다음숫자와의 차이의 (3-2) 1/4지점 (여기서는 2 + 0.25 = 2.25) 구한다.
+  * upper bound는 뒷부분의 반인 5에서 한칸 아래의 숫자인 4 더하기, 그 다음 숫자와의 차이의 (5와 4의 차이인 1) 3/4 지점을 (여기서는 4 + 0.75 = 4.75) 구한다.
+<code>
+> j3 <- c(7, 18, 5, 9, 12, 15)
+> j3s <- sort(j3)
+> j3s
+[1]  5  7  9 12 15 18
+> quantile(j3s)
+%   25%   50%   75%  100%
+.00  7.50 10.50 14.25 18.00
+>
+</code>
+median = (9+12)/2
+the 1st quartile = 7 + (9-7)*(1/4) = 7 + 0.5 = 7.5
+the 3rd quartile = 12 + (12-9)*(3/4) = 12 + 2.25 = 14.25
 ----
 in r