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quartile [2020/09/21 03:56] – [e.g. 1, Head First method] hkimscilquartile [2023/09/10 23:42] (current) – [r method] hkimscil
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 ====== Finding lower and upper quartile ====== ====== Finding lower and upper quartile ======
 ===== e.g. 1, Head First method ===== ===== e.g. 1, Head First method =====
- 
 <code>> k <- c(1:8) <code>> k <- c(1:8)
 > k > k
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-===== e.g. 2 ===== +===== r method =====
- +
-r uses a different method. See the next e.g. 2.  +
-https://stats.stackexchange.com/questions/134229/finding-quartiles-in-r +
- +
 in r in r
-<code>j <- c(6715363940414243, 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) > quantile(j)
   0%  25%  50%  75% 100%    0%  25%  50%  75% 100% 
- 6.0 25.40.0 42.5 49.0 </code> +      2    3    4     
- +>  
-Method r+</code> 
 +Odd number of elements 
   * Use the median to divide the ordered data set into two halves.   * 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) 
 + 
 <code> <code>
-jh1 <- c(6715363940+j2 <- c(1,2,3,4,5,6
-(15+36)/ +j2 <- sort(j2) 
-jh2 <- c(404142434749+quantile(j2) 
-(42+43)/2+</code> 
 +<code> 
 +> j2 <- c(1,2,3,4,5,6
 +> j2 <- sort(j2) 
 +> quantile(j2) 
 +  0%  25%  50%  75% 100%  
 +1.00 2.25 3.50 4.75 6.00  
 +>  
 +
 </code> </code>
  
-  * If there are an even number of data points in the original ordered data set, split this data set exactly in half. +Even number of elements 
-  * The lower quartile value is the median of the lower half of the dataThe upper quartile value is the median of the upper half of the data+  * 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  
-  * The values found by this method are also known as "Tukey's hinges."+  * 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> <code>
-<- c(6, 7, 1536394041, 42, 43, 47, 49, 50+j3 <- c(7, 18591215
-ks <- sort(k+j3s <- sort(j3
-ks +j3s 
- [1]   7 15 36 39 40 41 42 43 47 49 50 +[1]    9 12 15 18 
-> length(ks) +> quantile(j3s)
-[1] 12 +
-> quantile(ks)+
    0%   25%   50%   75%  100%     0%   25%   50%   75%  100% 
- 6.00 30.75 40.50 44.00 50.00 + 5.00  7.50 10.50 14.25 18.00 
  
 </code> </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
  
 ---- ----
quartile.1600660617.txt.gz · Last modified: by hkimscil

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