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quartile [2019/09/19 10:42] – [e.g. 1] hkimscilquartile [2023/09/11 08:42] (current) – [r method] hkimscil
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 ====== Finding lower and upper quartile ====== ====== Finding lower and upper quartile ======
-===== e.g. 1 ===== +===== e.g. 1, Head First method =====
 <code>> k <- c(1:8) <code>> k <- c(1:8)
 > k > k
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 위의 방법으로는  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 Ordered Data Set: 6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49
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 +===== r method =====
 +in r
 +<code>
 +j <- c(1,2,3,4,5)
 +j <- sort(j)
 +quantile(j)
 +</code>
  
-r uses a different method. See the next e.g. 2 +<code> 
-https://stats.stackexchange.com/questions/134229/finding-quartiles-in-r+> j <- c(1,2,3,4,5) 
 +> j <- sort(j) 
 +> quantile(j) 
 +  0%  25%  50%  75% 100%  
 +      2    3    4    5  
 +>  
 +</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. (가운데 숫자)
  
-===== e.g. 2 ===== 
  
- +<code> 
-in r +j2 <- c(1,2,3,4,5,6) 
-<code>> j <- c(67153639404142434749+j2 <- sort(j2) 
-> quantile(j)+quantile(j2) 
 +</code> 
 +<code> 
 +> j2 <- c(1,2,3,4,5,6) 
 +> j2 <- sort(j2
 +> quantile(j2)
   0%  25%  50%  75% 100%    0%  25%  50%  75% 100% 
- 6.0 25.5 40.0 42.5 49.0 </code>+1.00 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) 
 +   0%   25%   50%   75%  100%  
 + 5.00  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
  
-Method r 
-  * 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 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." 
 ---- ----
 in r in r
quartile.1568857320.txt.gz · Last modified: by hkimscil

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